Drorbn - User contributions [en]

User:C8sd

2021-01-01T21:35:05Z

C8sd: Totally necessary update

* Attended [[09-240]]
* Computer Science and Mathematics graduate (2012)
* Experienced with (media)wiki markup
* Learned some [http://en.wikibooks.org/wiki/LaTeX/Introduction LaTeX] for CSC236
* Working as a software engineer after graduating

Template:09-240/Navigation

2009-12-14T03:53:17Z

C8sd: Fix days of week.

{| cellpadding="0" cellspacing="0" style="clear: right; float: right"
|- align=right
|<div class="NavFrame"><div class="NavHead">[[09-240]]/[[Template:09-240/Navigation|Navigation Panel]]  </div>
<div class="NavContent">
{| border="1px" cellpadding="1" cellspacing="0" width="220" style="margin: 0 0 1em 0.5em; font-size: small"
|- align=center
|colspan=3 style="font-size: large; color: red;"|'''Additions to the MAT 240 web site may count towards [[09-240/Register of Good Deeds|good deed points]] until Monday December 14 at 9AM'''
|- align=left
!#
!Week of...
!Notes and Links
|- align=left
|align=center|1
|Sep 7
|<s>Tue</s>, [[09-240/About This Class|About]], [[09-240/Classnotes for Thursday September 10|Thu]]
|- align=left
|align=center|2
|Sep 14
|[[09-240/Classnotes for Tuesday September 15|Tue]], [[09-240:HW1|HW1]], [[09-240:HW1 Solution|HW1 Solution]], [[09-240/Classnotes for Thursday September 17|Thu]]
|- align=left
|align=center|3
|Sep 21
|[[09-240/Classnotes for Tuesday September 22|Tue]], [[09-240:HW2|HW2]], [[09-240:HW2 Solution|HW2 Solution]], [[09-240/Classnotes for Thursday September 24|Thu]], [[09-240/Class Photo|Photo]]
|- align=left
|align=center|4
|Sep 28
|[[09-240/Classnotes for Tuesday September 29|Tue]], [[09-240:HW3|HW3]], [[09-240:HW3 Solution|HW3 Solution]], [[09-240/Classnotes for Thursday October 1|Thu]]
|- align=left
|align=center|5
|Oct 5
|[[09-240/Classnotes for Tuesday October 6|Tue]], [[09-240:HW4|HW4]], [[09-240:HW4 Solution|HW4 Solution]], [[09-240/Classnotes for Thursday October 8|Thu]],
|- align=left
|align=center|6
|Oct 12
|[[09-240/Classnotes for Tuesday October 13|Tue]], [[09-240/Classnotes for Thursday October 15|Thu]]
|- align=left
|align=center|7
|Oct 19
|[[09-240/Classnotes for Tuesday October 20|Tue]], [[09-240:HW5|HW5]], [[09-240:HW5 Solution|HW5 Solution]], [[09-240/Term Test|Term Test on Thu]]
|- align=left
|align=center|8
|Oct 26
|[[09-240/Classnotes for Tuesday October 27|Tue]], [[09-240/Linear Algebra - Why We Care|Why LinAlg?]], [[09-240:HW6|HW6]], [[09-240:HW6 Solution|HW6 Solution]], [[09-240/Classnotes for Thursday October 29|Thu]]
|- align=left
|align=center|9
|Nov 2
|[[09-240/Classnotes for Tuesday November 3|Tue]], [[09-240/Useful links to the MIT linear algebra course|MIT LinAlg]], [[09-240/Classnotes for Thursday November 5|Thu]]
|- align=left
|align=center|10
|Nov 9
|[[09-240/Classnotes for Tuesday November 10|Tue]], [[09-240:HW7|HW7]], [[09-240:HW7 Solution|HW7 Solution]] <s>Thu</s>
|- align=left
|align=center|11
|Nov 16
|[[09-240/Classnotes for Tuesday November 17|Tue]], [[09-240:HW8|HW8]], [[09-240:HW8 Solution|HW8 Solution]], [[09-240/Classnotes for Thursday November 19|Thu]]
|- align=left
|align=center|12
|Nov 23
|[[09-240/Classnotes for Tuesday November 24|Tue]], [[09-240:HW9|HW9]], [[09-240:HW9 Solution|HW9 Solution]], [[09-240/Classnotes for Thursday November 26|Thu]]
|- align=left
|align=center|13
|Nov 30
|[[09-240/Classnotes for Tuesday December 1|Tue]], [[09-240/On The Final Exam|On the final]], [[09-240/Classnotes for Thursday December 3|Thu]]
|- align=left
|align=center|S
|Dec 7
|[[09-240/Final Exam Preparation Forum|Forum]], [[09-240/On The Final Exam|Office Hours]]
|- align=left
|align=center|F
|Dec 14
|[[09-240/The Final Exam|Final on Dec 16]]
|- align=left
|colspan=3 align=center|[[09-240/To do|To Do List]]
|- align=left
|colspan=3 align=center|[http://www.youtube.com/watch?v=UTby_e4-Rhg The Algebra Song!]
|- align=left
|colspan=3 align=center|[http://www.youtube.com/user/khanacademy Helpful Links to School Subjects including Linear Algebra!]
|- align=left
|colspan=3 align=center|[[09-240/Register of Good Deeds|Register of Good Deeds]]
|- align=left
|colspan=3 align=center|[[09-240/Misplaced Material|Misplaced Material]]
|- align=left
|colspan=3 align=center|[[Image:09-240-ClassPhoto.jpg|180px]] [[09-240/Class Photo|Add your name / see who's in!]]
|}
</div></div>
|}

09-240/Classnotes for Tuesday December 3

2009-12-14T03:52:38Z

C8sd: 09-240/Classnotes for Tuesday December 3 moved to 09-240/Classnotes for Thursday December 3: Fix day of week

#redirect [[09-240/Classnotes for Thursday December 3]]

09-240/Classnotes for Thursday December 3

2009-12-14T03:52:38Z

C8sd: 09-240/Classnotes for Tuesday December 3 moved to 09-240/Classnotes for Thursday December 3

{{09-240/Navigation}}
<gallery>
Image:ALA240-2009_-_December_3rd.pdf|A set of the December 3rd lecture notes given by Professor Dror Bar-Natan for the Fall Session of MAT240 at the University of Toronto. Please note that this set is missing a diagram that could not be converted to PDF form.
Image:Lecture Notes for Dec 3 Pg 1.JPG
Image:Lecture Notes for Dec 3 Pg 2.JPG
</gallery>

~In the above gallery, there is a copy of notes for the lecture given on December 3rd by Professor Natan (in PDF format). Please note that there is an important diagram missing from this file. This is because the diagram was too complex to coherently convert to PDF. Other notes must be referenced for this diagram (otherwise, the PDF is complete)

09-240/Classnotes for Tuesday November 26

2009-12-14T03:52:22Z

C8sd: 09-240/Classnotes for Tuesday November 26 moved to 09-240/Classnotes for Thursday November 26: Fix day of week

#redirect [[09-240/Classnotes for Thursday November 26]]

09-240/Classnotes for Thursday November 26

2009-12-14T03:52:22Z

C8sd: 09-240/Classnotes for Tuesday November 26 moved to 09-240/Classnotes for Thursday November 26

{{09-240/Navigation}}
<gallery>
Image:ALA240-2009_-_November_26th.pdf|A complete copy of notes for the lecture given on November 26th by Professor Natan (in PDF format)
Image:09-240-26.10-1.jpg
Image:09-240-26.10-2.jpg
Image:09-240-26.10-3.jpg
</gallery>
In the above gallery, there is a complete copy of notes for the lecture given on November 24th by Professor Natan (in PDF format).

Template:09-240/Navigation

2009-12-09T22:45:34Z

C8sd: HW9

{| cellpadding="0" cellspacing="0" style="clear: right; float: right"
|- align=right
|<div class="NavFrame"><div class="NavHead">[[09-240]]/[[Template:09-240/Navigation|Navigation Panel]]  </div>
<div class="NavContent">
{| border="1px" cellpadding="1" cellspacing="0" width="220" style="margin: 0 0 1em 0.5em; font-size: small"
|- align=left
!#
!Week of...
!Notes and Links
|- align=left
|align=center|1
|Sep 7
|<s>Tue</s>, [[09-240/About This Class|About]], [[09-240/Classnotes for Thursday September 10|Thu]]
|- align=left
|align=center|2
|Sep 14
|[[09-240/Classnotes for Tuesday September 15|Tue]], [[09-240:HW1|HW1]], [[09-240:HW1 Solution|HW1 Solution]], [[09-240/Classnotes for Thursday September 17|Thu]]
|- align=left
|align=center|3
|Sep 21
|[[09-240/Classnotes for Tuesday September 22|Tue]], [[09-240:HW2|HW2]], [[09-240:HW2 Solution|HW2 Solution]], [[09-240/Classnotes for Thursday September 24|Thu]], [[09-240/Class Photo|Photo]]
|- align=left
|align=center|4
|Sep 28
|[[09-240/Classnotes for Tuesday September 29|Tue]], [[09-240:HW3|HW3]], [[09-240:HW3 Solution|HW3 Solution]], [[09-240/Classnotes for Thursday October 1|Thu]]
|- align=left
|align=center|5
|Oct 5
|[[09-240/Classnotes for Tuesday October 6|Tue]], [[09-240:HW4|HW4]], [[09-240:HW4 Solution|HW4 Solution]], [[09-240/Classnotes for Thursday October 8|Thu]],
|- align=left
|align=center|6
|Oct 12
|[[09-240/Classnotes for Tuesday October 13|Tue]], [[09-240/Classnotes for Thursday October 15|Thu]]
|- align=left
|align=center|7
|Oct 19
|[[09-240/Classnotes for Tuesday October 20|Tue]], [[09-240:HW5|HW5]], [[09-240:HW5 Solution|HW5 Solution]], [[09-240/Term Test|Term Test on Thu]]
|- align=left
|align=center|8
|Oct 26
|[[09-240/Classnotes for Tuesday October 27|Tue]], [[09-240/Linear Algebra - Why We Care|Why LinAlg?]], [[09-240:HW6|HW6]], [[09-240/Classnotes for Thursday October 29|Thu]]
|- align=left
|align=center|9
|Nov 2
|[[09-240/Classnotes for Tuesday November 3|Tue]], [[09-240/Useful links to the MIT linear algebra course|MIT LinAlg]], [[09-240/Classnotes for Thursday November 5|Thu]]
|- align=left
|align=center|10
|Nov 9
|[[09-240/Classnotes for Tuesday November 10|Tue]], [[09-240:HW7|HW7]], [[09-240:HW7 Solution|HW7 Solution]] <s>Thu</s>
|- align=left
|align=center|11
|Nov 16
|[[09-240/Classnotes for Tuesday November 17|Tue]], [[09-240:HW8|HW8]], [[09-240:HW8 Solution|HW8 Solution]], [[09-240/Classnotes for Thursday November 19|Thu]]
|- align=left
|align=center|12
|Nov 23
|[[09-240/Classnotes for Tuesday November 24|Tue]], [[09-240:HW9|HW9]], [[09-240:HW9 Solution|HW9 Solution]], [[09-240/Classnotes for Tuesday November 26|Thu]]
|- align=left
|align=center|13
|Nov 30
|[[09-240/Classnotes for Tuesday December 1|Tue]], [[09-240/On The Final Exam|On the final]], [[09-240/Classnotes for Tuesday December 3|Thu]]
|- align=left
|align=center|S
|Dec 7
|[[09-240/Final Exam Preparation Forum|Forum]], [[09-240/On The Final Exam|Office Hours]]
|- align=left
|align=center|F
|Dec 14
|[[09-240/The Final Exam|Final on Dec 16]]
|- align=left
|colspan=3 align=center|[[09-240/To do|To Do List]]
|- align=left
|colspan=3 align=center|[http://www.youtube.com/watch?v=UTby_e4-Rhg The Algebra Song!]
|- align=left
|colspan=3 align=center|[[09-240/Register of Good Deeds|Register of Good Deeds]]
|- align=left
|colspan=3 align=center|[[09-240/Misplaced Material|Misplaced Material]]
|- align=left
|colspan=3 align=center|[[Image:09-240-ClassPhoto.jpg|180px]] [[09-240/Class Photo|Add your name / see who's in!]]
|}
</div></div>
|}

09-240/To do

2009-12-09T22:36:02Z

C8sd: update

# Upload solutions for [[09-240:HW6 Solution|HW6]] and [[09-240:HW9 Solution|HW9]].

Template:09-240/Navigation

2009-12-09T22:34:33Z

C8sd: Move link for consistency.

{| cellpadding="0" cellspacing="0" style="clear: right; float: right"
|- align=right
|<div class="NavFrame"><div class="NavHead">[[09-240]]/[[Template:09-240/Navigation|Navigation Panel]]  </div>
<div class="NavContent">
{| border="1px" cellpadding="1" cellspacing="0" width="220" style="margin: 0 0 1em 0.5em; font-size: small"
|- align=left
!#
!Week of...
!Notes and Links
|- align=left
|align=center|1
|Sep 7
|<s>Tue</s>, [[09-240/About This Class|About]], [[09-240/Classnotes for Thursday September 10|Thu]]
|- align=left
|align=center|2
|Sep 14
|[[09-240/Classnotes for Tuesday September 15|Tue]], [[09-240:HW1|HW1]], [[09-240:HW1 Solution|HW1 Solution]], [[09-240/Classnotes for Thursday September 17|Thu]]
|- align=left
|align=center|3
|Sep 21
|[[09-240/Classnotes for Tuesday September 22|Tue]], [[09-240:HW2|HW2]], [[09-240:HW2 Solution|HW2 Solution]], [[09-240/Classnotes for Thursday September 24|Thu]], [[09-240/Class Photo|Photo]]
|- align=left
|align=center|4
|Sep 28
|[[09-240/Classnotes for Tuesday September 29|Tue]], [[09-240:HW3|HW3]], [[09-240:HW3 Solution|HW3 Solution]], [[09-240/Classnotes for Thursday October 1|Thu]]
|- align=left
|align=center|5
|Oct 5
|[[09-240/Classnotes for Tuesday October 6|Tue]], [[09-240:HW4|HW4]], [[09-240:HW4 Solution|HW4 Solution]], [[09-240/Classnotes for Thursday October 8|Thu]],
|- align=left
|align=center|6
|Oct 12
|[[09-240/Classnotes for Tuesday October 13|Tue]], [[09-240/Classnotes for Thursday October 15|Thu]]
|- align=left
|align=center|7
|Oct 19
|[[09-240/Classnotes for Tuesday October 20|Tue]], [[09-240:HW5|HW5]], [[09-240:HW5 Solution|HW5 Solution]], [[09-240/Term Test|Term Test on Thu]]
|- align=left
|align=center|8
|Oct 26
|[[09-240/Classnotes for Tuesday October 27|Tue]], [[09-240/Linear Algebra - Why We Care|Why LinAlg?]], [[09-240:HW6|HW6]], [[09-240/Classnotes for Thursday October 29|Thu]]
|- align=left
|align=center|9
|Nov 2
|[[09-240/Classnotes for Tuesday November 3|Tue]], [[09-240/Useful links to the MIT linear algebra course|MIT LinAlg]], [[09-240/Classnotes for Thursday November 5|Thu]]
|- align=left
|align=center|10
|Nov 9
|[[09-240/Classnotes for Tuesday November 10|Tue]], [[09-240:HW7|HW7]], [[09-240:HW7 Solution|HW7 Solution]] <s>Thu</s>
|- align=left
|align=center|11
|Nov 16
|[[09-240/Classnotes for Tuesday November 17|Tue]], [[09-240:HW8|HW8]], [[09-240:HW8 Solution|HW8 Solution]], [[09-240/Classnotes for Thursday November 19|Thu]]
|- align=left
|align=center|12
|Nov 23
|[[09-240/Classnotes for Tuesday November 24|Tue]], [[09-240:HW9|HW9]], [[09-240/Classnotes for Tuesday November 26|Thu]]
|- align=left
|align=center|13
|Nov 30
|[[09-240/Classnotes for Tuesday December 1|Tue]], [[09-240/On The Final Exam|On the final]], [[09-240/Classnotes for Tuesday December 3|Thu]]
|- align=left
|align=center|S
|Dec 7
|[[09-240/Final Exam Preparation Forum|Forum]], [[09-240/On The Final Exam|Office Hours]]
|- align=left
|align=center|F
|Dec 14
|[[09-240/The Final Exam|Final on Dec 16]]
|- align=left
|colspan=3 align=center|[[09-240/To do|To Do List]]
|- align=left
|colspan=3 align=center|[http://www.youtube.com/watch?v=UTby_e4-Rhg The Algebra Song!]
|- align=left
|colspan=3 align=center|[[09-240/Register of Good Deeds|Register of Good Deeds]]
|- align=left
|colspan=3 align=center|[[09-240/Misplaced Material|Misplaced Material]]
|- align=left
|colspan=3 align=center|[[Image:09-240-ClassPhoto.jpg|180px]] [[09-240/Class Photo|Add your name / see who's in!]]
|}
</div></div>
|}

09-240/Classnotes for Tuesday December 1

2009-12-08T04:24:45Z

C8sd: Format, correct mistakes.

{{09-240/Navigation}}
<gallery>
Image:ALA240-2009_-_December_1st.pdf|A complete set of the December 1st lecture notes given by Professor Dror Bar-Natan for the Fall Session of MAT240 at the University of Toronto.
Image:Dec 1 lecture notes Pg 1.JPG|
Image:Dec 1 lecture notes Pg 2.JPG|
Image:Dec 1 lecture notes Pg 3.JPG|
Image:Dec 1 lecture notes Pg 4.JPG|
Image:Dec 1 lecture notes Pg 5.JPG|
Image:dec1-1.jpg|
Image:dec1-2.jpg|
Image:dec1-3.jpg|
Image:dec1-4.jpg|
Image:dec1-5.jpg|
Image:dec1-6.jpg|
Image:dec1-7.jpg|
Image:dec1-8.jpg|
</gallery>

~In the above gallery, there is a complete copy of notes for the lecture given on December 1st by Professor Natan (in PDF format).

--- Wiki Format ---

MAT240 – December 1st

Basic Properties of <math>\det : \mathbb M_{n \times n} \rightarrow F</math>:

(Note that det(''EA'') = det(''E'')·det(''A'') and that det(''A'') may be written as |''A''|.)

0. <math>\,\! \det(I) = 1</math>

1. <math>\det(E^1_{i,j}A) = -\det(A) ; |E^1_{i,j}|= -1</math>

: Exchanging two rows flips the sign.

2. <math>\det(E^2_{i,c}A) = c \cdot \det(A) ; |E^2_{i,c}| = c</math>

: These are "enough"!

3. <math>\det(E^3_{i,j,c}A) = \det(A) ; |E^3_{i,j,c}| = 1</math>

: Adding a multiple of one row to another does not change the determinant.

The determinant of any matrix can be calculated using the properties above.

'''Theorem''':

If <math>{\det}' : \mathbb M_{n \times n} \rightarrow F</math> satisfies properties 0-3 above, then <math>\det' = \det</math>

<math>\det(A) = \det'(A)</math>

Philosophical remark: Why not begin our inquiry with the properties above?

We must find an implied need for their use; thus, we must know whether a function <math>\det</math> exists first.

09-240/Term Test

2009-12-07T07:53:30Z

C8sd: /* Solution Set */ Format.

{{09-240/Navigation}}

==Announcement==

Our one and only Term Test is coming up. It will take place in class on Thursday October 22 2009, starting promptly at 1:10PM and ending at 3:00PM sharp, in our normal classroom, MP103. It will consist of 4-5 questions (each may have several parts) on everything that will be covered in class by October 16: the axiomatic definition of fields and some basic properties of fields, <math>{\mathbb C}</math> and other examples, a tiny bit on the field with <math>p</math> elements <math>F_p</math>, the axiomatic definition of vector spaces, basic properties and examples of vector spaces, spans, linear combinations and linear equations, linear dependence and independence, bases, the replacement lemma and its consequences, a bit about linear transformations and a few smaller topics that we touched but that do not deserve their own headers.

Note that there may be some computations, but nothing that will require a calculator. Note also that I may include some questions from the homework assignments verbatim or nearly verbatim.

;Will there be "proof questions"?
:Sure. What else have we done so far?
;Do we need to know the proofs from class?
:Sure. There's a reason why these proofs are in class to start with; if they weren't valuable, we wouldn't have covered them.

No electronic devices capable of displaying text or sounding speech will be allowed.

In style and spirit this exam will not be very different of the one I gave 3 years ago. See [[06-240/Term Test]].

==The Test==

===Front Page===

<center>

'''Do not turn this page until instructed.'''

Math 240 Algebra I - Term Test

University of Toronto, October 22, 2009

'''Solve the 5 problems on the other side of this page.'''

Each of the problems is worth 20 points.

You have an hour and 50 minutes.

'''Notes.'''

</center>

* No outside material other than stationary and a basic calculator is allowed.
* We will have an hour of discussion time right after this test.
* The final exam date was posted by the faculty --- it will take place on Wednesday December 16 from 9AM until noon at room BN2S of the Clara Benson Building, 320 Huron Street (south west of Harbord cross Huron, home of the Faculty of Physical Education and Health).

<center> '''Good Luck!''' </center>

===Questions Page===

'''Solve the following 5 problems.''' Each of the problems is worth 20 points. You have an hour and 50 minutes.

'''Problem 1.''' Let <math>V</math> be a vector space over a field <math>F</math>, let <math>c\in F</math> and let <math>v\in V</math>. Prove that if <math>cv=0</math>, then either <math>c=0</math> or <math>v=0</math>.

'''Problem 2.'''
# In the field <math>{\mathbb C}</math> of complex numbers, compute
<center><math>4i(1+i)</math>      and     <math>\frac{4i}{1+i}</math>.</center>
(To be precise, "compute" means "write in the form <math>a+ib</math>, where
<math>a,b\in{\mathbb R}</math>").
# In the field <math>{\mathbb C}</math> of complex numbers, find an element <math>z</math> so that <math>z^2=2i</math>.
# In the 11-element field <math>F_{11}</math> of remainders modulo 11, find
all solutions of the equation <math>x^2=-2</math>.

'''Problem 3.''' Let <math>V</math> be a vector space and let <math>W_1</math> and <math>W_2</math> be subspaces of <math>V</math>. Prove that <math>W_1\cup W_2</math> is a subspace of <math>V</math> iff <math>W_1\subset W_2</math> or <math>W_2\subset W_1</math>.

'''Problem 4.''' Let <math>u_1,\ldots,u_n</math> be vectors in a
vector space over the field with two element <math>F_2</math>. Show that the number of
elements in the set <math>\mbox{Span}\{u_1,\dots,u_n\}</math> is equal to <math>2^n</math> if and only
if <math>u_1,\ldots,u_n</math> are linearly independent..

'''Problem 5.''' Find a polynomial <math>f\in P_3({\mathbb R})</math> that
satisfies <math>f(-1)=5</math>, <math>f(0)=4</math>, <math>f(1)=3</math>, and <math>f(2)=8</math>.

<center> '''Good Luck!''' </center>

{{09-240/Results of the Term Test}}

==Solution Set==
Students are most welcome to post a solution set here.

'''Problem 1'''

If <math>c \cdot v = 0</math> and <math>c \neq 0,</math>

then <math>c^{-1} \cdot (c \cdot v) = 0;</math>

thus <math>(c^{-1} \cdot c) \cdot v = 0;</math>

thus <math>1 \cdot v = 0;</math>

thus <math>v = 0.</math>

It should be remembered that "or" means "one or the other, or both."

'''Problem 2'''

(1)

<math>4i(1 + i) = 4i + 4i^2 = -4 + 4i</math>

<math>\frac{4i}{1+i} = \frac{4i(1 - i)}{(1 + i)(1 - i)} = \frac{4 + 4i}{2} = 2 + 2i</math>

(2)

If <math>z^2 = 2i</math> then <math>z = 1 + i \Rightarrow (1 + i)^2 = 1 + 2i + i^2 = 2i</math>

(3)

<math>x = 8, 3</math>

'''Problem 3.''' First suppose <math>W_1\subset W_2</math> or <math>W_2\subset W_1</math>, then <math>W_1\cup W_2</math> equals <math>W_2</math> or <math>W_1</math>. Either case <math>W_1\cup W_2</math> is a subspace of <math>V</math>.

Now suppose <math>W_1\cup W_2</math> is a subspace of <math>V</math> and neither <math>W_1\subset W_2</math> nor <math>W_2\subset W_1</math>. It must follow that <math>\exists\ x\in W_1</math> s.t. <math>x</math> is not in <math>W_2</math> and <math>\exists\ y\in W_2</math> s.t. <math>y</math> is not in <math>W_1</math>. Since <math>x,y\in W_1\cup W_2</math>, <math>x+y\in W_1\cup W_2</math>. If <math>x+y\in W_1</math>, then <math>\exists\ -x\in W_1</math> s.t. <math>(-x)+x+y\in W_1</math> and <math>(-x)+x=0</math>, so <math>y\in W_1</math>, a contradiction. If <math>x+y\in W_2</math>, then <math>\exists\ -y\in W_2</math> s.t. <math>x+y+(-y)\in W_2</math> and <math>(-y)+y=0</math>, so <math>x\in W_2</math>, a contradiction. Therefore either <math>W_1\subset W_2</math> or <math>W_2\subset W_1</math>.

'''Problem 4'''

Let <math>S_1 = \{ u_1, \ldots, u_n \}</math> and <math>S_2 = \operatorname{span}(S_1)</math>.

Assume <math>S_1</math> is linearly independent. Then <math>S_1</math> is a basis, and by basis properties, <math>\forall x \in S_2, \exists</math> unique <math>a_1, \ldots, a_n \in F</math> such that <math>a_1 u_1 + \ldots + a_n u_n = x</math>. <math>\forall a \in F</math>, <math>a = 0</math> or <math>a = 1</math>, so there are two possibilities for every <math>a</math>. Therefore, <math>|S_2| = \underbrace{2 \times \ldots \times 2}_n = 2^n</math>, so <math>S_1</math> is linearly independent <math>\Rightarrow |S_2| = 2^n</math>.

Assume <math>S_1</math> is linearly dependent. Then <math>\exists</math> non-trivial <math>a_i, \ldots, a_n</math> such that <math>a_1 u_1 + \ldots + a_n u_n = 0</math>. Since 0 has at least two representations, <math>|S_2| < 2^n</math>. Hence, <math>S_1</math> is linearly dependent <math>\Rightarrow |S_2| \ne 2^n</math>. By contrapositive, <math>|S_2| = 2^n \Rightarrow S_1</math> is linearly independent.

Hence, <math>S_1</math> is linearly independent <math>\iff |S_2| = 2^n</math>.

'''Problem 5'''

Recall that<math> f_i(x) = \prod_{k = 0 k \neq i}^{n}{\frac{x - c_k}{c_i - c_k}}</math> and that <math>g(x) = \sum_{i = 0}^{n}{b_if_i}</math>

After tedious computation with the addition of eraser bits covering your paper, you should obtain the following:

<math>f_0(x) = -\frac{1}{6}(x^3 - 3x^2 + 2x)</math>

<math>f_1(x) = \frac{1}{2}(x^3 - 2x^2 - x + 2)</math>

<math>f_2(x) = -\frac{1}{2}(x^3 - x^2 - 2x)</math>

<math>f_3(x) = \frac{1}{6}(x^3 - x)</math>

Therefore,

<math>g(x) = 5 \cdot \left( -\frac{1}{6}(x^3 - 3x^2 + 2x) \right) + 4 \cdot \frac{1}{2}(x^3 - 2x^2 - x + 2) + 3 \cdot \left( -\frac{1}{2}(x^3 - x^2 - 2x) \right) + 8 \cdot \frac{1}{6}(x^3 - x)</math>
: <math>\,\! = x^3 - 2x + 4</math>

09-240/Classnotes for Tuesday October 20

2009-12-07T07:46:53Z

C8sd: Put second player on bottom.

{{09-240/Navigation}}
{{09-240/Class Notes Warning}}

'''Definition''': '''V''' and '''W''' are "isomorphic" if there exist linear transformations <math>\mathrm{T : V \rightarrow W}</math> and <math>\mathrm{S : W \rightarrow V}</math> such that <math>\mathrm{T \circ S} = I_\mathrm{W}</math> and <math>\mathrm{S \circ T} = I_\mathrm{V}</math>

'''Theorem''': If '''V''' and '''W''' are finite-dimensional over ''F'', then '''V''' is isomorphic to '''W''' iff dim('''V''') = dim('''W''')

'''Corollary''': If dim('''V''') = ''n'' then <math>\mathrm{V} \cong F^n</math>
:Note: <math>\cong</math> represents "is isomorphic to"

----

Two "mathematical structures" are "isomorphic" if there exists a "bijection" between their elements which preserves all relevant relations between such elements.

'''Example''': Plastic chess is "isomorphic" to ivory chess, but it is not isomorphic to checkers.

'''Example''': The game of 15. Players alternate drawing one card each.

Goal: To have exactly three of your cards add to 15.

Sample game:
* X picks 3
* O picks 7
* X picks 8
* O picks ''4''
* X picks 1
* O picks ''6''
* X picks 2
* O picks ''5''
* 4 + 6 + 5 = 15. O wins.

This game is isomorphic to Tic Tac Toe!

{| class="wikitable" border="0" cellpadding="2" cellspacing="0"
|-
| style="border-style: none solid solid none" | 4
| style="border-style: none solid solid solid" | 9
| style="border-style: none none solid solid" | 2
|-
| style="border-style: solid solid solid none" | 3
| style="border-style: solid solid solid solid" | 5
| style="border-style: solid none solid solid" | 7
|-
| style="border-style: solid solid none none" | 8
| style="border-style: solid solid none solid" | 1
| style="border-style: solid none none solid" | 6
|}

: X: 3, 8, 1, 2
: O: 7, ''4'', ''6'', ''5'' -- Wins!

Converts to:

{| class="wikitable" border="0" cellpadding="2" cellspacing="0"
|-
| style="border-style: none solid solid none" | O
| style="border-style: none solid solid solid" | 9
| style="border-style: none none solid solid" | X
|-
| style="border-style: solid solid solid none" | X
| style="border-style: solid solid solid solid" | O
| style="border-style: solid none solid solid" | O
|-
| style="border-style: solid solid none none" | X
| style="border-style: solid solid none solid" | X
| style="border-style: solid none none solid" | O
|}

: <math>\mathrm{S \circ T} = I_\mathrm{V}</math>
: <math>\mathrm{T \circ S} = I_\mathrm{W}</math>
: <math>\mathrm T(O_\mathrm{V}) = O_\mathrm{W}</math>

: <math>\mathrm T(x + y) = T(x) + T(y)</math>
: <math>\mathrm T(cv) = c\mathrm T(v)</math>
: Likewise for <math>\mathrm S</math>

: <math>z = x + y \Rightarrow \mathrm T(z) = \mathrm T(x) + \mathrm T(y)</math>
: <math>u = 7v \Rightarrow \mathrm T(u) = 7\mathrm T(v)</math>

Proof of Theorem <math>\iff</math> Assume dim('''V''') = dim('''W''') = ''n''
: There exists basis <math>\beta = \{u_1, \ldots, u_n\} \in \mathrm V</math>
: <math>\alpha = \{w_1, ..., w_n\} \in \mathrm W</math>
: by an earlier theorem, there exists a l.t. <math>\mathrm{T : V \rightarrow W}</math> such that <math>\mathrm T(u_i) = w_i</math>

<math>\mathrm T(\sum a_i u_i) = \sum a_i \mathrm T(u_i) = \sum a_i w_i</math>

There exists a l.t. <math>\mathrm{S : W \rightarrow V}</math> such that <math>\mathrm S(w_i) = u_i</math>

== Claim ==
: <math>\mathrm{S \circ T} = I_\mathrm{V}</math>
: <math>\mathrm{T \circ S} = I_\mathrm{W}</math>

== Proof ==
If u∈<math> \mathrm{V} </math> unto U=∑aiui
: (S∘T)(u)=S(T(u))=S(T(∑aiui))
: =S(∑aiwi)=∑aiui=u
: ⇒S∘T=Iv...
: ⇒Assume T&S as above exist
: Choose a basis β= (U1...Un) of V

== Claim ==
α=(W1=Tu1, W2=Tu2, ..., Wn=Tun)
: is a basis of W, so dim W=n

== Proof ==
α is lin. indep.
: T(0)=0=∑aiwi=∑aiTui=T(∑aiui)
: Apply S to both sides:
: 0=∑aiui
: So ∃iai=0 as β is a basis

α Spans W
: Given any w∈W let u=S(W)
: As β is a basis find ais in F s.t. v=∑aiui
Apply T to both sides: T(S(W))=T(u)=T(∑aiui)=∑aiT(ui)=∑aiWi
::: ∴ I win!!! (QED)

: T T
: V → W ⇔ V' → W'
: rank T=rank T'
Fix t:V→Wa l.t.<math>Insert formula here</math>

== Definition ==
# N(T) = ker(T) = {u∈V : Tu = 0W}
# R(T) = im(T) = {T(u) : u∈V}

== Prop/Def ==
# N(T) ⊂ V is a subspace of V-------nullity(T) := dim N(T)
# R(T) ⊂ W is a subspace of W--------rank(T) := dim R(T)

== Proof 1 ==
: x,y ∈N(T)⇒T(x)=0, T(y)=0
: T(x+y)=T9x)+T(y)=0+0=0
: x+y∈N(T)
::: ∴ I win!!! (QED)

== Proof 2 ==
: Let y∈R(T)⇒fix x s.t y=T(x),
: --------7y=7T(x)=T(7x)
: ----------⇒7y∈R(T)
::: ∴ I win!!! (QED)

== Examples ==
1.
: 0:V→W---------N(0)=V
: R(0)={0W}-----------nullity(0)=dim V
: --------------rank(0)=0
:: dim V+0=dimV
2.
:IV:V→V
:N(I)={0}
:nullity=0
:R(I)=dim V
:2'If T:V→W is an imorphism
:N(T)={0}
:nullity =0
:R(T)=W
:rank=dim W
::0+dim V=dim V
3.
:D:P7(R)→P7(R)
:Df=f'
::N(D)={C⊃C°: C∈R}=P0(R)
:R(D)⊂P6(R)
::nullity(D)=1
::basis:(1x°)
::rank(D)=7
:::7+1=8
4.
:3':D2:P7(R)
:D2f=f''
:W(D2)={ax+b: a,b∈R}=P1(R)
::nullity(D2)=2
::R(D2)=P5(R)
:::rank (D2)=6
::6+2=8

== Theorem ==
(rank-nullity Theorem, a.k.a. dimension Theorem)
:nullity(T)+rank(T)=dim V
:(for a l.t. T:V→W) when V is F.d.

== Proof ==
(To be continued next day)

09-240/Classnotes for Tuesday October 13

2009-12-07T07:26:27Z

C8sd: class notes warning

{{09-240/Navigation}}
{{09-240/Class Notes Warning}}

<gallery>
Image:Oct 13 notes page 1.JPG|[[User:Bright|Bright]] - 1
Image:Oct 13 notes page 2.JPG|2
Image:Oct 13 notes page 3.JPG|3
Image:Oct 13 notes page 4.JPG|4
</gallery>

== Replacement Theorem ==

...

<ol>
<li value="3">
dim('''V''') = ''n''
<ol style="list-style-type: lower-alpha">
<li>If ''G'' generates '''V''' then <math>|G| \ge n</math>. If also <math>\,\! |G| = n</math> then ''G'' is a basis.</li>
<li>If ''L'' is linearly dependent then <math>|L| \le n</math>. If also <math>\,\! |L| = n</math> then ''L'' is a basis. If also <math>\,\! |L| < n</math> then ''L'' can be extended to a basis.

Proofs
''a1.'' If ''G'' has a subset which is a basis then that subset has ''n'' elements, so <math>|G| \ge n</math>. 
''a2.'' Let <math>\beta</math> be a basis of '''V''', then <math>\,\! |B| = n</math>. Now use replacement with ''G'' & ''L'' = <math>\beta</math>. Hence, <math>|G| \ge |L| = |B| = n</math>. 
a. From a1 and a2, we know <math>|G| \ge n</math>. If <math>\,\! |G| = n</math> then ''G'' contains a basis <math>\beta</math>. But <math>\,\! |B| = n</math>, so <math>\,\! |G| = \beta</math>, and hence ''G'' is a basis. 
 
b. Use replacement with ''G' being some basis of '''V'''. |''G''| = ''n''. 
 
If <math>|L| = |G|</math> then <math>|R| = n = |G|</math>, so <math>R = G</math>, so <math>(G \backslash R) \cup L</math> generates, so ''L'' generates, so ''L'' is a basis since it is linearly independent. 
 
We have that ''L'' is basis. If <math>|L| < |G|</math> the nagain find <math>R \subset G</math> such that <math>|R| = |L|</math> and <math>(G \backslash R) \cup L</math> generates. 
 
1. <math>\beta</math> generates '''V'''. 
2. <math>|B| \le |G| - |R| + |L| = n</math>. So by part a, <math>\beta</math> is a basis. 
3.
</li>
</ol>
</li>
<li value="4">
If '''V''' is finite-dimensional (f.d.) and <math>\mathbf W \subset \mathbf V</math> is a subspace of '''V''', then '''W''' is also finite, and <math>\operatorname{dim}(\mathbf W) \le \operatorname{dim}(\mathbf V)</math>. If also dim('''W''') = dim('''V''') then '''W''' = '''V''' and if dim('''W''') < dim('''V''') then any basis of '''W''' can be extended to a basis of '''V'''. 
 
Proof: Assuming '''W''' is finite-dimensional, pick a basis <math>\beta</math> of '''W'''; <math>\beta</math> is linearly independent in '''V''' so by Corollary 3 of part b, <math>|B| \le \operatorname{dim}(\mathbf V) \Rightarrow \operatorname{dim}(\mathbf W) = |B| \le \operatorname{dim}(\mathbf V)</math>. So span(<math>\beta</math>) = '''V''' = '''W''' so '''V''' = '''W'''. ... 
 
Assume '''W''' is not finite-dimensional. <math>\mathbf W \ne \{0\}</math> so pick a <math>x_1 \in W</math> such that <math>x_1 \ne 0</math>. So <math>\{x_1\}</math> is linearly independent in '''W''', and <math>\operatorname{span}(\{x_1\}) \subsetneq \mathbf W</math>. Pick <math>x_2 \in W \backslash \operatorname{span}(\{x_1\})</math>. So <math>\{x_1, x_2\}</math> is linearly dependent and <math>\operatorname{span}(\{x_1, x_2\}) \subsetneq \mathbf W</math>. Pick <math>x_3 \in W \backslash \operatorname{span}(\{x_1, x_2\})</math> ... continue in this way to get a sequence <math>x_1, x_2, \ldots, x_{n+1}</math> where ''n'' = dim('''V''') and <math>\{x_1, \ldots, x_{n+1}\}</math> is linearly independent. There is a contradiction by Corollary 3.b.
</li>
</ol>

== The Lagrange Interpolation Formula ==

[Aside: <math>(a - x)(b - x) \ldots (z - x) = 0</math> because <math>x - x = 0</math>]

Where <math>1 \le i < n,</math>

Let <math>x_i</math> be distinct points in <math>\real</math>.

Let <math>y_i</math> be any points in <math>\real</math>.

Can you find a polynomial <math>P \in \mathbb P_n(\real)</math> such that <math>P(x_i) = y_i</math>? Is it unique?

'''Example''':

: <math>x_i = 0, 1, 3</math>
: <math>y_i = 5, 2, 2</math>

Can we find a <math>P \in \mathbb P_2</math> such that
: <math>P(0) = 5</math>
: <math>P(1) = 2</math>
: <math>P(2) = 2</math>?

Solution: Let <math>\tilde P_i(x) = \prod_{j=1, j \ne 1}^{n+1} (x - x_j) \in \mathbb P_n(\real)</math>. (Remember [http://en.wikipedia.org/wiki/Capital_pi_notation capital pi notation].)

Then <math>\tilde P_i(x_k) = \begin{cases}
0 & i \ne k \\
\prod_{i \ne j} (x_i - x_j) & i = k \\
\end{cases}</math>

: <math>\tilde P_1 = (x - x_2)(x - x_3) = (x - 1)(x - 3) = x^2 - 4x + 3</math>
: <math>\tilde P_2 = (x - 0)(x - 3) = x^2 - 3x</math>
: <math>\tilde P_3 = (x - 0)(x - 1) = x^2 - x</math>

<math>\begin{matrix}
\tilde P_1(0) = 3 & \tilde P_1(1) = 0 & \tilde P_1(3) = 0 \\
\tilde P_2(0) = 0 & \tilde P_2(1) = -2 & \tilde P_2(3) = 0 \\
\tilde P_3(0) = 0 & \tilde P_3(1) = 0 & \tilde P_3(3) = 6 \\
\end{matrix}</math>

Set <math>P_i(x) = \frac1{\tilde P_i(x)} \cdot \tilde P_i = \prod_{j \ne i} \frac{(x - x_j)}{(x_i - x_j)}</math>

Then <math>P_i(x_k) = \begin{cases}
0 & i \ne k \\
1 & i = k \\
\end{cases} </math>

: <math>P_1(x) = \frac13 x^2 - \frac43 x + 1</math>
: <math>P_2(x) = -\frac12 x^2 + \frac32 x</math>
: <math>P_3(x) = \frac16 x^2 - \frac16 x</math>

Let <math>P = \sum_{i=1}^{n+1} y_i P_i(x) \in \mathbb P_n(\real)</math>

: <math>P = 5 \cdot \left( \frac13 x^2 - \frac43 x + 1 \right) + 2 \cdot \left( -\frac12 x^2 + \frac32 x \right) + 2 \cdot \left( \frac16 x^2 - \frac16 x \right)</math>
: <math>\,\! = x^2 - 4x + 5</math>

09-240/Classnotes for Tuesday October 20

2009-12-07T07:25:41Z

C8sd: Add standard templates.

{{09-240/Navigation}}
{{09-240/Class Notes Warning}}

'''Definition''': '''V''' and '''W''' are "isomorphic" if there exist linear transformations <math>\mathrm{T : V \rightarrow W}</math> and <math>\mathrm{S : W \rightarrow V}</math> such that <math>\mathrm{T \circ S} = I_\mathrm{W}</math> and <math>\mathrm{S \circ T} = I_\mathrm{V}</math>

'''Theorem''': If '''V''' and '''W''' are finite-dimensional over ''F'', then '''V''' is isomorphic to '''W''' iff dim('''V''') = dim('''W''')

'''Corollary''': If dim('''V''') = ''n'' then <math>\mathrm{V} \cong F^n</math>
:Note: <math>\cong</math> represents "is isomorphic to"

----

Two "mathematical structures" are "isomorphic" if there exists a "bijection" between their elements which preserves all relevant relations between such elements.

'''Example''': Plastic chess is "isomorphic" to ivory chess, but it is not isomorphic to checkers.

'''Example''': The game of 15. Players alternate drawing one card each.

Goal: To have exactly three of your cards add to 15.

Sample game:
* X picks 3
* O picks 7
* X picks 8
* O picks ''4''
* X picks 1
* O picks ''6''
* X picks 2
* O picks ''5''
* 4 + 6 + 5 = 15. O wins.

This game is isomorphic to Tic Tac Toe!

{| class="wikitable" border="0" cellpadding="2" cellspacing="0"
|-
| style="border-style: none solid solid none" | 4
| style="border-style: none solid solid solid" | 9
| style="border-style: none none solid solid" | 2
|-
| style="border-style: solid solid solid none" | 3
| style="border-style: solid solid solid solid" | 5
| style="border-style: solid none solid solid" | 7
|-
| style="border-style: solid solid none none" | 8
| style="border-style: solid solid none solid" | 1
| style="border-style: solid none none solid" | 6
|}

: O: 7, ''4'', ''6'', ''5'' -- Wins!
: X: 3, 8, 1, 2

Converts to:

{| class="wikitable" border="0" cellpadding="2" cellspacing="0"
|-
| style="border-style: none solid solid none" | O
| style="border-style: none solid solid solid" | 9
| style="border-style: none none solid solid" | X
|-
| style="border-style: solid solid solid none" | X
| style="border-style: solid solid solid solid" | O
| style="border-style: solid none solid solid" | O
|-
| style="border-style: solid solid none none" | X
| style="border-style: solid solid none solid" | X
| style="border-style: solid none none solid" | O
|}

: <math>\mathrm{S \circ T} = I_\mathrm{V}</math>
: <math>\mathrm{T \circ S} = I_\mathrm{W}</math>
: <math>\mathrm T(O_\mathrm{V}) = O_\mathrm{W}</math>

: <math>\mathrm T(x + y) = T(x) + T(y)</math>
: <math>\mathrm T(cv) = c\mathrm T(v)</math>
: Likewise for <math>\mathrm S</math>

: <math>z = x + y \Rightarrow \mathrm T(z) = \mathrm T(x) + \mathrm T(y)</math>
: <math>u = 7v \Rightarrow \mathrm T(u) = 7\mathrm T(v)</math>

Proof of Theorem <math>\iff</math> Assume dim('''V''') = dim('''W''') = ''n''
: There exists basis <math>\beta = \{u_1, \ldots, u_n\} \in \mathrm V</math>
: <math>\alpha = \{w_1, ..., w_n\} \in \mathrm W</math>
: by an earlier theorem, there exists a l.t. <math>\mathrm{T : V \rightarrow W}</math> such that <math>\mathrm T(u_i) = w_i</math>

<math>\mathrm T(\sum a_i u_i) = \sum a_i \mathrm T(u_i) = \sum a_i w_i</math>

There exists a l.t. <math>\mathrm{S : W \rightarrow V}</math> such that <math>\mathrm S(w_i) = u_i</math>

== Claim ==
: <math>\mathrm{S \circ T} = I_\mathrm{V}</math>
: <math>\mathrm{T \circ S} = I_\mathrm{W}</math>

== Proof ==
If u∈<math> \mathrm{V} </math> unto U=∑aiui
: (S∘T)(u)=S(T(u))=S(T(∑aiui))
: =S(∑aiwi)=∑aiui=u
: ⇒S∘T=Iv...
: ⇒Assume T&S as above exist
: Choose a basis β= (U1...Un) of V

== Claim ==
α=(W1=Tu1, W2=Tu2, ..., Wn=Tun)
: is a basis of W, so dim W=n

== Proof ==
α is lin. indep.
: T(0)=0=∑aiwi=∑aiTui=T(∑aiui)
: Apply S to both sides:
: 0=∑aiui
: So ∃iai=0 as β is a basis

α Spans W
: Given any w∈W let u=S(W)
: As β is a basis find ais in F s.t. v=∑aiui
Apply T to both sides: T(S(W))=T(u)=T(∑aiui)=∑aiT(ui)=∑aiWi
::: ∴ I win!!! (QED)

: T T
: V → W ⇔ V' → W'
: rank T=rank T'
Fix t:V→Wa l.t.<math>Insert formula here</math>

== Definition ==
# N(T) = ker(T) = {u∈V : Tu = 0W}
# R(T) = im(T) = {T(u) : u∈V}

== Prop/Def ==
# N(T) ⊂ V is a subspace of V-------nullity(T) := dim N(T)
# R(T) ⊂ W is a subspace of W--------rank(T) := dim R(T)

== Proof 1 ==
: x,y ∈N(T)⇒T(x)=0, T(y)=0
: T(x+y)=T9x)+T(y)=0+0=0
: x+y∈N(T)
::: ∴ I win!!! (QED)

== Proof 2 ==
: Let y∈R(T)⇒fix x s.t y=T(x),
: --------7y=7T(x)=T(7x)
: ----------⇒7y∈R(T)
::: ∴ I win!!! (QED)

== Examples ==
1.
: 0:V→W---------N(0)=V
: R(0)={0W}-----------nullity(0)=dim V
: --------------rank(0)=0
:: dim V+0=dimV
2.
:IV:V→V
:N(I)={0}
:nullity=0
:R(I)=dim V
:2'If T:V→W is an imorphism
:N(T)={0}
:nullity =0
:R(T)=W
:rank=dim W
::0+dim V=dim V
3.
:D:P7(R)→P7(R)
:Df=f'
::N(D)={C⊃C°: C∈R}=P0(R)
:R(D)⊂P6(R)
::nullity(D)=1
::basis:(1x°)
::rank(D)=7
:::7+1=8
4.
:3':D2:P7(R)
:D2f=f''
:W(D2)={ax+b: a,b∈R}=P1(R)
::nullity(D2)=2
::R(D2)=P5(R)
:::rank (D2)=6
::6+2=8

== Theorem ==
(rank-nullity Theorem, a.k.a. dimension Theorem)
:nullity(T)+rank(T)=dim V
:(for a l.t. T:V→W) when V is F.d.

== Proof ==
(To be continued next day)

09-240/Classnotes for Tuesday October 20

2009-12-07T07:24:31Z

C8sd: Missed closing tag.

'''Definition''': '''V''' and '''W''' are "isomorphic" if there exist linear transformations <math>\mathrm{T : V \rightarrow W}</math> and <math>\mathrm{S : W \rightarrow V}</math> such that <math>\mathrm{T \circ S} = I_\mathrm{W}</math> and <math>\mathrm{S \circ T} = I_\mathrm{V}</math>

'''Theorem''': If '''V''' and '''W''' are finite-dimensional over ''F'', then '''V''' is isomorphic to '''W''' iff dim('''V''') = dim('''W''')

'''Corollary''': If dim('''V''') = ''n'' then <math>\mathrm{V} \cong F^n</math>
:Note: <math>\cong</math> represents "is isomorphic to"

----

Two "mathematical structures" are "isomorphic" if there exists a "bijection" between their elements which preserves all relevant relations between such elements.

'''Example''': Plastic chess is "isomorphic" to ivory chess, but it is not isomorphic to checkers.

'''Example''': The game of 15. Players alternate drawing one card each.

Goal: To have exactly three of your cards add to 15.

Sample game:
* X picks 3
* O picks 7
* X picks 8
* O picks ''4''
* X picks 1
* O picks ''6''
* X picks 2
* O picks ''5''
* 4 + 6 + 5 = 15. O wins.

This game is isomorphic to Tic Tac Toe!

{| class="wikitable" border="0" cellpadding="2" cellspacing="0"
|-
| style="border-style: none solid solid none" | 4
| style="border-style: none solid solid solid" | 9
| style="border-style: none none solid solid" | 2
|-
| style="border-style: solid solid solid none" | 3
| style="border-style: solid solid solid solid" | 5
| style="border-style: solid none solid solid" | 7
|-
| style="border-style: solid solid none none" | 8
| style="border-style: solid solid none solid" | 1
| style="border-style: solid none none solid" | 6
|}

: O: 7, ''4'', ''6'', ''5'' -- Wins!
: X: 3, 8, 1, 2

Converts to:

{| class="wikitable" border="0" cellpadding="2" cellspacing="0"
|-
| style="border-style: none solid solid none" | O
| style="border-style: none solid solid solid" | 9
| style="border-style: none none solid solid" | X
|-
| style="border-style: solid solid solid none" | X
| style="border-style: solid solid solid solid" | O
| style="border-style: solid none solid solid" | O
|-
| style="border-style: solid solid none none" | X
| style="border-style: solid solid none solid" | X
| style="border-style: solid none none solid" | O
|}

: <math>\mathrm{S \circ T} = I_\mathrm{V}</math>
: <math>\mathrm{T \circ S} = I_\mathrm{W}</math>
: <math>\mathrm T(O_\mathrm{V}) = O_\mathrm{W}</math>

: <math>\mathrm T(x + y) = T(x) + T(y)</math>
: <math>\mathrm T(cv) = c\mathrm T(v)</math>
: Likewise for <math>\mathrm S</math>

: <math>z = x + y \Rightarrow \mathrm T(z) = \mathrm T(x) + \mathrm T(y)</math>
: <math>u = 7v \Rightarrow \mathrm T(u) = 7\mathrm T(v)</math>

Proof of Theorem <math>\iff</math> Assume dim('''V''') = dim('''W''') = ''n''
: There exists basis <math>\beta = \{u_1, \ldots, u_n\} \in \mathrm V</math>
: <math>\alpha = \{w_1, ..., w_n\} \in \mathrm W</math>
: by an earlier theorem, there exists a l.t. <math>\mathrm{T : V \rightarrow W}</math> such that <math>\mathrm T(u_i) = w_i</math>

<math>\mathrm T(\sum a_i u_i) = \sum a_i \mathrm T(u_i) = \sum a_i w_i</math>

There exists a l.t. <math>\mathrm{S : W \rightarrow V}</math> such that <math>\mathrm S(w_i) = u_i</math>

== Claim ==
: <math>\mathrm{S \circ T} = I_\mathrm{V}</math>
: <math>\mathrm{T \circ S} = I_\mathrm{W}</math>

== Proof ==
If u∈<math> \mathrm{V} </math> unto U=∑aiui
: (S∘T)(u)=S(T(u))=S(T(∑aiui))
: =S(∑aiwi)=∑aiui=u
: ⇒S∘T=Iv...
: ⇒Assume T&S as above exist
: Choose a basis β= (U1...Un) of V

== Claim ==
α=(W1=Tu1, W2=Tu2, ..., Wn=Tun)
: is a basis of W, so dim W=n

== Proof ==
α is lin. indep.
: T(0)=0=∑aiwi=∑aiTui=T(∑aiui)
: Apply S to both sides:
: 0=∑aiui
: So ∃iai=0 as β is a basis

α Spans W
: Given any w∈W let u=S(W)
: As β is a basis find ais in F s.t. v=∑aiui
Apply T to both sides: T(S(W))=T(u)=T(∑aiui)=∑aiT(ui)=∑aiWi
::: ∴ I win!!! (QED)

: T T
: V → W ⇔ V' → W'
: rank T=rank T'
Fix t:V→Wa l.t.<math>Insert formula here</math>

== Definition ==
# N(T) = ker(T) = {u∈V : Tu = 0W}
# R(T) = im(T) = {T(u) : u∈V}

== Prop/Def ==
# N(T) ⊂ V is a subspace of V-------nullity(T) := dim N(T)
# R(T) ⊂ W is a subspace of W--------rank(T) := dim R(T)

== Proof 1 ==
: x,y ∈N(T)⇒T(x)=0, T(y)=0
: T(x+y)=T9x)+T(y)=0+0=0
: x+y∈N(T)
::: ∴ I win!!! (QED)

== Proof 2 ==
: Let y∈R(T)⇒fix x s.t y=T(x),
: --------7y=7T(x)=T(7x)
: ----------⇒7y∈R(T)
::: ∴ I win!!! (QED)

== Examples ==
1.
: 0:V→W---------N(0)=V
: R(0)={0W}-----------nullity(0)=dim V
: --------------rank(0)=0
:: dim V+0=dimV
2.
:IV:V→V
:N(I)={0}
:nullity=0
:R(I)=dim V
:2'If T:V→W is an imorphism
:N(T)={0}
:nullity =0
:R(T)=W
:rank=dim W
::0+dim V=dim V
3.
:D:P7(R)→P7(R)
:Df=f'
::N(D)={C⊃C°: C∈R}=P0(R)
:R(D)⊂P6(R)
::nullity(D)=1
::basis:(1x°)
::rank(D)=7
:::7+1=8
4.
:3':D2:P7(R)
:D2f=f''
:W(D2)={ax+b: a,b∈R}=P1(R)
::nullity(D2)=2
::R(D2)=P5(R)
:::rank (D2)=6
::6+2=8

== Theorem ==
(rank-nullity Theorem, a.k.a. dimension Theorem)
:nullity(T)+rank(T)=dim V
:(for a l.t. T:V→W) when V is F.d.

== Proof ==
(To be continued next day)

09-240/Classnotes for Tuesday October 20

2009-12-07T07:20:41Z

C8sd: Format first half.

'''Definition''': '''V''' and '''W''' are "isomorphic" if there exist linear transformations <math>\mathrm{T : V \rightarrow W}</math> and <math>\mathrm{S : W \rightarrow V}</math> such that <math>\mathrm{T \circ S} = I_\mathrm{W}</math> and <math>\mathrm{S \circ T} = I_\mathrm{V}</math>

'''Theorem''': If '''V''' and '''W''' are finite-dimensional over ''F'', then '''V''' is isomorphic to '''W''' iff dim('''V''') = dim('''W''')

'''Corollary''': If dim('''V''') = ''n'' then <math>\mathrm{V} \cong F^n</math>
:Note: <math>\cong</math> represents "is isomorphic to"

----

Two "mathematical structures" are "isomorphic" if there exists a "bijection" between their elements which preserves all relevant relations between such elements.

'''Example''': Plastic chess is "isomorphic" to ivory chess, but it is not isomorphic to checkers.

'''Example''': The game of 15. Players alternate drawing one card each.

Goal: To have exactly three of your cards add to 15.

Sample game:
* X picks 3
* O picks 7
* X picks 8
* O picks ''4''
* X picks 1
* O picks ''6''
* X picks 2
* O picks ''5''
* 4 + 6 + 5 = 15. O wins.

This game is isomorphic to Tic Tac Toe!

{| class="wikitable" border="0" cellpadding="2" cellspacing="0"
|-
| style="border-style: none solid solid none" | 4
| style="border-style: none solid solid solid" | 9
| style="border-style: none none solid solid" | 2
|-
| style="border-style: solid solid solid none" | 3
| style="border-style: solid solid solid solid" | 5
| style="border-style: solid none solid solid" | 7
|-
| style="border-style: solid solid none none" | 8
| style="border-style: solid solid none solid" | 1
| style="border-style: solid none none solid" | 6
|}

: O: 7, ''4'', ''6'', ''5'' -- Wins!
: X: 3, 8, 1, 2

Converts to:

{| class="wikitable" border="0" cellpadding="2" cellspacing="0"
|-
| style="border-style: none solid solid none" | O
| style="border-style: none solid solid solid" | 9
| style="border-style: none none solid solid" | X
|-
| style="border-style: solid solid solid none" | X
| style="border-style: solid solid solid solid" | O
| style="border-style: solid none solid solid" | O
|-
| style="border-style: solid solid none none" | X
| style="border-style: solid solid none solid" | X
| style="border-style: solid none none solid" | O
|}

: <math>\mathrm{S \circ T} = I_\mathrm{V}</math>
: <math>\mathrm{T \circ S} = I_\mathrm{W}</math>
: <math>\mathrm T(O_\mathrm{V}) = O_\mathrm{W}</math>

: <math>\mathrm T(x + y) = T(x) + T(y)</math>
: <math>\mathrm T(cv) = c\mathrm T(v)</math>
: Likewise for <math>\mathrm S</math>

: <math>z = x + y \Rightarrow \mathrm T(z) = \mathrm T(x) + \mathrm T(y)</math>
: <math>u = 7v \Rightarrow \mathrm T(u) = 7\mathrm T(v)

Proof of Theorem <math>\iff</math> Assume dim('''V''') = dim('''W''') = ''n''
: There exists basis <math>\beta = \{u_1, \ldots, u_n\} \in \mathrm V</math>
: <math>\alpha = \{w_1, ..., w_n\} \in \mathrm W</math>
: by an earlier theorem, there exists a l.t. <math>\mathrm{T : V \rightarrow W}</math> such that <math>\mathrm T(u_i) = w_i</math>

<math>\mathrm T(\sum a_i u_i) = \sum a_i \mathrm T(u_i) = \sum a_i w_i</math>

There exists a l.t. <math>\mathrm{S : W \rightarrow V}</math> such that <math>\mathrm S(w_i) = u_i</math>

== Claim ==
: <math>\mathrm{S \circ T} = I_\mathrm{V}</math>
: <math>\mathrm{T \circ S} = I_\mathrm{W}</math>

== Proof ==
If u∈<math> \mathrm{V} </math> unto U=∑aiui
: (S∘T)(u)=S(T(u))=S(T(∑aiui))
: =S(∑aiwi)=∑aiui=u
: ⇒S∘T=Iv...
: ⇒Assume T&S as above exist
: Choose a basis β= (U1...Un) of V

== Claim ==
α=(W1=Tu1, W2=Tu2, ..., Wn=Tun)
: is a basis of W, so dim W=n

== Proof ==
α is lin. indep.
: T(0)=0=∑aiwi=∑aiTui=T(∑aiui)
: Apply S to both sides:
: 0=∑aiui
: So ∃iai=0 as β is a basis

α Spans W
: Given any w∈W let u=S(W)
: As β is a basis find ais in F s.t. v=∑aiui
Apply T to both sides: T(S(W))=T(u)=T(∑aiui)=∑aiT(ui)=∑aiWi
::: ∴ I win!!! (QED)

: T T
: V → W ⇔ V' → W'
: rank T=rank T'
Fix t:V→Wa l.t.<math>Insert formula here</math>

== Definition ==
# N(T) = ker(T) = {u∈V : Tu = 0W}
# R(T) = im(T) = {T(u) : u∈V}

== Prop/Def ==
# N(T) ⊂ V is a subspace of V-------nullity(T) := dim N(T)
# R(T) ⊂ W is a subspace of W--------rank(T) := dim R(T)

== Proof 1 ==
: x,y ∈N(T)⇒T(x)=0, T(y)=0
: T(x+y)=T9x)+T(y)=0+0=0
: x+y∈N(T)
::: ∴ I win!!! (QED)

== Proof 2 ==
: Let y∈R(T)⇒fix x s.t y=T(x),
: --------7y=7T(x)=T(7x)
: ----------⇒7y∈R(T)
::: ∴ I win!!! (QED)

== Examples ==
1.
: 0:V→W---------N(0)=V
: R(0)={0W}-----------nullity(0)=dim V
: --------------rank(0)=0
:: dim V+0=dimV
2.
:IV:V→V
:N(I)={0}
:nullity=0
:R(I)=dim V
:2'If T:V→W is an imorphism
:N(T)={0}
:nullity =0
:R(T)=W
:rank=dim W
::0+dim V=dim V
3.
:D:P7(R)→P7(R)
:Df=f'
::N(D)={C⊃C°: C∈R}=P0(R)
:R(D)⊂P6(R)
::nullity(D)=1
::basis:(1x°)
::rank(D)=7
:::7+1=8
4.
:3':D2:P7(R)
:D2f=f''
:W(D2)={ax+b: a,b∈R}=P1(R)
::nullity(D2)=2
::R(D2)=P5(R)
:::rank (D2)=6
::6+2=8

== Theorem ==
(rank-nullity Theorem, a.k.a. dimension Theorem)
:nullity(T)+rank(T)=dim V
:(for a l.t. T:V→W) when V is F.d.

== Proof ==
(To be continued next day)

09-240/Classnotes for Tuesday October 20

2009-12-07T06:50:32Z

C8sd: Format.

== Definition ==
V & W are "isomorphic" if there exists a linear transformation T:V → W & S:W → V such that T∘S=IW and S∘T=IV

== Theorem ==
If V& W are field dimensions over F, then V is isomorphic to W iff dim V=dim W

== Corollary ==
If dim V = n then <math> \mathrm{V} \cong \mathrm{F^n} </math>
:Note: <math> \cong </math> represents isomorphism

Two "mathematical structures" are "isomorphic" if there's a "bijection" between their elements which preserves all relevant relations between such elements.

Example:
Plastic chess is "isomorphic" to ivory chess, but it is not isomorphic to checkers.

Ex:
The game of 15. Players alternate drawing one card each.
Goal: To have exactly three of your cards add to 15.

O: 7, ''4, 6, 5'' → Wins!
X: 3, 8, 1, 2

This game is isomorphic to Tic Tac Toe!

{| class="wikitable" border="0" cellpadding="2" cellspacing="0"
|-
| style="border-style: none solid solid none" | 4
| style="border-style: none solid solid solid" | 9
| style="border-style: none none solid solid" | 2
|-
| style="border-style: solid solid solid none" | 3
| style="border-style: solid solid solid solid" | 5
| style="border-style: solid none solid solid" | 7
|-
| style="border-style: solid solid none none" | 8
| style="border-style: solid solid none solid" | 1
| style="border-style: solid none none solid" | 6
|}

Converts to:

{| class="wikitable" border="0" cellpadding="2" cellspacing="0"
|-
| style="border-style: none solid solid none" | O
| style="border-style: none solid solid solid" | 9
| style="border-style: none none solid solid" | X
|-
| style="border-style: solid solid solid none" | X
| style="border-style: solid solid solid solid" | O
| style="border-style: solid none solid solid" | O
|-
| style="border-style: solid solid none none" | X
| style="border-style: solid solid none solid" | X
| style="border-style: solid none none solid" | O
|}

: S∘T=IV
: T∘S=IW
: T(OV)=OW

: T(x+y)=T(x)+T(y)
: T(cV)=cT(V)
: Likewise for <math> \mathrm{S} </math>

: z=x+y ⇒ T(z)=T(x)+T(y)
: u=7v ⇒ T(u)=7T(v)

Proof of Theorem <math> \Leftrightarrow </math> Assume dim V= dim W=n
: ∃ basis β= (U1...Un) of V
: α=(W1...Wn) of W
: by an earlier theorem, ∃ a l.t. T:V→W such that T(Ui)=Wi

(T(∑aiui)=∑aiT(ui)=∑aiui)

∃ a l.t. S:W→V s.t. S(Wi)=Ui

== Claim ==
: S∘T=Iv
: T∘S=Iw

== Proof ==
If u∈<math> \mathrm{V} </math> unto U=∑aiui
: (S∘T)(u)=S(T(u))=S(T(∑aiui))
: =S(∑aiwi)=∑aiui=u
: ⇒S∘T=Iv...
: ⇒Assume T&S as above exist
: Choose a basis β= (U1...Un) of V

== Claim ==
α=(W1=Tu1, W2=Tu2, ..., Wn=Tun)
: is a basis of W, so dim W=n

== Proof ==
α is lin. indep.
: T(0)=0=∑aiwi=∑aiTui=T(∑aiui)
: Apply S to both sides:
: 0=∑aiui
: So ∃iai=0 as β is a basis

α Spans W
: Given any w∈W let u=S(W)
: As β is a basis find ais in F s.t. v=∑aiui
Apply T to both sides: T(S(W))=T(u)=T(∑aiui)=∑aiT(ui)=∑aiWi
::: ∴ I win!!! (QED)

: T T
: V → W ⇔ V' → W'
: rank T=rank T'
Fix t:V→Wa l.t.<math>Insert formula here</math>

== Definition ==
# N(T) = ker(T) = {u∈V : Tu = 0W}
# R(T) = im(T) = {T(u) : u∈V}

== Prop/Def ==
# N(T) ⊂ V is a subspace of V-------nullity(T) := dim N(T)
# R(T) ⊂ W is a subspace of W--------rank(T) := dim R(T)

== Proof 1 ==
: x,y ∈N(T)⇒T(x)=0, T(y)=0
: T(x+y)=T9x)+T(y)=0+0=0
: x+y∈N(T)
::: ∴ I win!!! (QED)

== Proof 2 ==
: Let y∈R(T)⇒fix x s.t y=T(x),
: --------7y=7T(x)=T(7x)
: ----------⇒7y∈R(T)
::: ∴ I win!!! (QED)

== Examples ==
1.
: 0:V→W---------N(0)=V
: R(0)={0W}-----------nullity(0)=dim V
: --------------rank(0)=0
:: dim V+0=dimV
2.
:IV:V→V
:N(I)={0}
:nullity=0
:R(I)=dim V
:2'If T:V→W is an imorphism
:N(T)={0}
:nullity =0
:R(T)=W
:rank=dim W
::0+dim V=dim V
3.
:D:P7(R)→P7(R)
:Df=f'
::N(D)={C⊃C°: C∈R}=P0(R)
:R(D)⊂P6(R)
::nullity(D)=1
::basis:(1x°)
::rank(D)=7
:::7+1=8
4.
:3':D2:P7(R)
:D2f=f''
:W(D2)={ax+b: a,b∈R}=P1(R)
::nullity(D2)=2
::R(D2)=P5(R)
:::rank (D2)=6
::6+2=8

== Theorem ==
(rank-nullity Theorem, a.k.a. dimension Theorem)
:nullity(T)+rank(T)=dim V
:(for a l.t. T:V→W) when V is F.d.

== Proof ==
(To be continued next day)

09-240/Classnotes for Tuesday October 13

2009-12-07T06:39:14Z

C8sd: Add partial notes.

{{09-240/Navigation}}
<gallery>
Image:Oct 13 notes page 1.JPG|[[User:Bright|Bright]] - 1
Image:Oct 13 notes page 2.JPG|2
Image:Oct 13 notes page 3.JPG|3
Image:Oct 13 notes page 4.JPG|4
</gallery>

== Replacement Theorem ==

...

<ol>
<li value="3">
dim('''V''') = ''n''
<ol style="list-style-type: lower-alpha">
<li>If ''G'' generates '''V''' then <math>|G| \ge n</math>. If also <math>\,\! |G| = n</math> then ''G'' is a basis.</li>
<li>If ''L'' is linearly dependent then <math>|L| \le n</math>. If also <math>\,\! |L| = n</math> then ''L'' is a basis. If also <math>\,\! |L| < n</math> then ''L'' can be extended to a basis.

Proofs
''a1.'' If ''G'' has a subset which is a basis then that subset has ''n'' elements, so <math>|G| \ge n</math>. 
''a2.'' Let <math>\beta</math> be a basis of '''V''', then <math>\,\! |B| = n</math>. Now use replacement with ''G'' & ''L'' = <math>\beta</math>. Hence, <math>|G| \ge |L| = |B| = n</math>. 
a. From a1 and a2, we know <math>|G| \ge n</math>. If <math>\,\! |G| = n</math> then ''G'' contains a basis <math>\beta</math>. But <math>\,\! |B| = n</math>, so <math>\,\! |G| = \beta</math>, and hence ''G'' is a basis. 
 
b. Use replacement with ''G' being some basis of '''V'''. |''G''| = ''n''. 
 
If <math>|L| = |G|</math> then <math>|R| = n = |G|</math>, so <math>R = G</math>, so <math>(G \backslash R) \cup L</math> generates, so ''L'' generates, so ''L'' is a basis since it is linearly independent. 
 
We have that ''L'' is basis. If <math>|L| < |G|</math> the nagain find <math>R \subset G</math> such that <math>|R| = |L|</math> and <math>(G \backslash R) \cup L</math> generates. 
 
1. <math>\beta</math> generates '''V'''. 
2. <math>|B| \le |G| - |R| + |L| = n</math>. So by part a, <math>\beta</math> is a basis. 
3.
</li>
</ol>
</li>
<li value="4">
If '''V''' is finite-dimensional (f.d.) and <math>\mathbf W \subset \mathbf V</math> is a subspace of '''V''', then '''W''' is also finite, and <math>\operatorname{dim}(\mathbf W) \le \operatorname{dim}(\mathbf V)</math>. If also dim('''W''') = dim('''V''') then '''W''' = '''V''' and if dim('''W''') < dim('''V''') then any basis of '''W''' can be extended to a basis of '''V'''. 
 
Proof: Assuming '''W''' is finite-dimensional, pick a basis <math>\beta</math> of '''W'''; <math>\beta</math> is linearly independent in '''V''' so by Corollary 3 of part b, <math>|B| \le \operatorname{dim}(\mathbf V) \Rightarrow \operatorname{dim}(\mathbf W) = |B| \le \operatorname{dim}(\mathbf V)</math>. So span(<math>\beta</math>) = '''V''' = '''W''' so '''V''' = '''W'''. ... 
 
Assume '''W''' is not finite-dimensional. <math>\mathbf W \ne \{0\}</math> so pick a <math>x_1 \in W</math> such that <math>x_1 \ne 0</math>. So <math>\{x_1\}</math> is linearly independent in '''W''', and <math>\operatorname{span}(\{x_1\}) \subsetneq \mathbf W</math>. Pick <math>x_2 \in W \backslash \operatorname{span}(\{x_1\})</math>. So <math>\{x_1, x_2\}</math> is linearly dependent and <math>\operatorname{span}(\{x_1, x_2\}) \subsetneq \mathbf W</math>. Pick <math>x_3 \in W \backslash \operatorname{span}(\{x_1, x_2\})</math> ... continue in this way to get a sequence <math>x_1, x_2, \ldots, x_{n+1}</math> where ''n'' = dim('''V''') and <math>\{x_1, \ldots, x_{n+1}\}</math> is linearly independent. There is a contradiction by Corollary 3.b.
</li>
</ol>

== The Lagrange Interpolation Formula ==

[Aside: <math>(a - x)(b - x) \ldots (z - x) = 0</math> because <math>x - x = 0</math>]

Where <math>1 \le i < n,</math>

Let <math>x_i</math> be distinct points in <math>\real</math>.

Let <math>y_i</math> be any points in <math>\real</math>.

Can you find a polynomial <math>P \in \mathbb P_n(\real)</math> such that <math>P(x_i) = y_i</math>? Is it unique?

'''Example''':

: <math>x_i = 0, 1, 3</math>
: <math>y_i = 5, 2, 2</math>

Can we find a <math>P \in \mathbb P_2</math> such that
: <math>P(0) = 5</math>
: <math>P(1) = 2</math>
: <math>P(2) = 2</math>?

Solution: Let <math>\tilde P_i(x) = \prod_{j=1, j \ne 1}^{n+1} (x - x_j) \in \mathbb P_n(\real)</math>. (Remember [http://en.wikipedia.org/wiki/Capital_pi_notation capital pi notation].)

Then <math>\tilde P_i(x_k) = \begin{cases}
0 & i \ne k \\
\prod_{i \ne j} (x_i - x_j) & i = k \\
\end{cases}</math>

: <math>\tilde P_1 = (x - x_2)(x - x_3) = (x - 1)(x - 3) = x^2 - 4x + 3</math>
: <math>\tilde P_2 = (x - 0)(x - 3) = x^2 - 3x</math>
: <math>\tilde P_3 = (x - 0)(x - 1) = x^2 - x</math>

<math>\begin{matrix}
\tilde P_1(0) = 3 & \tilde P_1(1) = 0 & \tilde P_1(3) = 0 \\
\tilde P_2(0) = 0 & \tilde P_2(1) = -2 & \tilde P_2(3) = 0 \\
\tilde P_3(0) = 0 & \tilde P_3(1) = 0 & \tilde P_3(3) = 6 \\
\end{matrix}</math>

Set <math>P_i(x) = \frac1{\tilde P_i(x)} \cdot \tilde P_i = \prod_{j \ne i} \frac{(x - x_j)}{(x_i - x_j)}</math>

Then <math>P_i(x_k) = \begin{cases}
0 & i \ne k \\
1 & i = k \\
\end{cases} </math>

: <math>P_1(x) = \frac13 x^2 - \frac43 x + 1</math>
: <math>P_2(x) = -\frac12 x^2 + \frac32 x</math>
: <math>P_3(x) = \frac16 x^2 - \frac16 x</math>

Let <math>P = \sum_{i=1}^{n+1} y_i P_i(x) \in \mathbb P_n(\real)</math>

: <math>P = 5 \cdot \left( \frac13 x^2 - \frac43 x + 1 \right) + 2 \cdot \left( -\frac12 x^2 + \frac32 x \right) + 2 \cdot \left( \frac16 x^2 - \frac16 x \right)</math>
: <math>\,\! = x^2 - 4x + 5</math>

09-240/Classnotes for Tuesday September 29

2009-12-07T05:29:42Z

C8sd: /* Linear combinations */ Formatting.

{{09-240/Navigation}}
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== Vector subspaces ==

'''Definition'''. <math>\mathbf W \subset \mathbf V</math> is a "subspace" if it is a vector space under the operations it inherits from '''V'''.

'''Theorem'''. <math>\mathbf W \subset \mathbf V</math> is a subspace iff it is "closed under addition and vector multiplication by scalars", i.e. <math>x, y \in \mathbf W \Rightarrow x + y \in \mathbf W</math> and <math>a \in F, x \in \mathbf W \Rightarrow ax \in \mathbf W</math>.

'''Goal''': Every VS has a "basis", so while we don't ''have'' to use coordinates, we always can.

'''Examples''' of what is not a subspace (without diagrams):
# A unit circle is not closed under addition of scalar multiplication.
# The x-axis <math>\cup</math> y-axis is closed under scalar multiplication, but not under addition.
# A single quadrant of the Cartesian plane is closed under addition, but not under scalar multiplication.

'''Examples''' of subspaces:
# <math>\{0\}</math>
# Any VS (which is a subspace of itself)
# A line passing through the origin (if it does not pass through the origin, then it is not closed under scalar multiplication)
# A plane
# Let <math>\mathbf V = \mathbb M_{n \times n}(F)</math>. If <math>W = \{ A \in \mathbf V : A^\top = A \}</math>, then '''W''' is a subspace of '''V'''. ('''W''' is the set of "symmetric" matrices in '''V'''; ''A''T denotes the [http://en.wikipedia.org/wiki/Transpose transpose] of ''A''.)
# <math>\mathbf W = \{ A \in \mathbb M_{n \times n} : \operatorname{tr}(A) = 0 \}</math>
#: where <math>\operatorname{tr}(A) = \sum_{i=1}^n a_{ii}</math> is the "trace" of ''A''.
#: Properties of trace:
## <math>\operatorname{tr}(0 \cdot A) = 0</math>
## <math>\operatorname{tr}(cA) = c \cdot \operatorname{tr}(A)</math>
## <math>\operatorname{tr}(A + B) = \operatorname{tr}(A) + \operatorname{tr}(B)</math>
#: so '''W''' is indeed a subspace.

'''Claim''': If '''W1''' and '''W2''' are subspaces of '''V''', then
# <math>W_1 \cap W_2 = \{ x \in \mathbf V : x \in \mathbf W_1 \mbox{ and } \mathbf W_2 \}</math> is a subspace of '''V''', '''W1''', and '''W2'''.
# But <math>W_1 \cup W_2 = \{ x \in \mathbf V : x \in \mathbf W_1 \mbox{ or } x \in \mathbf W_2 \}</math> is a subspace of '''V''' iff <math>\mathbf W_1 \subset \mathbf W_2</math> or <math>\mathbf W_2 \subset \mathbf W_1</math>. (See [[09-240:HW2|HW2]] pp. 20-21, #19.)

== Linear combinations ==

'''Definition''': A vector ''u'' is a "linear combination" (l.c.) of vectors ''u''1, ..., ''u''n if there exists scalars ''a''1, ..., ''a''n such that
: <math>u = a_1 u_1 + a_2 u_2 + \ldots + a_n u_n</math>

'''Example''': <math>\mathbb P_n(F) = \{ \mbox{Polynomials of degree at most } n \mbox{ with coefficients in } F \}</math> 
<math>= \left\{ \sum_{i=0}^n a_i x^i : a_i \in F \right\}</math>

'''Definition''': A subset <math>S \subset \mathbf V</math> "generates" or "spans" '''V''' iff the set of linear combinations of elements of ''S'' is all of '''V'''.

'''Example''': Let <math>\mathbf V = \mathbb M_{n \times n}(\real)</math> 
Let <math>M_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, M_2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, M_3 = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, M_4 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}</math> 
Then <math>S = \{ M_1, M_2, M_3, M_4 \}</math> generates '''V'''.

Proof: Given <math>\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathbb M_{2 \times 2}(\real)</math>, write 
<math>\begin{pmatrix} a & b \\ c & d \end{pmatrix} = aM_1 + bM_2 + cM_3 + dM_4.</math>

'''Example''': Let <math>N_1 = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}, N_2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, N_3 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, N_4 = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}</math> 
Does <math>\{ N_1, N_2, N_3, N_4 \}</math> generate '''V'''? 
: <math>M_1 = -\frac23 N_1 + \frac13(N_2 + N_3 + N_4)</math>
: <math>M_2 = -\frac23 N_2 + \frac13(N_1 + N_3 + N_4)</math>
: <math>M_3 = -\frac23 N_3 + \frac13(N_1 + N_2 + N_4)</math>
: <math>M_4 = -\frac23 N_4 + \frac13(N_1 + N_2 + N_3)</math>
Then <math>\begin{pmatrix} a & b \\ c & d \end{pmatrix} = aM_1 + bM_2 + cM_3 + dM_4 =</math>
: <math>a \cdot \left( -\frac23 N_1 + \frac13(N_2 + N_3 + N_4) \right) + b \cdot \left( -\frac23 N_2 + \frac13(N_1 + N_3 + N_4) \right) + \ldots</math>

'''Theorem''': If <math>S \in \mathbf V</math>, then <math>\operatorname{span}(S) = </math> {all l.c. of elements of ''S''} is a subspace of '''V'''.

09-240/Classnotes for Tuesday September 29

2009-12-07T05:28:27Z

C8sd: Add partial notes.

{{09-240/Navigation}}
{{09-240/Class Notes Warning}}

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</gallery>

== Vector subspaces ==

'''Definition'''. <math>\mathbf W \subset \mathbf V</math> is a "subspace" if it is a vector space under the operations it inherits from '''V'''.

'''Theorem'''. <math>\mathbf W \subset \mathbf V</math> is a subspace iff it is "closed under addition and vector multiplication by scalars", i.e. <math>x, y \in \mathbf W \Rightarrow x + y \in \mathbf W</math> and <math>a \in F, x \in \mathbf W \Rightarrow ax \in \mathbf W</math>.

'''Goal''': Every VS has a "basis", so while we don't ''have'' to use coordinates, we always can.

'''Examples''' of what is not a subspace (without diagrams):
# A unit circle is not closed under addition of scalar multiplication.
# The x-axis <math>\cup</math> y-axis is closed under scalar multiplication, but not under addition.
# A single quadrant of the Cartesian plane is closed under addition, but not under scalar multiplication.

'''Examples''' of subspaces:
# <math>\{0\}</math>
# Any VS (which is a subspace of itself)
# A line passing through the origin (if it does not pass through the origin, then it is not closed under scalar multiplication)
# A plane
# Let <math>\mathbf V = \mathbb M_{n \times n}(F)</math>. If <math>W = \{ A \in \mathbf V : A^\top = A \}</math>, then '''W''' is a subspace of '''V'''. ('''W''' is the set of "symmetric" matrices in '''V'''; ''A''T denotes the [http://en.wikipedia.org/wiki/Transpose transpose] of ''A''.)
# <math>\mathbf W = \{ A \in \mathbb M_{n \times n} : \operatorname{tr}(A) = 0 \}</math>
#: where <math>\operatorname{tr}(A) = \sum_{i=1}^n a_{ii}</math> is the "trace" of ''A''.
#: Properties of trace:
## <math>\operatorname{tr}(0 \cdot A) = 0</math>
## <math>\operatorname{tr}(cA) = c \cdot \operatorname{tr}(A)</math>
## <math>\operatorname{tr}(A + B) = \operatorname{tr}(A) + \operatorname{tr}(B)</math>
#: so '''W''' is indeed a subspace.

'''Claim''': If '''W1''' and '''W2''' are subspaces of '''V''', then
# <math>W_1 \cap W_2 = \{ x \in \mathbf V : x \in \mathbf W_1 \mbox{ and } \mathbf W_2 \}</math> is a subspace of '''V''', '''W1''', and '''W2'''.
# But <math>W_1 \cup W_2 = \{ x \in \mathbf V : x \in \mathbf W_1 \mbox{ or } x \in \mathbf W_2 \}</math> is a subspace of '''V''' iff <math>\mathbf W_1 \subset \mathbf W_2</math> or <math>\mathbf W_2 \subset \mathbf W_1</math>. (See [[09-240:HW2|HW2]] pp. 20-21, #19.)

== Linear combinations ==

'''Definition''': A vector ''u'' is a "linear combination" (l.c.) of vectors ''u''1, ..., ''u''n if there exists scalars ''a''1, ..., ''a''n such that
: <math>u = a_1 u_1 + a_2 u_2 + \ldots + a_n u_n</math>

'''Example''': <math>\mathbb P_n(F) = \{ \mbox{Polynomials of degree at most } n \mbox{ with coefficients in } ''F'' \}</math> 
<math>= \{ \sum_{i=0}^n a_i x^i : a_i \in F \}</math>

'''Definition''': A subset <math>S \subset \mathbf V</math> "generates" or "spans" '''V''' iff the set of linear combinations of elements of ''S'' is all of ''V''.

'''Example''': Let <math>\mathbf V = \mathbb M_{n \times n}(\real)</math> 
Let <math>M_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, M_2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, M_3 = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, M_4 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}</math> 
Then <math>S = \{ M_1, M_2, M_3, M_4 \}</math> generates '''V'''.

Proof: Given <math>\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathbb M_{2 \times 2}(\real), write</math> 
<math>\begin{pmatrix} a & b \\ c & d \end{pmatrix} = aM_1 + bM_2 + cM_3 + dM_4.</math>

'''Example''': Let <math>N_1 = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}, N_2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, N_3 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, N_4 = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}</math> 
Does <math>\{ N_1, N_2, N_3, N_4 \}</math> generate '''V'''? 
: <math>M_1 = -\frac23 N_1 + \frac13(N_2 + N_3 + N_4)</math>
: <math>M_2 = -\frac23 N_2 + \frac13(N_1 + N_3 + N_4)</math>
: <math>M_3 = -\frac23 N_3 + \frac13(N_1 + N_2 + N_4)</math>
: <math>M_4 = -\frac23 N_4 + \frac13(N_1 + N_2 + N_3)</math>
Then <math>\begin{pmatrix} a & b \\ c & d \end{pmatrix} = aM_1 + bM_2 + cM_3 + dM_4 =</math>
: <math>a \cdot \left( -\frac23 N_1 + \frac13(N_2 + N_3 + N_4) \right) + b \cdot \left( -\frac23 N_2 + \frac13(N_1 + N_3 + N_4) \right) + \ldots</math>

'''Theorem''': If <math>S \in \mathbf V</math>, then <math>\operatorname{span}(S) = </math> {all l.c. of elements of ''S''} is a subspace of '''V'''.

09-240:HW2

2009-12-07T03:18:03Z

C8sd: 117,648 = 7^6 - 1

{{09-240/Navigation}}

Read sections 1.1 through 1.3 in our textbook, and solve the following problems:

* Problems 3a and 3bcd on page 6, problems 1, 7, 18, 19 and 21 on pages 14-16 and problems 8, 9, 11 and 19 on pages 20-21. You need to submit only the underlined problems.

* Note that the numbers <math>1^6-1=0</math>, <math>2^6-1=63</math>, <math>3^6-1=728</math>, <math>4^6-1=4,095</math>, <math>5^6-1=15,624</math> and <math>6^6-1=46,655</math> are all divisible by <math>7</math>. The following four part exercise explains that this is not a coincidence. But first, let <math>p</math> be some odd prime number and let <math>{\mathbb F}_p</math> be the field with p elements as defined in class.
*# Prove that the product <math>b:=1\cdot 2\cdot\ldots\cdot(p-2)\cdot(p-1)</math> is a non-zero element of <math>{\mathbb F}_p</math>.
*# Let <math>a</math> be a non-zero element of <math>{\mathbb F}_p</math>. Prove that the sets <math>\{1,2,\ldots,(p-1)\}</math> and <math>\{1a,2a,\ldots,(p-1)a\}</math> are the same (though their elements may be listed here in a different order).
*# With <math>a</math> and <math>b</math> as in the previous two parts, show that <math>ba^{p-1}=b</math> in <math>{\mathbb F}_p</math>, and therefore <math>a^{p-1}=1</math> in <math>{\mathbb F}_p</math>.
*# How does this explain the fact that <math>4^6-1</math> is divisible by <math>7</math>?
You don't need to submit this exercise at all, but you will learn a lot by doing it!

This assignment is due at the tutorials on Thursday October 1. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if the teaching assistants will be having hard time deciphering your work they will give up and assume it is wrong.

09-240/Term Test

2009-12-03T18:02:38Z

C8sd: /* Solution Set */ Q4

{{09-240/Navigation}}

==Announcement==

Our one and only Term Test is coming up. It will take place in class on Thursday October 22 2009, starting promptly at 1:10PM and ending at 3:00PM sharp, in our normal classroom, MP103. It will consist of 4-5 questions (each may have several parts) on everything that will be covered in class by October 16: the axiomatic definition of fields and some basic properties of fields, <math>{\mathbb C}</math> and other examples, a tiny bit on the field with <math>p</math> elements <math>F_p</math>, the axiomatic definition of vector spaces, basic properties and examples of vector spaces, spans, linear combinations and linear equations, linear dependence and independence, bases, the replacement lemma and its consequences, a bit about linear transformations and a few smaller topics that we touched but that do not deserve their own headers.

Note that there may be some computations, but nothing that will require a calculator. Note also that I may include some questions from the homework assignments verbatim or nearly verbatim.

;Will there be "proof questions"?
:Sure. What else have we done so far?
;Do we need to know the proofs from class?
:Sure. There's a reason why these proofs are in class to start with; if they weren't valuable, we wouldn't have covered them.

No electronic devices capable of displaying text or sounding speech will be allowed.

In style and spirit this exam will not be very different of the one I gave 3 years ago. See [[06-240/Term Test]].

==The Test==

===Front Page===

<center>

'''Do not turn this page until instructed.'''

Math 240 Algebra I - Term Test

University of Toronto, October 22, 2009

'''Solve the 5 problems on the other side of this page.'''

Each of the problems is worth 20 points.

You have an hour and 50 minutes.

'''Notes.'''

</center>

* No outside material other than stationary and a basic calculator is allowed.
* We will have an hour of discussion time right after this test.
* The final exam date was posted by the faculty --- it will take place on Wednesday December 16 from 9AM until noon at room BN2S of the Clara Benson Building, 320 Huron Street (south west of Harbord cross Huron, home of the Faculty of Physical Education and Health).

<center> '''Good Luck!''' </center>

===Questions Page===

'''Solve the following 5 problems.''' Each of the problems is worth 20 points. You have an hour and 50 minutes.

'''Problem 1.''' Let <math>V</math> be a vector space over a field <math>F</math>, let <math>c\in F</math> and let <math>v\in V</math>. Prove that if <math>cv=0</math>, then either <math>c=0</math> or <math>v=0</math>.

'''Problem 2.'''
# In the field <math>{\mathbb C}</math> of complex numbers, compute
<center><math>4i(1+i)</math>      and     <math>\frac{4i}{1+i}</math>.</center>
(To be precise, "compute" means "write in the form <math>a+ib</math>, where
<math>a,b\in{\mathbb R}</math>").
# In the field <math>{\mathbb C}</math> of complex numbers, find an element <math>z</math> so that <math>z^2=2i</math>.
# In the 11-element field <math>F_{11}</math> of remainders modulo 11, find
all solutions of the equation <math>x^2=-2</math>.

'''Problem 3.''' Let <math>V</math> be a vector space and let <math>W_1</math> and <math>W_2</math> be subspaces of <math>V</math>. Prove that <math>W_1\cup W_2</math> is a subspace of <math>V</math> iff <math>W_1\subset W_2</math> or <math>W_2\subset W_1</math>.

'''Problem 4.''' Let <math>u_1,\ldots,u_n</math> be vectors in a
vector space over the field with two element <math>F_2</math>. Show that the number of
elements in the set <math>\mbox{Span}\{u_1,\dots,u_n\}</math> is equal to <math>2^n</math> if and only
if <math>u_1,\ldots,u_n</math> are linearly independent..

'''Problem 5.''' Find a polynomial <math>f\in P_3({\mathbb R})</math> that
satisfies <math>f(-1)=5</math>, <math>f(0)=4</math>, <math>f(1)=3</math>, and <math>f(2)=8</math>.

<center> '''Good Luck!''' </center>

{{09-240/Results of the Term Test}}

==Solution Set==
Students are most welcome to post a solution set here.

'''Problem 1'''

If <math>c \cdot v = 0</math> and <math>c \neq 0,</math>

then <math>c^{-1} \cdot (c \cdot v) = 0;</math>

thus <math>(c^{-1} \cdot c) \cdot v = 0;</math>

thus <math>1 \cdot v = 0;</math>

thus <math>v = 0.</math>

It should be remembered that "or" means "one or the other, or both."

'''Problem 2'''

(1)

<math>4i(1 + i) = 4i + 4i^2 = -4 + 4i</math>

<math>\frac{4i}{1+i} = \frac{4i(1 - i)}{(1 + i)(1 - i)} = \frac{4 + 4i}{2} = 2 + 2i</math>

(2)

If <math>z^2 = 2i</math> then <math>z = 1 + i \Rightarrow (1 + i)^2 = 1 + 2i + i^2 = 2i</math>

(3)

<math>x = 8, 3</math>

'''Problem 3.''' First suppose <math>W_1\subset W_2</math> or <math>W_2\subset W_1</math>, then <math>W_1\cup W_2</math> equals <math>W_2</math> or <math>W_1</math>. Either case <math>W_1\cup W_2</math> is a subspace of <math>V</math>.

Now suppose <math>W_1\cup W_2</math> is a subspace of <math>V</math> and neither <math>W_1\subset W_2</math> nor <math>W_2\subset W_1</math>. It must follow that <math>\exists\ x\in W_1</math> s.t. <math>x</math> is not in <math>W_2</math> and <math>\exists\ y\in W_2</math> s.t. <math>y</math> is not in <math>W_1</math>. Since <math>x,y\in W_1\cup W_2</math>, <math>x+y\in W_1\cup W_2</math>. If <math>x+y\in W_1</math>, then <math>\exists\ -x\in W_1</math> s.t. <math>(-x)+x+y\in W_1</math> and <math>(-x)+x=0</math>, so <math>y\in W_1</math>, a contradiction. If <math>x+y\in W_2</math>, then <math>\exists\ -y\in W_2</math> s.t. <math>x+y+(-y)\in W_2</math> and <math>(-y)+y=0</math>, so <math>x\in W_2</math>, a contradiction. Therefore either <math>W_1\subset W_2</math> or <math>W_2\subset W_1</math>.

'''Problem 4'''

Let <math>S_1 = \{ u_1, \ldots, u_n \}</math> and <math>S_2 = \operatorname{span}(S_1)</math>.

Assume <math>S_1</math> is linearly independent. Then <math>S_1</math> is a basis, and by basis properties, <math>\forall x \in S_2, \exists</math> unique <math>a_1, \ldots, a_n \in F</math> such that <math>a_1 u_1 + \ldots + a_n u_n = x</math>. <math>\forall a \in F</math>, <math>a = 0</math> or <math>a = 1</math>, so there are two possibilities for every <math>a</math>. Therefore, <math>|S_2| = \underbrace{2 \times \ldots \times 2}_n = 2^n</math>, so <math>S_1</math> is linearly independent <math>\Rightarrow |S_2| = 2^n</math>.

Assume <math>S_1</math> is linearly dependent. Then <math>\exists</math> non-trivial <math>a_i, \ldots, a_n</math> such that <math>a_1 u_1 + \ldots + a_n u_n = 0</math>. Since 0 has at least two representations, <math>|S_2| < 2^n</math>. Hence, <math>S_1</math> is linearly dependent <math>\Rightarrow |S_2| \ne 2^n</math>. By contrapositive, <math>|S_2| = 2^n \Rightarrow S_1</math> is linearly independent.

Hence, <math>S_1</math> is linearly independent <math>\iff |S_2| = 2^n</math>.

'''Problem 5'''

Recall that<math> f_i(x) = \prod_{k = 0 k \neq i}^{n}{\frac{x - c_k}{c_i - c_k}}</math> and that <math>g(x) = \sum_{i = 0}^{n}{b_if_i}</math>

After tedious computation with the addition of eraser bits covering your paper, you should obtain the following:

<math>f_0(x) = -\frac{1}{6}(x^3 - 3x^2 + 2x)</math>

<math>f_1(x) = \frac{1}{2}(x^3 - 2x^2 - x + 2)</math>

<math>f_2(x) = -\frac{1}{2}(x^3 - x^2 - 2x)</math>

<math>f_3(x) = \frac{1}{6}(x^3 - x)</math>

Therefore,

<math>g(x) = 5 \cdot -\frac{1}{6}(x^3 - 3x^2 + 2x) + 4 \cdot \frac{1}{2}(x^3 - 2x^2 - x + 2) + 3 \cdot -\frac{1}{2}(x^3 - x^2 - 2x) + 8 \cdot \frac{1}{6}(x^3 - x)</math>

Infinitely many eraser bits later...

<math>g(x) = x^3 - 2x + 4</math>

09-240/Term Test

2009-10-28T02:16:39Z

C8sd: /* Solution Set */ #3 copypasta

{{09-240/Navigation}}

==Announcement==

Our one and only Term Test is coming up. It will take place in class on Thursday October 22 2009, starting promptly at 1:10PM and ending at 3:00PM sharp, in our normal classroom, MP103. It will consist of 4-5 questions (each may have several parts) on everything that will be covered in class by October 16: the axiomatic definition of fields and some basic properties of fields, <math>{\mathbb C}</math> and other examples, a tiny bit on the field with <math>p</math> elements <math>F_p</math>, the axiomatic definition of vector spaces, basic properties and examples of vector spaces, spans, linear combinations and linear equations, linear dependence and independence, bases, the replacement lemma and its consequences, a bit about linear transformations and a few smaller topics that we touched but that do not deserve their own headers.

Note that there may be some computations, but nothing that will require a calculator. Note also that I may include some questions from the homework assignments verbatim or nearly verbatim.

;Will there be "proof questions"?
:Sure. What else have we done so far?
;Do we need to know the proofs from class?
:Sure. There's a reason why these proofs are in class to start with; if they weren't valuable, we wouldn't have covered them.

No electronic devices capable of displaying text or sounding speech will be allowed.

In style and spirit this exam will not be very different of the one I gave 3 years ago. See [[06-240/Term Test]].

==The Test==

===Front Page===

<center>

'''Do not turn this page until instructed.'''

Math 240 Algebra I - Term Test

University of Toronto, October 22, 2009

'''Solve the 5 problems on the other side of this page.'''

Each of the problems is worth 20 points.

You have an hour and 50 minutes.

'''Notes.'''

</center>

* No outside material other than stationary and a basic calculator is allowed.
* We will have an hour of discussion time right after this test.
* The final exam date was posted by the faculty --- it will take place on Wednesday December 16 from 9AM until noon at room BN2S of the Clara Benson Building, 320 Huron Street (south west of Harbord cross Huron, home of the Faculty of Physical Education and Health).

<center> '''Good Luck!''' </center>

===Questions Page===

'''Solve the following 5 problems.''' Each of the problems is worth 20 points. You have an hour and 50 minutes.

'''Problem 1.''' Let <math>V</math> be a vector space over a field <math>F</math>, let <math>c\in F</math> and let <math>v\in V</math>. Prove that if <math>cv=0</math>, then either <math>c=0</math> or <math>v=0</math>.

'''Problem 2.'''
# In the field <math>{\mathbb C}</math> of complex numbers, compute
<center><math>4i(1+i)</math>      and     <math>\frac{4i}{1+i}</math>.</center>
(To be precise, "compute" means "write in the form <math>a+ib</math>, where
<math>a,b\in{\mathbb R}</math>").
# In the field <math>{\mathbb C}</math> of complex numbers, find an element <math>z</math> so that <math>z^2=2i</math>.
# In the 11-element field <math>F_{11}</math> of remainders modulo 11, find
all solutions of the equation <math>x^2=-2</math>.

'''Problem 3.''' Let <math>V</math> be a vector space and let <math>W_1</math> and <math>W_2</math> be subspaces of <math>V</math>. Prove that <math>W_1\cup W_2</math> is a subspace of <math>V</math> iff <math>W_1\subset W_2</math> or <math>W_2\subset W_1</math>.

'''Problem 4.''' Let <math>u_1,\ldots,u_n</math> be vectors in a
vector space over the field with two element <math>F_2</math>. Show that the number of
elements in the set <math>\mbox{Span}\{u_1,\dots,u_n\}</math> is equal to <math>2^n</math> if and only
if <math>u_1,\ldots,u_n</math> are linearly independent..

'''Problem 5.''' Find a polynomial <math>f\in P_3({\mathbb R})</math> that
satisfies <math>f(-1)=5</math>, <math>f(0)=4</math>, <math>f(1)=3</math>, and <math>f(2)=8</math>.

<center> '''Good Luck!''' </center>

{{09-240/Results of the Term Test}}

==Solution Set==
Students are most welcome to post a solution set here.

'''Problem 1'''

If <math>a \cdot b = 0</math> and <math>a \neq 0,</math>

then <math>a^{-1} \cdot (a \cdot b) = 0;</math>

thus <math>(a^{-1} \cdot a) \cdot b = 0;</math>

thus <math>1 \cdot b = 0;</math>

thus <math>b = 0.</math>

It should be remembered that "or" means "one or the other, or both."

'''Problem 2'''

(1)

<math>4i(1 + i) = 4i + 4i^2 = -4 + 4i</math>

<math>\frac{4i}{1+i} = \frac{4i(1 - i)}{(1 + i)(1 - i)} = \frac{4 + 4i}{2} = 2 + 2i</math>

(2)

If <math>z^2 = 2i</math> then <math>z = 1 + i \Rightarrow (1 + i)^2 = 1 + 2i + i^2 = 2i</math>

(3)

<math>x = 8, 3</math>

'''Problem 3.''' First suppose <math>W_1\subset W_2</math> or <math>W_2\subset W_1</math>, then <math>W_1\cup W_2</math> equals <math>W_2</math> or <math>W_1</math>. Either case <math>W_1\cup W_2</math> is a subspace of <math>V</math>.

Now suppose <math>W_1\cup W_2</math> is a subspace of <math>V</math> and neither <math>W_1\subset W_2</math> nor <math>W_2\subset W_1</math>. It must follow that <math>\exists\ x\in W_1</math> s.t. <math>x</math> is not in <math>W_2</math> and <math>\exists\ y\in W_2</math> s.t. <math>y</math> is not in <math>W_1</math>. Since <math>x,y\in W_1\cup W_2</math>, <math>x+y\in W_1\cup W_2</math>. If <math>x+y\in W_1</math>, then <math>\exists\ -x\in W_1</math> s.t. <math>(-x)+x+y\in W_1</math> and <math>(-x)+x=0</math>, so <math>y\in W_1</math>, a contradiction. If <math>x+y\in W_2</math>, then <math>\exists\ -y\in W_2</math> s.t. <math>x+y+(-y)\in W_2</math> and <math>(-y)+y=0</math>, so <math>x\in W_2</math>, a contradiction. Therefore either <math>W_1\subset W_2</math> or <math>W_2\subset W_1</math>.

'''Problem 4'''

'''Problem 5'''

Recall that<math> f_i(x) = \prod_{k = 0 k \neq i}^{n}{\frac{x - c_k}{c_i - c_k}}</math> and that <math>g(x) = \sum_{i = 0}^{n}{b_if_i}</math>

After tedious computation with the addition of eraser bits covering your paper, you should obtain the following:

<math>f_0(x) = -\frac{1}{6}(x^3 - 3x^2 + 2x)</math>

<math>f_1(x) = \frac{1}{2}(x^3 - 2x^2 - x + 2)</math>

<math>f_2(x) = -\frac{1}{2}(x^3 - x^2 - 2x)</math>

<math>f_3(x) = \frac{1}{6}(x^3 - x)</math>

Therefore,

<math>g(x) = 5 \cdot -\frac{1}{6}(x^3 - 3x^2 + 2x) + 4 \cdot \frac{1}{2}(x^3 - 2x^2 - x + 2) + 3 \cdot -\frac{1}{2}(x^3 - x^2 - 2x) + 8 \cdot \frac{1}{6}(x^3 - x)</math>

Infinitely many eraser bits later...

<math>g(x) = x^3 - 2x + 4</math>

06-240/Term Test

2009-10-22T03:49:46Z

C8sd: /* Solution Set */ grammar

{{06-240/Navigation}}

==The Test==

===Front Page===

<center>

'''Do not turn this page until instructed.'''

Math 240 Algebra I - Term Test

University of Toronto, October 24, 2006

'''Solve the 5 problems on the other side of this page.'''

Each of the problems is worth 20 points.

You have an hour and 45 minutes.

'''Notes.'''

</center>

* No outside material other than stationary and a basic calculator is allowed.
* We will have an extra hour of class time in our regular class room on Thursday, replacing the first tutorial hour.
* The final exam date was posted by the faculty - it will take place on Wednesday December 13 from 2PM until 5PM at room 3 of the Clara Benson Building, 320 Huron Street (south west of Harbord cross Huron, home of the Faculty of Physical Education and Health).

<center> '''Good Luck!''' </center>

===Questions Page===

'''Solve the following 5 problems.''' Each of the problems is worth 20 points. You have an hour and 45 minutes.

'''Problem 1.''' Let <math>F</math> be a field with zero element <math>0_F</math>, let <math>V</math> be a vector space with zero element <math>0_V</math> and let <math>v\in V</math> be some vector. Using only the axioms of fields and vector spaces, prove that <math>0_F\cdot v=0_V</math>.

'''Problem 2.'''
# In the field <math>{\mathbb C}</math> of complex numbers, compute <center><math>\frac{1}{2+3i}+\frac{1}{2-3i}</math>      and     <math>\frac{1}{2+3i}-\frac{1}{2-3i}</math>.</center>
# Working in the field <math>{\mathbb Z}/7</math> of integers modulo 7, make a table showing the values of <math>a^{-1}</math> for every <math>a\neq 0</math>.

'''Problem 3.''' Let <math>V</math> be a vector space and let <math>W_1</math> and <math>W_2</math> be subspaces of <math>V</math>. Prove that <math>W_1\cup W_2</math> is a subspace of <math>V</math> iff <math>W_1\subset W_2</math> or <math>W_2\subset W_1</math>.

'''Problem 4.''' In the vector space <math>M_{2\times 2}({\mathbb Q})</math>, decide if the matrix <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}</math> is a linear combination of the elements of <math>S=\left\{\begin{pmatrix}1&0\\-1&0\end{pmatrix},\ \begin{pmatrix}0&1\\0&1\end{pmatrix},\ \begin{pmatrix}1&1\\0&0\end{pmatrix}\right\}</math>.

'''Problem 5.''' Let <math>V</math> be a finite dimensional vector space and let <math>W_1</math> and <math>W_2</math> be subspaces of <math>V</math> for which <math>W_1\cap W_2=\{0\}</math>. Denote the linear span of <math>W_1\cup W_2</math> by <math>W_1+W_2</math>. Prove that <math>\dim(W_1+W_2)=\dim W_1 + \dim W_2</math>.

<center> '''Good Luck!''' </center>

{{06-240/Results of the Term Test}}

==Solution Set==
Students are most welcome to post a solution set here.

<div style="color: red;">WARNING: The solution set below, written for students and by students, is provided "as is", with absolutely no warranty. It can not be assumed to be complete, correct, reliable or relevant. If you don't like it, don't read it.

Visit this pages' history tab to see who added what and when.</div>

'''Problem 1.''' <math>0_F\cdot v=(0_F+0_F)\cdot v</math> (by F3)

<math>(0_F+0_F)\cdot v=0_F\cdot v+0_F\cdot v</math> (by VS8)

By VS4, <math>\exists\ (0_F\cdot v)' s.t. (0_F\cdot v)+(0_F\cdot v)'=0_V</math>

Add <math>(0_F\cdot v)'</math> to both sides of <math>0_F\cdot v=0_F\cdot v+0_F\cdot v</math>

<math>(0_F\cdot v)'+(0_F\cdot v)=[(0_F\cdot v)'+0_F\cdot v]+0_F\cdot v</math>

<math>0_V=0_V+0_F\cdot v</math> (by construction)

<math>0_V=0_F\cdot v</math> (by VS3)

'''Problem 2.''' 1) <math>\frac{1}{2+3i}+\frac{1}{2-3i}=\frac{(2-3i)+(2+3i)}{(2+3i)(2-3i)}=\frac{4}{2^2+3^3}=\frac{4}{13}</math>

<math>\frac{1}{2+3i}-\frac{1}{2-3i}=\frac{(2-3i)-(2+3i)}{(2+3i)(2-3i)}=\frac{-6i}{2^2+3^3}=-\frac{6}{13}i</math>

2) <math>1^{-1}=1</math>

<math>2^{-1}=4</math>

<math>3^{-1}=5</math>

<math>4^{-1}=2</math>

<math>5^{-1}=3</math>

<math>6^{-1}=6</math>

'''Problem 3.''' First suppose <math>W_1\subset W_2</math> or <math>W_2\subset W_1</math>, then <math>W_1\cup W_2</math> equals <math>W_2</math> or <math>W_1</math>. Either case <math>W_1\cup W_2</math> is a subspace of <math>V</math>.

Now suppose <math>W_1\cup W_2</math> is a subspace of <math>V</math> and neither <math>W_1\subset W_2</math> nor <math>W_2\subset W_1</math>. It must follow that <math>\exists\ x\in W_1</math> s.t. <math>x</math> is not in <math>W_2</math> and <math>\exists\ y\in W_2</math> s.t. <math>y</math> is not in <math>W_1</math>. Since <math>x,y\in W_1\cup W_2</math>, <math>x+y\in W_1\cup W_2</math>. If <math>x+y\in W_1</math>, then <math>\exists\ -x\in W_1</math> s.t. <math>(-x)+x+y\in W_1</math> and <math>(-x)+x=0</math>, so <math>y\in W_1</math>, a contradiction. If <math>x+y\in W_2</math>, then <math>\exists\ -y\in W_2</math> s.t. <math>x+y+(-y)\in W_2</math> and <math>(-y)+y=0</math>, so <math>x\in W_2</math>, a contradiction. Therefore either <math>W_1\subset W_2</math> or <math>W_2\subset W_1</math>.

'''Problem 4.''' Suppose <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}</math> is a linear combination of <math>\begin{pmatrix}1&0\\-1&0\end{pmatrix}</math>, <math>\begin{pmatrix}0&1\\0&1\end{pmatrix}</math>, and <math>\begin{pmatrix}1&1\\0&0\end{pmatrix}</math>, then <math>\exists\ a_1, a_2, a_3\in F</math> s.t. <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}=a_1\begin{pmatrix}1&0\\-1&0\end{pmatrix}+a_2\begin{pmatrix}0&1\\0&1\end{pmatrix}+a_3\begin{pmatrix}1&1\\0&0\end{pmatrix}</math>.

<math>\mbox{Need to solve}\begin{cases}
1=a_1^{}+a_3^{}\\
2=a_2^{}+a_3^{}\\
-3=-a_1^{}\\
4=a_2^{}\end{cases}</math>

Solving the equations yields <math>a_1=3</math>, <math>a_2=4</math>, <math>a_3=-2</math>, so <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}</math> is a linear combination of <math>\begin{pmatrix}1&0\\-1&0\end{pmatrix}</math>, <math>\begin{pmatrix}0&1\\0&1\end{pmatrix}</math>, and <math>\begin{pmatrix}1&1\\0&0\end{pmatrix}</math>. Specifically, <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}=3\begin{pmatrix}1&0\\-1&0\end{pmatrix}+4\begin{pmatrix}0&1\\0&1\end{pmatrix}-2\begin{pmatrix}1&1\\0&0\end{pmatrix}</math>.

'''Problem 5.''' Since <math>V</math> is finite-dimensional, then so are <math>W_1</math> and <math>W_2</math> and their basis. Let <math>{a_1, a_2,..., a_m}</math> be the basis of <math>W_1</math> and <math>{b_1, b_2,..., b_n}</math> be the basis of <math>W_2</math> so <math>\dim W_1=m</math> and <math>\dim W_2=n</math>.

We know <math>{a_1, a_2,..., a_m}</math> and <math>{b_1, b_2,..., b_n}</math> are linearly independent and clearly <math>{a_1, a_2,..., a_m, b_1, b_2,..., b_n}</math> spans <math>W_1+W_2</math>. If <math>{a_1, a_2,..., a_m, b_1, b_2,..., b_n}</math> is linearly dependent, then there exist not all zero coefficients <math>c_1, c_2,..., c_{m+n}\in F</math> s.t. <math>c_1 a_1+c_2 a_2+...+c_m a_m+c_{m+1} b_1+c_{m+2} b_2+...+c_{m+n} b_n=0</math>. Then some linear combinations of <math>{a_1, a_2,..., a_m}</math> with not all zero coefficients can be expressed in linear combinations of <math>{b_1, b_2,..., b_n}</math>, but this would imply <math>W_1\cap W_2\neq \{0\}</math>, a contradiction. Therefore <math>{a_1, a_2,..., a_m, b_1, b_2,..., b_n}</math> is a linearly independent set that spans <math>W_1+W_2</math>, it's the basis of <math>W_1+W_2</math>. We have <math>\dim (W_1+W_2)=m+n=\dim W_1+\dim W_2</math>.

Template:09-240/Navigation

2009-10-16T16:43:09Z

C8sd: Add space after comma.

09-240/Class Photo

2009-10-09T21:25:37Z

C8sd: /* Who We Are... */ abc order

Our class on September 24, 2009:

[[Image:09-240-ClassPhoto.jpg|thumb|centre|500px|Class Photo: click to enlarge]]
{{09-240/Navigation}}

Please identify yourself in this photo! There are two ways to do that:

* [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.
* Send [[User:Drorbn|Dror]] an email message with this information.

The first option is more fun but less private.

===Who We Are...===

{| align=center border=1 cellspacing=0
|-
!First name
!Last name
!UserID
!Email
!In the photo
!Comments
{{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn@ math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank. For better line-breaking, leave a space next to the "@" in email addresses.}}
{{Photo Entry|last=Bahadur|first=Jared|userid=bahadurj|email=jared.bahadur@ utoronto.ca| location=left index fingernail visible on handrail at Bar-Natan's right|comments=}}
{{Photo Entry|last=Barkley|first=Max|userid=SpectralWolf|email=max.barkley@ utoronto.ca|location=fourth guy from the left in the front row, beside the guy with the huge mouth|comments=}}
{{Photo Entry|last=Binder|first=Polina|userid=polyacat|email=polina.binder@ utoronto.ca|location= The girl in blue, to the right of the stairs at the top.|comments=}}
{{Photo Entry|last=Chen|first=Yi Le|userid=AlecC|email=alec.chen@ utoronto.ca|location= the guy in black T-shirt at 2nd to the left, on the 2nd standing row, with glasses|comments=}}
{{Photo Entry|last=Chou|first=Daniel|userid=Danielchou|email=daniel.chou@ utoronto.ca|location=the guy above the guy with the yellow shirt.|comments=Hi People.}}
{{Photo Entry|last=Chung|first=Ha Yoon|userid=babo|email=hayoon.chung@ utoronto.ca|location= the guy above k.ott|comments=I'm hungry.}}
{{Photo Entry|last=Chung|first=Wen-Jian|userid=cwjian90|email=cwjian90@ yahoo.com|location= top right corner, guy with Einstein's face on his shirt.|comments=}}
{{Photo Entry|last=Desmarais|first=Eric|userid=EricD|email=eric.desmarais@ utoronto.ca|location=back row, two to the right of the centre lamp post with the beard and glasses and the "has the picture been taken?" look.|comments=}}
{{Photo Entry|last=Dranovski|first=Anne|userid=anne.d|email=a.dranovski@ utoronto.ca|location=third row, sitting, green and happy.|comments=}}
{{Photo Entry|last=Gafar|first=Jonathan|userid=JG89|email=jonathan.gafar@ utoronto.ca|location=the guy in the last row to the right of the big pole.|comments=}}
{{Photo Entry|last=Gu|first=Hyungmo|userid=Hmgu7|email=moe.gu@ utoronto.ca|location=the guy with brown
shirt in the fifth row, third from the right.|comments=}}
{{Photo Entry|last=Ivanov|first=Vesselin|userid=Gungrave|email=vesselin.ivanov@ utoronto.ca|location=Last row, 7th from the right, with a shiny dog-tag.|comments=}}
{{Photo Entry|last=Jeffery|first=Travis|userid=Travisjeffery|email=t.jeffery@ utoronto.ca|location=Blue sweater, headphones around neck on the left.|comments=Ask her to wait a moment--I am almost done.}}
{{Photo Entry|last=Klambauer|first=Max|userid=Max.k|email=maximilian. klambauer@ utoronto.ca|location=Last row, almost as far right as possible; I'm the guy with long hair hidden behind the other guy with long, black hair.|comments=}}
{{Photo Entry|last=Lee|first=Joonhan|userid=Joonhan86|email=tomko_lee@ hotmail.com|location= one small single picture at top right hand side|comments=}}
{{Photo Entry|last=Li|first=Zhao|userid=lzh8571|email=lzh8571@ live.com|location=at the 4th from right in the top right picture, with glasses and jacket|comments=}}
{{Photo Entry|last=Lindquist|first=Emma|userid=Elindquist|email=emmalindquist@ utoronto.ca|location=Far left of front row|comments=}}
{{Photo Entry|last=Makarov|first=Serhei|userid=Serhei|email=serhei.makarov@ utoronto.ca|location=gray shirt, standing on the step under the guy with fingers on the rail (i.e. bahadurj)|comments=}}
{{Photo Entry|last=Mann|first=Alex|userid=Mannimal|email=alexander.mann@ utoronto.ca|location=last row, third guy to the right of the guy on the rail|comments=240>157}}
{{Photo Entry|last=Mantynen|first=Paul|userid=mantynen|email=paul.mantynen@ utoronto.ca| location= 4th row, third from the right in a black shirt|comments=}}
{{Photo Entry|last=McTaggart|first=Raymond|userid=Raymct|email=raymond.mctaggart@ gmail.com|location=Flying in the top-right sky. The one with the black t-shirt|comments= be <math> \mathbb{R} </math>}}
{{Photo Entry|last=Miao|first=Ying|userid=Miaoying|email=ying.miao@ utoronto.ca|location=girl beside the guy who was wearing blue cap and smiling happily.|comments=}}
{{Photo Entry|last=Michaud|first=Adam|userid=Invariel|email=invariel@ gmail.com|location=Back row, black coat, dark hair, full, dark beard, fifth from the right.|comments=Braaaaaaains...}}
{{Photo Entry|last=Milcak|first=Juraj|userid=Milcak|email=j.milcak@ utoronto.ca|location=2nd row, 1st from the right, green shirt|comments=}}
{{Photo Entry|last=Nikitakis|first=George|userid=GeorgeN|email=george.nikitakis@ utoronto.ca|location=Part of the group floating in the top right corner. Second from the left|comments=}}
{{Photo Entry|last=Ott|first=Kristian|userid=k.ott|email=k.ott@ utoronto.ca| location= 4th guy from the right in the second row, wearing a green striped sweater|comments=}}
{{Photo Entry|last=Park|first=Sanghee|userid=SangheeP|email=shee.park@ utoronto.ca|location=glasses, a girl, first row, third from right, holding starbucks."|comments=}}
{{Photo Entry|last=Pistone|first=Jamie|userid=JPistone|email=jamie.pistone@ utoronto.ca|location=Guy in the second row on the outside of the left railing, guarded by the two guys with their arms crossed.|comments=}}
{{Photo Entry|last=Ramdhayan|first=Dinesh|userid=Dinesh®|email=dinesh.ramdhayan@ utoronto.ca|location=top left corner, 4th from the left|comments=}}
{{Photo Entry|last=Simmons|first=Olivia|userid=OSimmons|email=olivia.simmons@ yahoo.ca|location=girl on the second top rown two people from the right|comments=}}
{{Photo Entry|last=Simpson|first=Evan|userid=ESimpson|email=evan.simpson@ utoronto.ca|location=2nd Front Row, 2nd from the left, Green Shirt, Glasses|comments=Math is addictive don't try it.}}
{{Photo Entry|last=Sinn|first=Daniel|userid=c8sd|email=|location=second-last row, fourth from left|comments=}}
{{Photo Entry|last=Valdez|first=Wilbur|userid=Valdez84|email=wilbur.valdez@ utoronto.ca|location= Third row yellow shirt guy :)|comments=}}
{{Photo Entry|last=VanZanten|first=Johan|userid=jvzanten|email=j.vanzanten@ utoronto.ca|location=first guy in blue shirt standing on the left of left railing |comments=}}
{{Photo Entry|last=Wang|first=Yu|userid=Bright|email=bright_wangca@ hotmail.com|location=6th row (i think), 5th person from right, white shirt. Behind the guy with a plaid shirt and glasses.|comments=}}
{{Photo Entry|last=Zapf-Belanger|first=Erik|userid=erikzb|email=erikzb@ gmail.com|location=the guy third from the left in the front with the gigantic mouth|comments=Live long and prosper.}}
{{Photo Entry|last=Zhao|first=Yi|userid=zy861|email=zy861100@ hotmail.com|location=guy at the 6th row and 4th from right, with grey shirt and black glasses|comments=}}

09-240/Class Photo

2009-10-08T17:06:25Z

C8sd: /* Who We Are... */ Z > V

Our class on September 24, 2009:

[[Image:09-240-ClassPhoto.jpg|thumb|centre|500px|Class Photo: click to enlarge]]
{{09-240/Navigation}}

Please identify yourself in this photo! There are two ways to do that:

* [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.
* Send [[User:Drorbn|Dror]] an email message with this information.

The first option is more fun but less private.

===Who We Are...===

{| align=center border=1 cellspacing=0
|-
!First name
!Last name
!UserID
!Email
!In the photo
!Comments
{{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn@ math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank. For better line-breaking, leave a space next to the "@" in email addresses.}}
{{Photo Entry|last=Bahadur|first=Jared|userid=bahadurj|email=jared.bahadur@ utoronto.ca| location=left index fingernail visible on handrail at Bar-Natan's right|comments=}}
{{Photo Entry|last=Barkley|first=Max|userid=SpectralWolf|email=max.barkley@ utoronto.ca|location=fourth guy from the left in the front row, beside the guy with the huge mouth|comments=}}
{{Photo Entry|last=Binder|first=Polina|userid=polyacat|email=polina.binder@ utoronto.ca|location= The girl in blue, to the right of the stairs at the top.|comments=}}
{{Photo Entry|last=Chen|first=Yi Le|userid=AlecC|email=alec.chen@ utoronto.ca|location= the guy in black T-shirt at 2nd to the left, on the 2nd standing row, with glasses|comments=}}
{{Photo Entry|last=Chou|first=Daniel|userid=Danielchou|email=daniel.chou@ utoronto.ca|location=the guy above the guy with the yellow shirt.|comments=Hi People.}}
{{Photo Entry|last=Chung|first=Wen-Jian|userid=cwjian90|email=cwjian90@ yahoo.com|location= top right corner, guy with Einstein's face on his shirt.|comments=}}
{{Photo Entry|last=Desmarais|first=Eric|userid=EricD|email=eric.desmarais@ utoronto.ca|location=back row, two to the right of the centre lamp post with the beard and glasses and the "has the picture been taken?" look.|comments=}}
{{Photo Entry|last=Dranovski|first=Anne|userid=anne.d|email=a.dranovski@ utoronto.ca|location=third row, sitting, green and happy.|comments=}}
{{Photo Entry|last=Gafar|first=Jonathan|userid=JG89|email=jonathan.gafar@ utoronto.ca|location=the guy in the last row to the right of the big pole.|comments=}}
{{Photo Entry|last=Gu|first=Hyungmo|userid=Hmgu7|email=moe.gu@ utoronto.ca|location=the guy with brown
shirt in the fifth row, third from the right.|comments=}}
{{Photo Entry|last=Ivanov|first=Vesselin|userid=Gungrave|email=vesselin.ivanov@ utoronto.ca|location=Last row, 7th from the right, with a shiny dog-tag.|comments=}}
{{Photo Entry|last=Jeffery|first=Travis|userid=Travisjeffery|email=t.jeffery@ utoronto.ca|location=Blue sweater, headphones around neck on the left.|comments=Ask her to wait a moment--I am almost done.}}
{{Photo Entry|last=Klambauer|first=Max|userid=Max.k|email=maximilian.klambauer@ utoronto.ca|location=Last row, almost as far right as possible; I'm the guy with long hair hidden behind the other guy with long, black hair.|comments=}}
{{Photo Entry|last=Li|first=Zhao|userid=lzh8571|email=lzh8571@ live.com|location=at the 4th from right in the top right picture, with glasses and jacket|comments=}}
{{Photo Entry|last=Lindquist|first=Emma|userid=Elindquist|email=emmalindquist@ utoronto.ca|location=Far left of front row|comments=}}
{{Photo Entry|last=Makarov|first=Serhei|userid=Serhei|email=serhei.makarov@ utoronto.ca|location=gray shirt, standing on the step under the guy with fingers on the rail (i.e. bahadurj)|comments=}}
{{Photo Entry|last=Mann|first=Alex|userid=Mannimal|email=alexander.mann@ utoronto.ca|location=last row, third guy to the right of the guy on the rail|comments=240>157}}
{{Photo Entry|last=Mantynen|first=Paul|userid=mantynen|email=paul.mantynen@ utoronto.ca| location= 4th row, third from the right in a black shirt|comments=}}
{{Photo Entry|last=McTaggart|first=Raymond|userid=Raymct|email=raymond.mctaggart@ gmail.com|location=Flying in the top-right sky. The one with the black t-shirt|comments= be <math> \mathbb{R} </math>}}
{{Photo Entry|last=Miao|first=Ying|userid=Miaoying|email=ying.miao@ utoronto.ca|location=girl beside the guy who was wearing blue cap and smiling happily.|comments=}}
{{Photo Entry|last=Michaud|first=Adam|userid=Invariel|email=invariel@ gmail.com|location=Back row, black coat, dark hair, full, dark beard, fifth from the right.|comments=Braaaaaaains...}}
{{Photo Entry|last=Milcak|first=Juraj|userid=Milcak|email=j.milcak@ utoronto.ca|location=2nd row, 1st from the right, green shirt|comments=}}
{{Photo Entry|last=Nikitakis|first=George|userid=GeorgeN|email=george.nikitakis@ utoronto.ca|location=Part of the group floating in the top right corner. Second from the left|comments=}}
{{Photo Entry|last=Ott|first=Kristian|userid=k.ott|email=k.ott@ utoronto.ca| location= 4th guy from the right in the second row, wearing a green striped sweater|comments=}}
{{Photo Entry|last=Park|first=Sanghee|userid=SangheeP|email=shee.park@ utoronto.ca|location=glasses, a girl, first row, third from right, holding starbucks."|comments=}}
{{Photo Entry|last=Pistone|first=Jamie|userid=JPistone|email=jamie.pistone@ utoronto.ca|location=Guy in the second row on the outside of the left railing, guarded by the two guys with their arms crossed.|comments=}}
{{Photo Entry|last=Ramdhayan|first=Dinesh|userid=Dinesh®|email=dinesh.ramdhayan@ utoronto.ca|location=top left corner, 4th from the left|comments=}}
{{Photo Entry|last=Simmons|first=Olivia|userid=OSimmons|email=olivia.simmons@ yahoo.ca|location=girl on the second top rown two people from the right|comments=}}
{{Photo Entry|last=Simpson|first=Evan|userid=ESimpson|email=evan.simpson@ utoronto.ca|location=2nd Front Row, 2nd from the left, Green Shirt, Glasses|comments=Math is addictive don't try it.}}
{{Photo Entry|last=Sinn|first=Daniel|userid=c8sd|email=|location=second-last row, fourth from left|comments=}}
{{Photo Entry|last=Valdez|first=Wilbur|userid=Valdez84|email=wilbur.valdez@ utoronto.ca|location= Third row yellow shirt guy :)|comments=}}
{{Photo Entry|last=VanZanten|first=Johan|userid=jvzanten|email=j.vanzanten@ utoronto.ca|location=first guy in blue shirt standing on the left of left railing |comments=}}
{{Photo Entry|last=Wang|first=Yu|userid=?|email=bright_wangca@ hotmail.com|location=6th row (i think), 5th person from right, white shirt. Behind the guy with a plaid shirt and glasses.|comments=}}
{{Photo Entry|last=Zapf-Belanger|first=Erik|userid=erikzb|email=erikzb@ gmail.com|location=the guy third from the left in the front with the gigantic mouth|comments=Live long and prosper.}}
{{Photo Entry|last=Zhao|first=Yi|userid=zy861|email=zy861100@ hotmail.com|location=guy at the 6th row and 4th from right, with grey shirt and black glasses|comments=}}

09-240/Class Photo

2009-10-08T17:03:30Z

C8sd: /* Who We Are... */ Fur => For

Our class on September 24, 2009:

[[Image:09-240-ClassPhoto.jpg|thumb|centre|500px|Class Photo: click to enlarge]]
{{09-240/Navigation}}

Please identify yourself in this photo! There are two ways to do that:

* [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.
* Send [[User:Drorbn|Dror]] an email message with this information.

The first option is more fun but less private.

===Who We Are...===

{| align=center border=1 cellspacing=0
|-
!First name
!Last name
!UserID
!Email
!In the photo
!Comments
{{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn@ math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank. For better line-breaking, leave a space next to the "@" in email addresses.}}
{{Photo Entry|last=Bahadur|first=Jared|userid=bahadurj|email=jared.bahadur@ utoronto.ca| location=left index fingernail visible on handrail at Bar-Natan's right|comments=}}
{{Photo Entry|last=Barkley|first=Max|userid=SpectralWolf|email=max.barkley@ utoronto.ca|location=fourth guy from the left in the front row, beside the guy with the huge mouth|comments=}}
{{Photo Entry|last=Binder|first=Polina|userid=polyacat|email=polina.binder@ utoronto.ca|location= The girl in blue, to the right of the stairs at the top.|comments=}}
{{Photo Entry|last=Chen|first=Yi Le|userid=AlecC|email=alec.chen@ utoronto.ca|location= the guy in black T-shirt at 2nd to the left, on the 2nd standing row, with glasses|comments=}}
{{Photo Entry|last=Chou|first=Daniel|userid=Danielchou|email=daniel.chou@ utoronto.ca|location=the guy above the guy with the yellow shirt.|comments=Hi People.}}
{{Photo Entry|last=Chung|first=Wen-Jian|userid=cwjian90|email=cwjian90@ yahoo.com|location= top right corner, guy with Einstein's face on his shirt.|comments=}}
{{Photo Entry|last=Desmarais|first=Eric|userid=EricD|email=eric.desmarais@ utoronto.ca|location=back row, two to the right of the centre lamp post with the beard and glasses and the "has the picture been taken?" look.|comments=}}
{{Photo Entry|last=Dranovski|first=Anne|userid=anne.d|email=a.dranovski@ utoronto.ca|location=third row, sitting, green and happy.|comments=}}
{{Photo Entry|last=Gafar|first=Jonathan|userid=JG89|email=jonathan.gafar@ utoronto.ca|location=the guy in the last row to the right of the big pole.|comments=}}
{{Photo Entry|last=Gu|first=Hyungmo|userid=Hmgu7|email=moe.gu@ utoronto.ca|location=the guy with brown
shirt in the fifth row, third from the right.|comments=}}
{{Photo Entry|last=Ivanov|first=Vesselin|userid=Gungrave|email=vesselin.ivanov@ utoronto.ca|location=Last row, 7th from the right, with a shiny dog-tag.|comments=}}
{{Photo Entry|last=Jeffery|first=Travis|userid=Travisjeffery|email=t.jeffery@ utoronto.ca|location=Blue sweater, headphones around neck on the left.|comments=Ask her to wait a moment--I am almost done.}}
{{Photo Entry|last=Klambauer|first=Max|userid=Max.k|email=maximilian.klambauer@ utoronto.ca|location=Last row, almost as far right as possible; I'm the guy with long hair hidden behind the other guy with long, black hair.|comments=}}
{{Photo Entry|last=Li|first=Zhao|userid=lzh8571|email=lzh8571@ live.com|location=at the 4th from right in the top right picture, with glasses and jacket|comments=}}
{{Photo Entry|last=Lindquist|first=Emma|userid=Elindquist|email=emmalindquist@ utoronto.ca|location=Far left of front row|comments=}}
{{Photo Entry|last=Makarov|first=Serhei|userid=Serhei|email=serhei.makarov@ utoronto.ca|location=gray shirt, standing on the step under the guy with fingers on the rail (i.e. bahadurj)|comments=}}
{{Photo Entry|last=Mann|first=Alex|userid=Mannimal|email=alexander.mann@ utoronto.ca|location=last row, third guy to the right of the guy on the rail|comments=240>157}}
{{Photo Entry|last=Mantynen|first=Paul|userid=mantynen|email=paul.mantynen@ utoronto.ca| location= 4th row, third from the right in a black shirt|comments=}}
{{Photo Entry|last=McTaggart|first=Raymond|userid=Raymct|email=raymond.mctaggart@ gmail.com|location=Flying in the top-right sky. The one with the black t-shirt|comments= be <math> \mathbb{R} </math>}}
{{Photo Entry|last=Miao|first=Ying|userid=Miaoying|email=ying.miao@ utoronto.ca|location=girl beside the guy who was wearing blue cap and smiling happily.|comments=}}
{{Photo Entry|last=Michaud|first=Adam|userid=Invariel|email=invariel@ gmail.com|location=Back row, black coat, dark hair, full, dark beard, fifth from the right.|comments=Braaaaaaains...}}
{{Photo Entry|last=Milcak|first=Juraj|userid=Milcak|email=j.milcak@ utoronto.ca|location=2nd row, 1st from the right, green shirt|comments=}}
{{Photo Entry|last=Nikitakis|first=George|userid=GeorgeN|email=george.nikitakis@ utoronto.ca|location=Part of the group floating in the top right corner. Second from the left|comments=}}
{{Photo Entry|last=Ott|first=Kristian|userid=k.ott|email=k.ott@ utoronto.ca| location= 4th guy from the right in the second row, wearing a green striped sweater|comments=}}
{{Photo Entry|last=Park|first=Sanghee|userid=SangheeP|email=shee.park@ utoronto.ca|location=glasses, a girl, first row, third from right, holding starbucks."|comments=}}
{{Photo Entry|last=Pistone|first=Jamie|userid=JPistone|email=jamie.pistone@ utoronto.ca|location=Guy in the second row on the outside of the left railing, guarded by the two guys with their arms crossed.|comments=}}
{{Photo Entry|last=Ramdhayan|first=Dinesh|userid=Dinesh®|email=dinesh.ramdhayan@ utoronto.ca|location=top left corner, 4th from the left|comments=}}
{{Photo Entry|last=Simmons|first=Olivia|userid=OSimmons|email=olivia.simmons@ yahoo.ca|location=girl on the second top rown two people from the right|comments=}}
{{Photo Entry|last=Simpson|first=Evan|userid=ESimpson|email=evan.simpson@ utoronto.ca|location=2nd Front Row, 2nd from the left, Green Shirt, Glasses|comments=Math is addictive don't try it.}}
{{Photo Entry|last=Sinn|first=Daniel|userid=c8sd|email=|location=second-last row, fourth from left|comments=}}
{{Photo Entry|last=VanZanten|first=Johan|userid=jvzanten|email=j.vanzanten@ utoronto.ca|location=first guy in blue shirt standing on the left of left railing |comments=}}
{{Photo Entry|last=Wang|first=Yu|userid=?|email=bright_wangca@ hotmail.com|location=6th row (i think), 5th person from right, white shirt. Behind the guy with a plaid shirt and glasses.|comments=}}
{{Photo Entry|last=Zapf-Belanger|first=Erik|userid=erikzb|email=erikzb@ gmail.com|location=the guy third from the left in the front with the gigantic mouth|comments=Live long and prosper.}}
{{Photo Entry|last=Zhao|first=Yi|userid=zy861|email=zy861100@ hotmail.com|location=guy at the 6th row and 4th from right, with grey shirt and black glasses|comments=}}
{{Photo Entry|last=Valdez|first=Wilbur|userid=Valdez84|email=wilbur.valdez@ utoronto.ca|location= Third row yellow shirt guy :)|comments=}}

09-240/Class Photo

2009-10-07T22:11:49Z

C8sd: /* Who We Are... */ abc order

Our class on September 24, 2009:

[[Image:09-240-ClassPhoto.jpg|thumb|centre|500px|Class Photo: click to enlarge]]
{{09-240/Navigation}}

Please identify yourself in this photo! There are two ways to do that:

* [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.
* Send [[User:Drorbn|Dror]] an email message with this information.

The first option is more fun but less private.

===Who We Are...===

{| align=center border=1 cellspacing=0
|-
!First name
!Last name
!UserID
!Email
!In the photo
!Comments
{{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn@math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank}}
{{Photo Entry|last=Bahadur|first=Jared|userid=bahadurj|email=jared.bahadur@utoronto.ca| location=left index fingernail visible on handrail at Bar-Natan's right|comments=}}
{{Photo Entry|last=Barkley|first=Max|userid=SpectralWolf|email=max.barkley@utoronto.ca|location=fourth guy from the left in the front row, beside the guy with the huge mouth|comments=}}
{{Photo Entry|last=Binder|first=Polina|userid=polyacat|email=polina.binder@utoronto.ca|location= The girl in blue, to the right of the stairs at the top.|comments=}}
{{Photo Entry|last=Chen|first=Yi Le|userid=AlecC|email=alec.chen@utoronto.ca|location= the guy in black T-shirt at 2nd to the left, on the 2nd standing row, with glasses|comments=}}
{{Photo Entry|last=Chou|first=Daniel|userid=Danielchou|email=daniel.chou@utoronto.ca|location=the guy above the guy with the yellow shirt.|comments=Hi People.}}
{{Photo Entry|last=Chung|first=Wen-Jian|userid=cwjian90|email=cwjian90@yahoo.com|location= top right corner, guy with Einstein's face on his shirt.|comments=}}
{{Photo Entry|last=Desmarais|first=Eric|userid=EricD|email=eric.desmarais@utoronto.ca|location=back row, two to the right of the centre lamp post with the beard and glasses and the "has the picture been taken?" look.|comments=}}
{{Photo Entry|last=Dranovski|first=Anne|userid=anne.d|email=a.dranovski@utoronto.ca|location=third row, sitting, green and happy.|comments=}}
{{Photo Entry|last=Ivanov|first=Vesselin|userid=Gungrave|email=vesselin.ivanov@utoronto.ca|location=Last row, 7th from the right, with a shiny dog-tag.|comments=}}
{{Photo Entry|last=Jeffery|first=Travis|userid=Travisjeffery|email=t.jeffery@utoronto.ca|location=Blue sweater, headphones around neck on the left.|comments=Ask her to wait a moment--I am almost done.}}
{{Photo Entry|last=Klambauer|first=Max|userid=Max.k|email=maximilian.klambauer@utoronto.ca|location=Last row, almost as far right as possible; I'm the guy with long hair hidden behind the other guy with long, black hair.|comments=}}
{{Photo Entry|last=Li|first=Zhao|userid=lzh8571|email=lzh8571@live.com|location=at the 4th from right in the top right picture, with glasses and jacket|comments=}}
{{Photo Entry|last=Lindquist|first=Emma|userid=Elindquist|email=emmalindquist@utoronto.ca|location=Far left of front row|comments=}}
{{Photo Entry|last=Makarov|first=Serhei|userid=Serhei|email=serhei.makarov@utoronto.ca|location=gray shirt, standing on the step under the guy with fingers on the rail (i.e. bahadurj)|comments=}}
{{Photo Entry|last=Mann|first=Alex|userid=Mannimal|email=alexander.mann@utoronto.ca|location=last row, third guy to the right of the guy on the rail|comments=240>157}}
{{Photo Entry|last=Mantynen|first=Paul|userid=mantynen|email=paul.mantynen@utoronto.ca| location= 4th row, third from the right in a black shirt|comments=}}
{{Photo Entry|last=McTaggart|first=Raymond|userid=Raymct|email=raymond.mctaggart@gmail.com|location=Flying in the top-right sky. The one with the black t-shirt|comments= be <math> \mathbb{R} </math>}}
{{Photo Entry|last=Michaud|first=Adam|userid=Invariel|email=invariel@gmail.com|location=Back row, black coat, dark hair, full, dark beard, fifth from the right.|comments=Braaaaaaains...}}
{{Photo Entry|last=Milcak|first=Juraj|userid=Milcak|email=j.milcak@utoronto.ca|location=2nd row, 1st from the right, green shirt|comments=}}
{{Photo Entry|last=Ott|first=Kristian|userid=k.ott|email=k.ott@utoronto.ca| location= 4th guy from the right in the second row, wearing a green striped sweater|comments=}}
{{Photo Entry|last=Park|first=Sanghee|userid=SangheeP|email=shee.park@utoronto.ca|location=glasses, a girl, first row, third from right, holding starbucks."|comments=}}
{{Photo Entry|last=Ramdhayan|first=Dinesh|userid=Dinesh®|email=dinesh.ramdhayan@utoronto.ca|location=top left corner, 4th from the left|comments=}}
{{Photo Entry|last=Simmons|first=Olivia|userid=OSimmons|email=olivia.simmons@yahoo.ca|location=girl on the second top rown two people from the right|comments=}}
{{Photo Entry|last=Simpson|first=Evan|userid=ESimpson|email=evan.simpson@utoronto.ca|location=2nd Front Row, 2nd from the left, Green Shirt, Glasses|comments=Math is addictive don't try it.}}
{{Photo Entry|last=Sinn|first=Daniel|userid=c8sd|email=|location=second-last row, fourth from left|comments=}}
{{Photo Entry|last=VanZanten|first=Johan|userid=jvzanten|email=j.vanzanten@utoronto.ca|location=first guy in blue shirt standing on the left of left railing |comments=}}
{{Photo Entry|last=Zapf-Belanger|first=Erik|userid=erikzb|email=erikzb@gmail.com|location=the guy third from the left in the front with the gigantic mouth|comments=Live long and prosper.}}
{{Photo Entry|last=Zhao|first=Yi|userid=zy861|email=zy861100@hotmail.com|location=guy at the 6th row and 4th from right, with grey shirt and black glasses|comments=}}

09-240/Class Photo

2009-10-02T22:31:25Z

C8sd: /* Who We Are... */ add myself

09-240/Class Photo

2009-10-02T22:28:29Z

C8sd: /* Who We Are... */ sort

09-240/Classnotes for Thursday September 24

2009-10-02T22:25:45Z

C8sd: /* Examples */ Fix indentation, forced png.

{{09-240/Navigation}}

{{09-240/Class Notes Warning}}

{| style="border: solid 1px black"
|-
| Convention for today:
: <math>x, y, z \in \mathbf V</math>
: <math>a, b, c \in F</math>
|}

A V.S. over <math>F: V, 0, +, \times</math> s.t.

VS1 <math>\forall x, y \in \mathbf V, x + y = y + x</math> 
VS2 <math>\ldots (x + y) + z = x + (y + z)</math> 
VS3 <math>\ldots x + 0 = x</math> 
VS4 <math>\forall x, \exists y \mbox{ s.t. } x + y = 0</math> 
VS5 <math>1 \cdot x = x</math> 
VS6 <math>a \cdot (b \cdot x) = (a \cdot b) \cdot x</math> 
VS7 <math>a \cdot (x + y) = ax + ay</math> 
VS8 <math>(a + b) \cdot x = ax + bx</math>

=== Examples ===

# <math>\left\{ \begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix} : a_i \in F \right\} = F^n</math>
# <math>\left\{ \begin{pmatrix}
a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & \cdots & a_{mn}
\end{pmatrix} \right\} = \mathrm M_{m \times n}(F)</math>
# Let ''S'' be a set (''F'' is some field)
#: <math>\mathcal F(S, F) = \{f: S \rightarrow F\}</math>
#: ''S'' = Primary colours = {red, green, blue}
#: ''F'' = ''F''2 = {0, 1}
#: <math>\mathcal F(S, F) = \left\{ \begin{matrix}
f_1(red) = 0 & f_1(green) = 1 & f_1(blue) = 0 \\
\cdots \\
f_2 \begin{pmatrix} \mbox{red} \\ \mbox{green} \\ \mbox{blue} \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} & \cdots
\end{matrix} \right\}</math>

----

<math>S = \mathbb N = \{ 1, 2, 3, 4, \ldots \} F = \mathbb R</math>

<math>\mathcal F(\mathbb N, \mathbb R) = \left\{ \begin{matrix}
1 & 2 & 3 & 4 & \ldots \\
6 & 6 & 6 & 6 & \ldots \\
\pi & 2\pi & e & 62 & \ldots \\
\end{matrix} \right\} = \{ \mbox{sequences} \}</math>

<math>S = \begin{pmatrix} \vdots \\ \vdots \end{pmatrix} n \Rightarrow \mathcal F(S, F) = F^n</math>

# <math>O_{\mathcal F(S, F)}(\sigma) = 0_F \forall \sigma \in S</math>
# <math>f, g \in \mathcal F(S, F)</math> <math>\,\! (f + g)(\sigma) = f(\sigma) + g(\sigma)</math> <math>f \in \mathcal F(S, F)</math> <math>a \in F \Rightarrow \forall \sigma \in S, S(af)(\sigma) = a \cdot (f(\sigma))</math>

'''Claim''': + is associative. Given <math>f, g, h \in \mathcal F(S, F), (f + g) + h = f + (g + h) \forall \sigma</math>
: <math>((f + g) + h)(\sigma) = (f + g)(\sigma) + h(\sigma)</math>
: <math>= (f(\sigma) + g(\sigma)) + h(\sigma)</math>
: <math>= f(\sigma) + g(\sigma) + h(\sigma) \mbox{ (by F2)}</math>
: <math>(f + (g + h))(\sigma) = f(\sigma) + (g + h)(\sigma)</math>
: <math>= f(\sigma) + (g(\sigma) + h(\sigma))</math>
: <math>= f(\sigma) + g(\sigma) + h(\sigma)</math>

<ol start="4">
<li><math>\mathbb C \mbox{ is a V.S. over } \mathbb R</math></li>
<li><math>\mathbb R \mbox{ is a V.S. over } \mathbb Q</math></li>
<li><math>\mathbb R \mbox{ is a V.S. over } \mathbb R</math></li>
<li><math>\,\! \{0\} \mbox{ is a V.S. over } F</math></li>
</ol>

=== Dull theorem ===

# Cancellation: <math>x + y = x + z \Rightarrow y = z</math> (add ''w'' to both sides s.t. ''x'' + ''w'' = 0)
# 0 is unique
# Negatives are unique: <math>x + y = 0 = x + z \Rightarrow y = 0</math>
# <math>0x = 0. a \cdot 0 = 0</math>
# <math>(-a) \cdot x = a \cdot (-x) = -(ax)</math>

09-240/Classnotes for Thursday September 24

2009-10-02T22:23:34Z

C8sd: /* Examples */ Forgot <math> tags.

{{09-240/Navigation}}

{{09-240/Class Notes Warning}}

{| style="border: solid 1px black"
|-
| Convention for today:
: <math>x, y, z \in \mathbf V</math>
: <math>a, b, c \in F</math>
|}

A V.S. over <math>F: V, 0, +, \times</math> s.t.

VS1 <math>\forall x, y \in \mathbf V, x + y = y + x</math> 
VS2 <math>\ldots (x + y) + z = x + (y + z)</math> 
VS3 <math>\ldots x + 0 = x</math> 
VS4 <math>\forall x, \exists y \mbox{ s.t. } x + y = 0</math> 
VS5 <math>1 \cdot x = x</math> 
VS6 <math>a \cdot (b \cdot x) = (a \cdot b) \cdot x</math> 
VS7 <math>a \cdot (x + y) = ax + ay</math> 
VS8 <math>(a + b) \cdot x = ax + bx</math>

=== Examples ===

# <math>\left\{ \begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix} : a_i \in F \right\} = F^n</math>
# <math>\left\{ \begin{pmatrix}
a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & \cdots & a_{mn}
\end{pmatrix} \right\} = \mathrm M_{m \times n}(F)</math>
# Let ''S'' be a set (''F'' is some field)
#: <math>\mathcal F(S, F) = \{f: S \rightarrow F\}</math>
#: ''S'' = Primary colours = {red, green, blue}
#: ''F'' = ''F''2 = {0, 1}
#: <math>\mathcal F(S, F) = \left\{ \begin{matrix}
f_1(red) = 0 & f_1(green) = 1 & f_1(blue) = 0 \\
\cdots \\
f_2 \begin{pmatrix} \mbox{red} \\ \mbox{green} \\ \mbox{blue} \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} & \cdots
\end{matrix} \right\}</math>

----

<math>S = \mathbb N = \{ 1, 2, 3, 4, \ldots \} F = \mathbb R</math>

<math>\mathcal F(\mathbb N, \mathbb R) = \left\{ \begin{matrix}
1 & 2 & 3 & 4 & \ldots \\
6 & 6 & 6 & 6 & \ldots \\
\pi & 2\pi & e & 62 & \ldots \\
\end{matrix} \right\} = \{ \mbox{sequences} \}</math>

<math>S = \begin{pmatrix} \vdots \\ \vdots \end{pmatrix} n \Rightarrow \mathcal F(S, F) = F^n</math>

# <math>O_{\mathcal F(S, F)}(\sigma) = 0_F \forall \sigma \in S</math>
# <math>f, g \in \mathcal F(S, F)</math>
#: <math>(f + g)(\sigma) = f(\sigma) + g(\sigma)</math>
#: <math>f \in \mathcal F(S, F)</math>
#: <math>a \in F \Rightarrow \forall \sigma \in S, S(af)(\sigma) = a \cdot (f(\sigma))</math>

'''Claim''': + is associative. Given <math>f, g, h \in \mathcal F(S, F), (f + g) + h = f + (g + h) \forall \sigma</math>
: <math>((f + g) + h)(\sigma) = (f + g)(\sigma) + h(\sigma)</math>
: <math>= (f(\sigma) + g(\sigma)) + h(\sigma)</math>
: <math>= f(\sigma) + g(\sigma) + h(\sigma) \mbox{ (by F2)}</math>
: <math>(f + (g + h))(\sigma) = f(\sigma) + (g + h)(\sigma)</math>
: <math>= f(\sigma) + (g(\sigma) + h(\sigma))</math>
: <math>= f(\sigma) + g(\sigma) + h(\sigma)</math>

<ol start="4">
<li><math>\mathbb C \mbox{ is a V.S. over } \mathbb R</math></li>
<li><math>\mathbb R \mbox{ is a V.S. over } \mathbb Q</math></li>
<li><math>\mathbb R \mbox{ is a V.S. over } \mathbb R</math></li>
<li><math>\{0\} \mbox{ is a V.S. over } F</math></li>
</ol>

=== Dull theorem ===

# Cancellation: <math>x + y = x + z \Rightarrow y = z</math> (add ''w'' to both sides s.t. ''x'' + ''w'' = 0)
# 0 is unique
# Negatives are unique: <math>x + y = 0 = x + z \Rightarrow y = 0</math>
# <math>0x = 0. a \cdot 0 = 0</math>
# <math>(-a) \cdot x = a \cdot (-x) = -(ax)</math>

Template:09-240/Navigation

2009-10-02T22:23:01Z

C8sd: Link 09/29 notes.

09-240/Classnotes for Tuesday September 29

2009-10-02T22:21:33Z

C8sd: Yangjiay's notes.

09-240/Classnotes for Thursday September 24

2009-10-02T22:18:36Z

C8sd:

{{09-240/Navigation}}

{{09-240/Class Notes Warning}}

{| style="border: solid 1px black"
|-
| Convention for today:
: <math>x, y, z \in \mathbf V</math>
: <math>a, b, c \in F</math>
|}

A V.S. over <math>F: V, 0, +, \times</math> s.t.

VS1 <math>\forall x, y \in \mathbf V, x + y = y + x</math> 
VS2 <math>\ldots (x + y) + z = x + (y + z)</math> 
VS3 <math>\ldots x + 0 = x</math> 
VS4 <math>\forall x, \exists y \mbox{ s.t. } x + y = 0</math> 
VS5 <math>1 \cdot x = x</math> 
VS6 <math>a \cdot (b \cdot x) = (a \cdot b) \cdot x</math> 
VS7 <math>a \cdot (x + y) = ax + ay</math> 
VS8 <math>(a + b) \cdot x = ax + bx</math>

=== Examples ===

# <math>\left\{ \begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix} : a_i \in F \right\} = F^n</math>
# <math>\left\{ \begin{pmatrix}
a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{m1} & \cdots & a_{mn}
\end{pmatrix} \right\} = \mathrm M_{m \times n}(F)</math>
# Let ''S'' be a set (''F'' is some field)
#: <math>\mathcal F(S, F) = \{f: S \rightarrow F\}</math>
#: ''S'' = Primary colours = {red, green, blue}
#: ''F'' = ''F''2 = {0, 1}
#: <math>\mathcal F(S, F) = \left\{ \begin{matrix}
f_1(red) = 0 & f_1(green) = 1 & f_1(blue) = 0 \\
\cdots \\
f_2 \begin{pmatrix} \mbox{red} \\ \mbox{green} \\ \mbox{blue} \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} & \cdots
\end{matrix} \right\}</math>

----

<math>S = \mathbb N = \{ 1, 2, 3, 4, \ldots \} F = \mathbb R</math>

<math>\mathcal F(\mathbb N, \mathbb R) = \left\{ \begin{matrix}
1 & 2 & 3 & 4 & \ldots \\
6 & 6 & 6 & 6 & \ldots \\
\pi & 2\pi & e & 62 & \ldots \\
\end{matrix} \right\} = \{ \mbox{sequences} \}</math>

<math>S = \begin{pmatrix} \vdots \\ \vdots \end{pmatrix} n \Rightarrow \mathcal F(S, F) = F^n</math>

# <math>O_{\mathcal F(S, F)}(\sigma) = 0_F \forall \sigma \in S</math>
# <math>f, g \in \mathcal F(S, F)</math>
#: <math>(f + g)(\sigma) = f(\sigma) + g(\sigma)</math>
#: <math>f \in \mathcal F(S, F)</math>
#: <math>a \in F \Rightarrow \forall \sigma \in S, S(af)(\sigma) = a \cdot (f(\sigma))</math>

'''Claim''': + is associative. Given <math>f, g, h \in \mathcal F(S, F), (f + g) + h = f + (g + h) \forall \sigma</math>
: ((f + g) + h)(\sigma) = (f + g)(\sigma) + h(\sigma)
: = (f(\sigma) + g(\sigma)) + h(\sigma)
: = f(\sigma) + g(\sigma) + h(\sigma) \mbox{ (by F2)}
: (f + (g + h))(\sigma) = f(\sigma) + (g + h)(\sigma)
: = f(\sigma) + (g(\sigma) + h(\sigma))
: = f(\sigma) + g(\sigma) + h(\sigma)

<ol start="4">
<li><math>\mathbb C \mbox{ is a V.S. over } \mathbb R</math></li>
<li><math>\mathbb R \mbox{ is a V.S. over } \mathbb Q</math></li>
<li><math>\mathbb R \mbox{ is a V.S. over } \mathbb R</math></li>
<li><math>\{0\} \mbox{ is a V.S. over } F</math></li>
</ol>

=== Dull theorem ===

# Cancellation: <math>x + y = x + z \Rightarrow y = z</math> (add ''w'' to both sides s.t. ''x'' + ''w'' = 0)
# 0 is unique
# Negatives are unique: <math>x + y = 0 = x + z \Rightarrow y = 0</math>
# <math>0x = 0. a \cdot 0 = 0</math>
# <math>(-a) \cdot x = a \cdot (-x) = -(ax)</math>

09-240/Class Photo

2009-09-26T23:04:47Z

C8sd: I think sweaters are striped.

Template:09-240/Navigation

2009-09-25T04:12:05Z

C8sd: Link (unwritten) notes.

09-240/Classnotes for Tuesday September 22

2009-09-24T21:49:35Z

C8sd: /* Class notes for today */ Format.

{{09-240/Navigation}}

==Some links==
* {{Pensieve Link|2009-09/nb/09-240-TheComplexField.pdf|The Complex Numbers by Computer}}.
* Dori Eldar's work on "mechanical computations": {{Home Link|People/Eldar/thesis/linkfunc.htm|Machines as Calculating Devices}} and {{Home Link|People/Eldar/thesis/squaring.htm|Computing the function <math>W=Z^2</math> the hard way}}.
* The "Dimensions" video on "Nombres complexes", is at http://dimensions-math.org/Dim_reg_AM.htm (and then go to "Dimensions_5".

{{09-240/Class Notes Warning}}
==Class notes for today==

<gallery>
Image:September 22 2009 lecture notes page 1.jpg|[[User:Yangjiay|Yangjiay]] - Page 1
Image:September 22 2009 lecture notes page 2.jpg|Page 2
Image:September 22 2009 lecture notes page 3.jpg|Page 3
Image:September 22 2009 lecture notes page 4.jpg|Page 4
Image:September 22 2009 lecture notes page 5.jpg|Page 5
</gallery>

Vectors:
# can be added
# can be multiplied by a number (not another vector)

Let ''F'' be a field. A vector space ''V'' over the field ''F'' is a set ''V'' (of vectors) with a special element 0''V'', a binary operation + : ''V'' × ''V'' → ''V'', a binary operation • : ''F'' × ''V'' → ''V''.

{| style="border: solid 1px black"
|-
| Convention for today:
: <math>x, y, z \in \mathbf V</math>
: <math>a, b, c \in F</math>
|}

VS1 <math>\forall x, y \in \mathbf V, x + y = y + x</math> 
VS2 <math>\cdots (x + y) + z = x + (y + z)</math> 
VS3 <math>\cdots x + 0 = x</math> 
VS4 <math>\forall x, \exists y \mbox{ s.t. } x + y = 0</math> 
VS5 <math>1 \cdot x = x</math> 
VS6 <math>a \cdot (b \cdot x) = (a \cdot b) \cdot x</math> 
VS7 <math>a \cdot (x + y) = ax + ay</math> 
VS8 <math>(a + b) \cdot x = ax + bx</math>

=== Proof of VS4 ===

Take an arbitrary <math>x = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} \in F^n</math>

Set <math>y = \begin{pmatrix} -a_1 \\ -a_2 \\ \vdots \\ -a_n \end{pmatrix}</math> and note
: <math>x + y = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} + \begin{pmatrix} -a_1 \\ -a_2 \\ \vdots \\ -a_n \end{pmatrix} = \begin{pmatrix} a_1 + (-a_1) \\ a_2 + (-a_2) \\ \vdots \\ a_n + (-a_n) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix} = 0_{F^n}</math>

=== Examples ===

# <math>F^n \mbox{ for } n \in \mathbb N</math>
#: <math>F^n = \left\{ \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} : a_i \in F \right\}</math>
#: <math>\begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ \vdots \\ a_n + b_n \end{pmatrix}</math>
#: <math>a \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix} = \begin{pmatrix} ab_1 \\ ab_2 \\ \vdots \\ ab_n \end{pmatrix}</math>
#: ...
# <math>\mathrm M_{m \times n}(F)</math>
#: ...
# <math>\mathcal F(S, F)</math>
# Polynomials
# <math>...</math>

=== Food for thought ===

What is wrong with setting

<math>
\begin{pmatrix}
2 & 3 \\
4 & 5 \\
\end{pmatrix} \cdot \begin{pmatrix}
6 & 7 \\
8 & 9 \\
\end{pmatrix} = \begin{pmatrix}
2 \cdot 6 & 3 \cdot 7 \\
4 \cdot 8 & 5 \cdot 9 \\
\end{pmatrix} = \begin{pmatrix}
12 & 21 \\
32 & 45 \\
\end{pmatrix} ?
</math>

# Unnecessary for a V.S.
# This is useless, since it does not describe reality. For example, a mathematical theory with 46 dimensions can be perfect and mathematically elegant, but if the only solution to it is a universe in which life cannot form it is not reality, hence we have no use for it.

09-240/Classnotes for Tuesday September 22

2009-09-24T17:52:27Z

C8sd: /* Class notes for today */ Format.

{{09-240/Navigation}}

==Some links==
* {{Pensieve Link|2009-09/nb/09-240-TheComplexField.pdf|The Complex Numbers by Computer}}.
* Dori Eldar's work on "mechanical computations": {{Home Link|People/Eldar/thesis/linkfunc.htm|Machines as Calculating Devices}} and {{Home Link|People/Eldar/thesis/squaring.htm|Computing the function <math>W=Z^2</math> the hard way}}.
* The "Dimensions" video on "Nombres complexes", is at http://dimensions-math.org/Dim_reg_AM.htm (and then go to "Dimensions_5".

{{09-240/Class Notes Warning}}
==Class notes for today==

<gallery>
Image:September 22 2009 lecture notes page 1.jpg|[[User:Yangjiay|Yangjiay]] - Page 1
Image:September 22 2009 lecture notes page 2.jpg|Page 2
Image:September 22 2009 lecture notes page 3.jpg|Page 3
Image:September 22 2009 lecture notes page 4.jpg|Page 4
Image:September 22 2009 lecture notes page 5.jpg|Page 5
</gallery>

Vectors:
# can be added
# can be multiplied by a number (not another vector)

Let ''F'' be a field. A vector space '''V''' over the field ''F'' is a set '''V''' (of vectors) with a special element 0'''V''', a binary operation + : '''V''' × '''V''' → '''V''', a binary operation • : ''F'' × '''V''' → '''V'''.

{| style="border: solid 1px black"
|-
| Convention for today:
: <math>x, y, z \in \mathbf V</math>
: <math>a, b, c \in F</math>
|}

VS1 <math>\forall x, y \in \mathbf V, x + y = y + x</math> 
VS2 <math>\cdots (x + y) + z = x + (y + z)</math> 
VS3 <math>\cdots x + 0 = x</math> 
VS4 <math>\forall x, \exists y \mbox{ s.t. } x + y = 0</math> 
VS5 <math>1 \cdot x = x</math> 
VS6 <math>a \cdot (b \cdot x) = (a \cdot b) \cdot x</math> 
VS7 <math>a \cdot (x + y) = ax + ay</math> 
VS8 <math>(a + b) \cdot x = ax + bx</math>

=== Proof of VS4 ===

Take an arbitrary <math>x = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} \in F^n</math>

Set <math>y = \begin{pmatrix} -a_1 \\ -a_2 \\ \vdots \\ -a_n \end{pmatrix}</math> and note
: <math>x + y = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} + \begin{pmatrix} -a_1 \\ -a_2 \\ \vdots \\ -a_n \end{pmatrix} = \begin{pmatrix} a_1 + (-a_1) \\ a_2 + (-a_2) \\ \vdots \\ a_n + (-a_n) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix} = 0_{F^n}</math>

=== Examples ===

# <math>F^n \mbox{ for } n \in \mathbb N</math>
#: <math>F^n = \left\{ \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} : a_i \in F \right\}</math>
#: <math>\begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ \vdots \\ a_n + b_n \end{pmatrix}</math>
#: <math>a \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix} = \begin{pmatrix} ab_1 \\ ab_2 \\ \vdots \\ ab_n \end{pmatrix}</math>
#: ...
# <math>\mathrm M_{m \times n}(F)</math>
#: ...
# <math>\mathcal F(S, F)</math>
# Polynomials
# <math>...</math>

=== Food for thought ===

What is wrong with setting

<math>
\begin{pmatrix}
2 & 3 \\
4 & 5 \\
\end{pmatrix} \cdot \begin{pmatrix}
6 & 7 \\
8 & 9 \\
\end{pmatrix} = \begin{pmatrix}
2 \cdot 6 & 3 \cdot 7 \\
4 \cdot 8 & 5 \cdot 9 \\
\end{pmatrix} = \begin{pmatrix}
12 & 21 \\
32 & 45 \\
\end{pmatrix} ?
</math>

# Unnecessary for a V.S.
# This is useless, since it does not describe reality. For example, a mathematical theory with 46 dimensions can be perfect and mathematically elegant, but if the only solution to it is a universe in which life cannot form it is not reality, hence we have no use for it.

09-240/Classnotes for Tuesday September 22

2009-09-23T02:46:32Z

C8sd: /* Class notes for today */ Yangijay's notes

{{09-240/Navigation}}

==Some links==
* {{Pensieve Link|2009-09/nb/09-240-TheComplexField.pdf|The Complex Numbers by Computer}}.
* Dori Eldar's work on "mechanical computations": {{Home Link|People/Eldar/thesis/linkfunc.htm|Machines as Calculating Devices}} and {{Home Link|People/Eldar/thesis/squaring.htm|Computing the function <math>W=Z^2</math> the hard way}}.
* The "Dimensions" video on "Nombres complexes", is at http://dimensions-math.org/Dim_reg_AM.htm (and then go to "Dimensions_5".

{{09-240/Class Notes Warning}}
==Class notes for today==

<gallery>
Image:September 22 2009 lecture notes page 1.jpg|[[User:Yangjiay|Yangjiay]] - Page 1
Image:September 22 2009 lecture notes page 2.jpg|Page 2
Image:September 22 2009 lecture notes page 3.jpg|Page 3
Image:September 22 2009 lecture notes page 4.jpg|Page 4
Image:September 22 2009 lecture notes page 5.jpg|Page 5
</gallery>

Vectors:
# can be added
# can be multiplied by a number (not another vector)

Let <math>\mathcal F</math> be a field. A vector space <math>\mathbf V</math> over the field <math>\mathcal F</math> is a set <math>\mathbf V</math> (of vectors) with a special element <math>0_V</math>, a binary operation <math>+ : \mathbf V \times \mathbf V \rightarrow \mathbf V</math>, a binary operation <math>\cdot : \mathcal F \times \mathbf V \rightarrow \mathbf V</math>.

{| style="border: solid 1px black"
|-
| Convention for today:
: <math>x, y, z \in \mathbf V</math>
: <math>a, b, c \in \mathcal F</math>
|}

VS1 <math>\forall x, y \in \mathbf V, x + y = y + x</math> 
VS2 <math>\cdots (x + y) + z = x + (y + z)</math> 
VS3 <math>\cdots x + 0 = x</math> 
VS4 <math>\forall x, \exists y \mbox{ s.t. } x + y = 0</math> 
VS5 <math>1 \cdot x = x</math> 
VS6 <math>a \cdot (b \cdot x) = (a \cdot b) \cdot x</math> 
VS7 <math>a \cdot (x + y) = ax + ay</math> 
VS8 <math>(a + b) \cdot x = ax + bx</math>

=== Proof of VS4 ===

Take an arbitrary <math>x = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} \in F^n</math>

Set <math>y = \begin{pmatrix} -a_1 \\ -a_2 \\ \vdots \\ -a_n \end{pmatrix}</math> and note
: <math>x + y = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} + \begin{pmatrix} -a_1 \\ -a_2 \\ \vdots \\ -a_n \end{pmatrix} = \begin{pmatrix} a_1 + (-a_1) \\ a_2 + (-a_2) \\ \vdots \\ a_n + (-a_n) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix} = 0_{F^n}</math>

=== Examples ===

# <math>F^n \mbox{ for } n \in \mathbb N</math>
#: <math>F^n = \left\{ \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} : a_i \in F \right\}</math>
#: <math>\begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ \vdots \\ a_n + b_n \end{pmatrix}</math>
#: <math>a \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix} = \begin{pmatrix} ab_1 \\ ab_2 \\ \vdots \\ ab_n \end{pmatrix}</math>
#: ...
# <math>\mathrm M_{m \times n}(F)</math>
#: ...
# <math>\mathcal F(S, F)</math>
# Polynomials
# <math>...</math>

=== Food for thought ===

What is wrong with setting

<math>
\begin{pmatrix}
2 & 3 \\
4 & 5 \\
\end{pmatrix} \cdot \begin{pmatrix}
6 & 7 \\
8 & 9 \\
\end{pmatrix} = \begin{pmatrix}
2 \cdot 6 & 3 \cdot 7 \\
4 \cdot 8 & 5 \cdot 9 \\
\end{pmatrix} = \begin{pmatrix}
12 & 21 \\
32 & 45 \\
\end{pmatrix} ?
</math>

# Unnecessary for a V.S.
# This is useless, since it does not describe reality. For example, a mathematical theory with 46 dimensions can be perfect and mathematically elegant, but if the only solution to it is a universe in which life cannot form it is not reality, hence we have no use for it.

09-240/Classnotes for Tuesday September 22

2009-09-22T21:50:10Z

C8sd: /* Examples */

{{09-240/Navigation}}

==Some links==
* {{Pensieve Link|2009-09/nb/09-240-TheComplexField.pdf|The Complex Numbers by Computer}}.
* Dori Eldar's work on "mechanical computations": {{Home Link|People/Eldar/thesis/linkfunc.htm|Machines as Calculating Devices}} and {{Home Link|People/Eldar/thesis/squaring.htm|Computing the function <math>W=Z^2</math> the hard way}}.
* The "Dimensions" video on "Nombres complexes", is at http://dimensions-math.org/Dim_reg_AM.htm (and then go to "Dimensions_5".

{{09-240/Class Notes Warning}}
==Class notes for today==

Vectors:
# can be added
# can be multiplied by a number (not another vector)

Let <math>\mathcal F</math> be a field. A vector space <math>\mathbf V</math> over the field <math>\mathcal F</math> is a set <math>\mathbf V</math> (of vectors) with a special element <math>0_V</math>, a binary operation <math>+ : \mathbf V \times \mathbf V \rightarrow \mathbf V</math>, a binary operation <math>\cdot : \mathcal F \times \mathbf V \rightarrow \mathbf V</math>.

{| style="border: solid 1px black"
|-
| Convention for today:
: <math>x, y, z \in \mathbf V</math>
: <math>a, b, c \in \mathcal F</math>
|}

VS1 <math>\forall x, y \in \mathbf V, x + y = y + x</math> 
VS2 <math>\cdots (x + y) + z = x + (y + z)</math> 
VS3 <math>\cdots x + 0 = x</math> 
VS4 <math>\forall x, \exists y \mbox{ s.t. } x + y = 0</math> 
VS5 <math>1 \cdot x = x</math> 
VS6 <math>a \cdot (b \cdot x) = (a \cdot b) \cdot x</math> 
VS7 <math>a \cdot (x + y) = ax + ay</math> 
VS8 <math>(a + b) \cdot x = ax + bx</math>

=== Proof of VS4 ===

Take an arbitrary <math>x = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} \in F^n</math>

Set <math>y = \begin{pmatrix} -a_1 \\ -a_2 \\ \vdots \\ -a_n \end{pmatrix}</math> and note
: <math>x + y = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} + \begin{pmatrix} -a_1 \\ -a_2 \\ \vdots \\ -a_n \end{pmatrix} = \begin{pmatrix} a_1 + (-a_1) \\ a_2 + (-a_2) \\ \vdots \\ a_n + (-a_n) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix} = 0_{F^n}</math>

=== Examples ===

# <math>F^n \mbox{ for } n \in \mathbb N</math>
#: <math>F^n = \left\{ \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} : a_i \in F \right\}</math>
#: <math>\begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ \vdots \\ a_n + b_n \end{pmatrix}</math>
#: <math>a \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix} = \begin{pmatrix} ab_1 \\ ab_2 \\ \vdots \\ ab_n \end{pmatrix}</math>
#: ...
# <math>\mathrm M_{m \times n}(F)</math>
#: ...
# <math>\mathcal F(S, F)</math>
# Polynomials
# <math>...</math>

=== Food for thought ===

What is wrong with setting

<math>
\begin{pmatrix}
2 & 3 \\
4 & 5 \\
\end{pmatrix} \cdot \begin{pmatrix}
6 & 7 \\
8 & 9 \\
\end{pmatrix} = \begin{pmatrix}
2 \cdot 6 & 3 \cdot 7 \\
4 \cdot 8 & 5 \cdot 9 \\
\end{pmatrix} = \begin{pmatrix}
12 & 21 \\
32 & 45 \\
\end{pmatrix} ?
</math>

# Unnecessary for a V.S.
# This is useless

09-240/Classnotes for Tuesday September 22

2009-09-22T21:46:22Z

C8sd: /* Proof of VS4 */ Incomplete examples, and food for thought

{{09-240/Navigation}}

==Some links==
* {{Pensieve Link|2009-09/nb/09-240-TheComplexField.pdf|The Complex Numbers by Computer}}.
* Dori Eldar's work on "mechanical computations": {{Home Link|People/Eldar/thesis/linkfunc.htm|Machines as Calculating Devices}} and {{Home Link|People/Eldar/thesis/squaring.htm|Computing the function <math>W=Z^2</math> the hard way}}.
* The "Dimensions" video on "Nombres complexes", is at http://dimensions-math.org/Dim_reg_AM.htm (and then go to "Dimensions_5".

{{09-240/Class Notes Warning}}
==Class notes for today==

Vectors:
# can be added
# can be multiplied by a number (not another vector)

Let <math>\mathcal F</math> be a field. A vector space <math>\mathbf V</math> over the field <math>\mathcal F</math> is a set <math>\mathbf V</math> (of vectors) with a special element <math>0_V</math>, a binary operation <math>+ : \mathbf V \times \mathbf V \rightarrow \mathbf V</math>, a binary operation <math>\cdot : \mathcal F \times \mathbf V \rightarrow \mathbf V</math>.

{| style="border: solid 1px black"
|-
| Convention for today:
: <math>x, y, z \in \mathbf V</math>
: <math>a, b, c \in \mathcal F</math>
|}

VS1 <math>\forall x, y \in \mathbf V, x + y = y + x</math> 
VS2 <math>\cdots (x + y) + z = x + (y + z)</math> 
VS3 <math>\cdots x + 0 = x</math> 
VS4 <math>\forall x, \exists y \mbox{ s.t. } x + y = 0</math> 
VS5 <math>1 \cdot x = x</math> 
VS6 <math>a \cdot (b \cdot x) = (a \cdot b) \cdot x</math> 
VS7 <math>a \cdot (x + y) = ax + ay</math> 
VS8 <math>(a + b) \cdot x = ax + bx</math>

=== Proof of VS4 ===

Take an arbitrary <math>x = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} \in F^n</math>

Set <math>y = \begin{pmatrix} -a_1 \\ -a_2 \\ \vdots \\ -a_n \end{pmatrix}</math> and note
: <math>x + y = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} + \begin{pmatrix} -a_1 \\ -a_2 \\ \vdots \\ -a_n \end{pmatrix} = \begin{pmatrix} a_1 + (-a_1) \\ a_2 + (-a_2) \\ \vdots \\ a_n + (-a_n) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix} = 0_{F^n}</math>

=== Examples ===

# <math>F^n \mbox{ for } n \in \mathbb N</math>
#: <math>F^n = \left\{ \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} : a_i \in F \right\}</math>
#: <math>\begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ \vdots \\ a_n + b_n \end{pmatrix}</math>
#: <math>a \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix} = \begin{pmatrix} ab_1 \\ ab_2 \\ \vdots \\ ab_n \end{pmatrix}</math>
#: ...
# <math>\mathrm M_{m \times n}(F)</math>
#: ...

=== Food for thought ===

What is wrong with setting

<math>
\begin{pmatrix}
2 & 3 \\
4 & 5 \\
\end{pmatrix} \cdot \begin{pmatrix}
6 & 7 \\
8 & 9 \\
\end{pmatrix} = \begin{pmatrix}
2 \cdot 6 & 3 \cdot 7 \\
4 \cdot 8 & 5 \cdot 9 \\
\end{pmatrix} = \begin{pmatrix}
12 & 21 \\
32 & 45 \\
\end{pmatrix} ?
</math>

# Unnecessary for a V.S.
# This is useless

09-240/Classnotes for Tuesday September 22

2009-09-22T21:24:52Z

C8sd: /* Class notes for today */ Add vector section before examples.

09-240/Classnotes for Tuesday September 15

2009-09-17T23:02:42Z

C8sd: 27-n = 27*23? Where does the subtraction and variable come from?

{{09-240/Navigation}}
{{09-240/Class Notes Warning}}

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The real numbers A set <math>\mathbb R</math> with two binary operators and two special elements <math>0, 1 \in \mathbb R</math> s.t.

: <math>F1.\quad \forall a, b \in \mathbb R, a + b = b + a \mbox{ and } a \cdot b = b \cdot a</math>
: <math>F2.\quad \forall a, b, c, (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)</math>
: <math>\mbox{(So for any real numbers } a_1, a_2, ..., a_n, \mbox{ one can sum them in any order and achieve the same result.}</math>
: <math>F3.\quad \forall a, a + 0 = a \mbox{ and } a \cdot 1 = a</math>
: <math>F4.\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1</math>
: <math>\mbox{So } a + (-a) = 0 \mbox{ and } a \cdot a^{-1} = 1</math>
: <math>\mbox{(So } (a + b) \cdot (a - b) = a^2 - b^2)</math>
: <math>\forall a, \exists x, x \cdot x = a \mbox{ or } a + x \cdot x = 0</math>
: Note: '''or''' means '''inclusive or''' in math.
: <math>F5.\quad (a + b) \cdot c = a \cdot c + b \cdot c</math>

Definition: A '''field''' is a set ''F'' with two binary operators <math>\,\!+</math>: ''F''×''F'' → ''F'', <math>\times\,\!</math>: ''F''×''F'' → ''F'' and two elements <math>0, 1 \in \mathbb R</math> s.t.
: <math>F1\quad \mbox{Commutativity } a + b = b + a \mbox{ and } a \cdot b = b \cdot a \forall a, b \in F</math>
: <math>F2\quad \mbox{Associativity } (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)</math>
: <math>F3\quad a + 0 = a, a \cdot 1 = a</math>
: <math>F4\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1</math>
: <math>F5\quad \mbox{Distributivity } (a + b) \cdot c = a \cdot c + b \cdot c</math>

== Examples ==

# <math>F = \mathbb R</math>
# <math>F = \mathbb Q</math>
# <math>\mathbb C = \{ a + bi : a, b \in \mathbb R \}</math>
#: <math>i = \sqrt{-1}</math>
#: <math>\,\!(a + bi) + (c + di) = (a + c) + (b + d)i</math>
#: <math>\,\!0 = 0 + 0i, 1 = 1 + 0i</math>
# <math>\,\!F_2 = \{ 0, 1 \}</math>
# <math>\,\!F_7 = \{ 0, 1,2,3,4,5,6 \}</math>
# <math>\,\!F_6 = \{ 0, 1,2,3,4,5 \}</math> is not a field because not every element has a multiplicative inverse.
#: Let <math>a = 2.</math>
#: Then <math>a \cdot 0 = 0, a \cdot 1 = 2, a \cdot 3 = 0, a \cdot 4 = 2, a \cdot 5 = 4</math>
#: Therefore F4 fails; there is '''no''' number ''b'' in ''F''6 s.t. ''a · b'' = 1

{|
|
{| border="1" cellspacing="0"
|+ Ex. 4
|-
! + !! 0 !! 1
|-
! 0
| 0 || 1
|-
! 1
| 1 || 0
|-
|}

|
{| border="1" cellspacing="0"
|+ Ex. 4
|-
! × !! 0 !! 1
|-
! 0
| 0 || 0
|-
! 1
| 0 || 1
|}

|-
|
{| border="1" cellspacing="0"
|+ Ex. 5
|-
! + !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
|-
! 0
| 0 || 1 || 2 || 3 || 4 || 5 || 6
|-
! 1
| 1 || 2 || 3 || 4 || 5 || 6 || 0
|-
! 2
| 2 || 3 || 4 || 5 || 6 || 0 || 1
|-
! 3
| 3 || 4 || 5 || 6 || 0 || 1 || 2
|-
! 4
| 4 || 5 || 6 || 0 || 1 || 2 || 3
|-
! 5
| 5 || 6 || 0 || 1 || 2 || 3 || 4
|-
! 6
| 6 || 0 || 1 || 2 || 3 || 4 || 5
|}

|
{|border="1" cellspacing="0"
|+ Ex. 5
|-
! × !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
|-
! 0
| 0 || 0 || 0 || 0 || 0 || 0 || 0
|-
! 1
| 0 || 1 || 2 || 3 || 4 || 5 || 6
|-
! 2
| 0 || 2 || 4 || 6 || 1 || 3 || 5
|-
! 3
| 0 || 3 || 6 || 2 || 5 || 1 || 4
|-
! 4
| 0 || 4 || 1 || 5 || 2 || 6 || 3
|-
! 5
| 0 || 5 || 3 || 1 || 6 || 4 || 2
|-
! 6
| 0 || 6 || 5 || 4 || 3 || 2 || 1
|}
|}

'''Theorem''': ''F''2 is a field.

In order to prove that the associative property holds, make a table (similar to a [http://en.wikipedia.org/wiki/Truth_table truth table]) for ''a'', ''b'' and ''c''.

{| border="1" cellspacing="0" style="text-align: center;"
! a !! b !! c !!  
|-
| 0 || 0 || 0 ||  
|-
| 0 || 0 || 1 ||  
|-
| 0 || 1 || 0 ||  
|-
| 0 || 1 || 1 || (0 + 1) + 1 =? 0 + (1 + 1) 1 + 1 =? 0 + 0 0 = 0
|-
| 1 || 0 || 0 ||  
|-
| 1 || 0 || 1 ||  
|-
| 1 || 1 || 0 ||  
|-
| 1 || 1 || 1 ||  
|}

'''Theorem''': <math>\,\! F_p </math> for <math>p > 1</math> is a field ''iff'' ([http://en.wikipedia.org/wiki/If_and_only_if if and only if]) <math>p</math> is a prime number

Proof:

Given a finite set with <math>m</math> elements in <math>\mathbb Z</math>, an element <math>a</math> will have a multiplicative inverse '''iff''' <math>gcd(a,m) = 1</math>

This can be shown using [http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity Bézout's identity]:

: <math>\exists x, y \mbox{ s.t. } ax + my = 1</math>
: <math>\left(ax + my\right) \pmod{m} = 1\pmod{m}</math>
: <math>ax = 1</math>
: <math>x = a^{-1}</math>

We have shown that <math>a</math> has a multiplicative inverse if <math>a</math> and <math>m</math> are relatively prime. It is therefore a natural conclusion that if <math>m</math> is prime all elements in the set will satisfy <math>gcd(a, m) = 1</math>

----

Multiplication is repeated addition.

<math>23 \times 27 = \begin{matrix} 27 \\ \overbrace{23 + 23 + 23 + \cdots + 23} \end{matrix} = 621</math>

<math>27 \times 23 = \begin{matrix} 23 \\ \overbrace{27 + 27 + 27 + \cdots + 27} \end{matrix} = 621</math>

One may interpret this as counting the units in a 23×27 rectangle; one may choose to count along either 23 rows or 27 columns, but both ways lead to the same answer.

Exponentiation is repeated multiplication, but it does not have the same properties as multiplication; 23 = 8, but 32 = 9.

== Tedious Theorem ==
# <math>a + b = c + d \Rightarrow a = c </math> "cancellation property"
#: Proof:
#: By F4, <math>\exists d \mbox{ s.t. } b + d = 0</math>
#: <math>\,\! (a + b) + d = (c + b) + d</math>
#: <math>\Rightarrow a + (b + d) = c + (b + d)</math> by F2
#: <math>\Rightarrow a + 0 = c + 0</math> by choice of ''d''
#: <math>\Rightarrow a = c</math> by F3
# <math> a \cdot b = c \cdot b , (b \ne 0) \Rightarrow a = c </math>
# <math>a + O' = a \Rightarrow O' = 0</math>
#: Proof:
#: <math>\,\! a + O' = a</math>
#: <math>\Rightarrow a + O' = a + 0</math> by F3
#: <math>\Rightarrow O' = 0</math> by adding the additive inverse of ''a'' to both sides
# <math>a \cdot l' = a, a \ne 0 \Rightarrow l' = 1</math>
# <math>a + b = 0 = a + b' \Rightarrow b = b'</math>
# <math>a \cdot b = 1 = a \cdot b' \Rightarrow b = b' = a^{-1}</math>
#: <math>\,\! \mbox{Aside: } a - b = a + (-b)</math>
#: <math>\frac ab = a \cdot b^{-1}</math>
# <math>\,\! -(-a) = a, (a^{-1})^{-1}</math>
# <math>a \cdot 0 = 0</math>
#: Proof:
#: <math>a \cdot 0 = a(0 + 0)</math> by F3
#: <math>= a \cdot 0 + a \cdot 0</math> by F5
#: <math>= 0 = a \cdot 0</math>
# <math>\forall b, 0 \cdot b \ne 1</math>
#: So there is no 0−1
# <math>(-a) \cdot b = a \cdot (-b) = -(a \cdot b)</math>
# <math>(-a) \cdot (-b) = a \cdot b</math>
# (Bonus) <math>\,\! (a + b)(a - b) = a^2 - b^2</math>

== Quotation of the Day ==
......

09-240/Classnotes for Tuesday September 15

2009-09-17T02:20:51Z

C8sd: oops, that's later

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The real numbers A set <math>\mathbb R</math> with two binary operators and two special elements <math>0, 1 \in \mathbb R</math> s.t.

: <math>F1.\quad \forall a, b \in \mathbb R, a + b = b + a \mbox{ and } a \cdot b = b \cdot a</math>
: <math>F2.\quad \forall a, b, c, (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)</math>
: <math>\mbox{(So for any real numbers } a_1, a_2, ..., a_n, \mbox{ one can sum them in any order and achieve the same result.}</math>
: <math>F3.\quad \forall a, a + 0 = a \mbox{ and } a \cdot 1 = a</math>
: <math>F4.\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1</math>
: <math>\mbox{So } a + (-a) = 0 \mbox{ and } a \cdot a^{-1} = 1</math>
: <math>\mbox{(So } (a + b) \cdot (a - b) = a^2 - b^2)</math>
: <math>\forall a, \exists x, x \cdot x = a \mbox{ or } a + x \cdot x = 0</math>
: Note: '''or''' means '''inclusive or''' in math.
: <math>F5.\quad (a + b) \cdot c = a \cdot c + b \cdot c</math>

Definition: A '''field''' is a set ''F'' with two binary operators <math>\,\!+</math>: ''F''×''F'' → ''F'', <math>\times\,\!</math>: ''F''×''F'' → ''F'' and two elements <math>0, 1 \in \mathbb R</math> s.t.
: <math>F1\quad \mbox{Commutativity } a + b = b + a \mbox{ and } a \cdot b = b \cdot a \forall a, b \in F</math>
: <math>F2\quad \mbox{Associativity } (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)</math>
: <math>F3\quad a + 0 = a, a \cdot 1 = a</math>
: <math>F4\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1</math>
: <math>F5\quad \mbox{Distributivity } (a + b) \cdot c = a \cdot c + b \cdot c</math>

== Examples ==

# <math>F = \mathbb R</math>
# <math>F = \mathbb Q</math>
# <math>\mathbb C = \{ a + bi : a, b \in \mathbb R \}</math>
#: <math>i = \sqrt{-1}</math>
#: <math>\,\!(a + bi) + (c + di) = (a + c) + (b + d)i</math>
#: <math>\,\!0 = 0 + 0i, 1 = 1 + 0i</math>
# <math>\,\!F_2 = \{ 0, 1 \}</math>
# <math>\,\!F_7 = \{ 0, 1,2,3,4,5,6 \}</math>
# <math>\,\!F_6 = \{ 0, 1,2,3,4,5 \}</math> is not a field because not every element has a multiplicative inverse.
#: Let <math>a = 2.</math>
#: Then <math>a \cdot 0 = 0, a \cdot 1 = 2, a \cdot 3 = 0, a \cdot 4 = 2, a \cdot 5 = 4</math>
#: Therefore F4 fails; there is '''no''' number ''b'' in ''F''6 s.t. ''a · b'' = 1

{|
|
{| border="1" cellspacing="0"
|+ Ex. 4
|-
! + !! 0 !! 1
|-
! 0
| 0 || 1
|-
! 1
| 1 || 0
|-
|}

|
{| border="1" cellspacing="0"
|+ Ex. 4
|-
! × !! 0 !! 1
|-
! 0
| 0 || 0
|-
! 1
| 0 || 1
|}

|-
|
{| border="1" cellspacing="0"
|+ Ex. 5
|-
! + !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
|-
! 0
| 0 || 1 || 2 || 3 || 4 || 5 || 6
|-
! 1
| 1 || 2 || 3 || 4 || 5 || 6 || 0
|-
! 2
| 2 || 3 || 4 || 5 || 6 || 0 || 1
|-
! 3
| 3 || 4 || 5 || 6 || 0 || 1 || 2
|-
! 4
| 4 || 5 || 6 || 0 || 1 || 2 || 3
|-
! 5
| 5 || 6 || 0 || 1 || 2 || 3 || 4
|-
! 6
| 6 || 0 || 1 || 2 || 3 || 4 || 5
|}

|
{|border="1" cellspacing="0"
|+ Ex. 5
|-
! × !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
|-
! 0
| 0 || 0 || 0 || 0 || 0 || 0 || 0
|-
! 1
| 0 || 1 || 2 || 3 || 4 || 5 || 6
|-
! 2
| 0 || 2 || 4 || 6 || 1 || 3 || 5
|-
! 3
| 0 || 3 || 6 || 2 || 5 || 1 || 4
|-
! 4
| 0 || 4 || 1 || 5 || 2 || 6 || 3
|-
! 5
| 0 || 5 || 3 || 1 || 6 || 4 || 2
|-
! 6
| 0 || 6 || 5 || 4 || 3 || 2 || 1
|}
|}

'''Theorem''': ''F''2 is a field.

In order to prove that the associative property holds, make a table (similar to a [http://en.wikipedia.org/wiki/Truth_table truth table]) for ''a'', ''b'' and ''c''.

{| border="1" cellspacing="0" style="text-align: center;"
! a !! b !! c !!  
|-
| 0 || 0 || 0 ||  
|-
| 0 || 0 || 1 ||  
|-
| 0 || 1 || 0 ||  
|-
| 0 || 1 || 1 || (0 + 1) + 1 =? 0 + (1 + 1) 1 + 1 =? 0 + 0 0 = 0
|-
| 1 || 0 || 0 ||  
|-
| 1 || 0 || 1 ||  
|-
| 1 || 1 || 0 ||  
|-
| 1 || 1 || 1 ||  
|}

'''Theorem''': <math>\,\! F_p </math> for <math>p > 1</math> is a field ''iff'' ([http://en.wikipedia.org/wiki/If_and_only_if if and only if]) <math>p</math> is a prime number

Proof:

Given a finite set with <math>m</math> elements in <math>\mathbb Z</math>, an element <math>a</math> will have a multiplicative inverse '''iff''' <math>gcd(a,m) = 1</math>

This can be shown using [http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity Bézout's identity]:

: <math>\exists x, y \mbox{ s.t. } ax + my = 1</math>
: <math>\left(ax + my\right) \pmod{m} = 1\pmod{m}</math>
: <math>ax = 1</math>
: <math>x = a^{-1}</math>

We have shown that <math>a</math> has a multiplicative inverse if <math>a</math> and <math>m</math> are relatively prime. It is therefore a natural conclusion that if <math>m</math> is prime all elements in the set will satisfy <math>gcd(a, m) = 1</math>

----

Multiplication is repeated addition.

<math>23 \times 27 = \begin{matrix} 27 \\ \overbrace{23 + 23 + 23 + \cdots + 23} \end{matrix} = 621</math>

<math>27 \times 23 = \begin{matrix} 23 \\ \overbrace{27 + 27 + 27 + \cdots + 27} \end{matrix} = 621</math>

One may interpret this as counting the units in a 23×27 rectangle; one may choose to count along either 23 rows or 27 columns, but both ways lead to the same answer.

Exponentiation is repeated multiplication, but it does not have the same properties as multiplication; 23 = 8, but 32 = 9.

== Tedious Theorem ==
# <math>a + b = c + d \Rightarrow a = c </math> "cancellation property"
#: Proof:
#: By F4, <math>\exists d \mbox{ s.t. } b + d = 0</math>
#: <math>\,\! (a + b) + d = (c + b) + d</math>
#: <math>\Rightarrow a + (b + d) = c + (b + d)</math> by F2
#: <math>\Rightarrow a + 0 = c + 0</math> by choice of ''d''
#: <math>\Rightarrow a = c</math> by F3
# <math> a \cdot b = c \cdot b , (b \ne 0) \Rightarrow a = c </math>
# <math>a + O' = a \Rightarrow O' = 0</math>
#: Proof:
#: <math>\,\! a + O' = a</math>
#: <math>\Rightarrow a + O' = a + 0</math> by F3
#: <math>\Rightarrow O' = 0</math> by adding the additive inverse of ''a'' to both sides
# <math>a \cdot l' = a, a \ne 0 \Rightarrow l' = 1</math>
# <math>a + b = 0 = a + b' \Rightarrow b = b'</math>
# <math>a \cdot b = 1 = a \cdot b' \Rightarrow b = b' = a^{-1}</math>
#: <math>\,\! \mbox{Aside: } a - b = a + (-b)</math>
#: <math>\frac ab = a \cdot b^{-1}</math>
# <math>\,\! -(-a) = a, (a^{-1})^{-1}</math>
# <math>a \cdot 0 = 0</math>
#: Proof:
#: <math>a \cdot 0 = a(0 + 0)</math> by F3
#: <math>= a \cdot 0 + a \cdot 0</math> by F5
#: <math>= 0 = a \cdot 0</math>
# <math>\forall b, 0 \cdot b \ne 1</math>
#: So there is no 0−1
# <math>(-a) \cdot b = a \cdot (-b) = -(a \cdot b)</math>
# <math>(-a) \cdot (-b) = a \cdot b</math>
# (Bonus) <math>\,\! (a + b)(a - b) = a^2 - b^2</math>

== Quotation of the Day ==
......

09-240/Classnotes for Tuesday September 15

2009-09-16T05:49:09Z

C8sd: /* Examples */

{{09-240/Navigation}}

<gallery>
Image:09-240 Classnotes for Tuesday September 15 2009 page 1.jpg|[[User:Yangjiay|Yangjiay]] - Page 1
Image:09-240 Classnotes for Tuesday September 15 2009 page 2.jpg|Page 2
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Image:09-240 Classnotes for Tuesday September 15 2009 page 4.jpg|Page 4
Image:09-240 Classnotes for Tuesday September 15 2009 page 5.jpg|Page 5
</gallery>

The real numbers A set <math>\mathbb R</math> with two binary operators and two special elements <math>0, 1 \in \mathbb R</math> s.t.

: <math>F1.\quad \forall a, b \in \mathbb R, a + b = b + a \mbox{ and } a \cdot b = b \cdot a</math>
: <math>F2.\quad \forall a, b, c, (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)</math>
: <math>\mbox{(So for any real numbers } a_1, a_2, ..., a_n, \mbox{ one can sum them in any order and achieve the same result.}</math>
: <math>F3.\quad \forall a, a + 0 = a \mbox{ and } a \cdot 0 = 0 \mbox{ and } a \cdot 1 = a</math>
: <math>F4.\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1</math>
: <math>\mbox{So } a + (-a) = 0 \mbox{ and } a \cdot a^{-1} = 1</math>
: <math>\mbox{(So } (a + b) \cdot (a - b) = a^2 - b^2)</math>
: <math>\forall a, \exists x, x \cdot x = a \mbox{ or } a + x \cdot x = 0</math>
: Note: '''or''' means '''inclusive or''' in math.
: <math>F5.\quad (a + b) \cdot c = a \cdot c + b \cdot c</math>

Definition: A '''field''' is a set ''F'' with two binary operators <math>\,\!+</math>: ''F''×''F'' → ''F'', <math>\times\,\!</math>: ''F''×''F'' → ''F'' and two elements <math>0, 1 \in \mathbb R</math> s.t.
: <math>F1\quad \mbox{Commutativity } a + b = b + a \mbox{ and } a \cdot b = b \cdot a \forall a, b \in F</math>
: <math>F2\quad \mbox{Associativity } (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)</math>
: <math>F3\quad a + 0 = a, a \cdot 1 = a</math>
: <math>F4\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1</math>
: <math>F5\quad \mbox{Distributivity } (a + b) \cdot c = a \cdot c + b \cdot c</math>

== Examples ==

# <math>F = \mathbb R</math>
# <math>F = \mathbb Q</math>
# <math>\mathbb C = \{ a + bi : a, b \in \mathbb R \}</math>
#: <math>i = \sqrt{-1}</math>
#: <math>\,\!(a + bi) + (c + di) = (a + c) + (b + d)i</math>
#: <math>\,\!0 = 0 + 0i, 1 = 1 + 0i</math>
# <math>\,\!F_2 = \{ 0, 1 \}</math>
# <math>\,\!F_7 = \{ 0, 1,2,3,4,5,6 \}</math>
# <math>\,\!F_6 = \{ 0, 1,2,3,4,5 \}</math> is not a field because not every element has a multiplicative inverse.
#: Let <math>a = 2.</math>
#: Then <math>a \cdot 0 = 0, a \cdot 1 = 2, a \cdot 3 = 0, a \cdot 4 = 2, a \cdot 5 = 4</math>
#: Therefore F4 fails; there is '''no''' number ''b'' in ''F''6 s.t. ''a · b'' = 1

{|
|
{| border="1" cellspacing="0"
|+ Ex. 4
|-
! + !! 0 !! 1
|-
! 0
| 0 || 1
|-
! 1
| 1 || 0
|-
|}

|
{| border="1" cellspacing="0"
|+ Ex. 4
|-
! × !! 0 !! 1
|-
! 0
| 0 || 0
|-
! 1
| 0 || 1
|}

|-
|
{| border="1" cellspacing="0"
|+ Ex. 5
|-
! + !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
|-
! 0
| 0 || 1 || 2 || 3 || 4 || 5 || 6
|-
! 1
| 1 || 2 || 3 || 4 || 5 || 6 || 0
|-
! 2
| 2 || 3 || 4 || 5 || 6 || 0 || 1
|-
! 3
| 3 || 4 || 5 || 6 || 0 || 1 || 2
|-
! 4
| 4 || 5 || 6 || 0 || 1 || 2 || 3
|-
! 5
| 5 || 6 || 0 || 1 || 2 || 3 || 4
|-
! 6
| 6 || 0 || 1 || 2 || 3 || 4 || 5
|}

|
{|border="1" cellspacing="0"
|+ Ex. 5
|-
! × !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
|-
! 0
| 0 || 0 || 0 || 0 || 0 || 0 || 0
|-
! 1
| 0 || 1 || 2 || 3 || 4 || 5 || 6
|-
! 2
| 0 || 2 || 4 || 6 || 1 || 3 || 5
|-
! 3
| 0 || 3 || 6 || 2 || 5 || 1 || 4
|-
! 4
| 0 || 4 || 1 || 5 || 2 || 6 || 3
|-
! 5
| 0 || 5 || 3 || 1 || 6 || 4 || 2
|-
! 6
| 0 || 6 || 5 || 4 || 3 || 2 || 1
|}
|}

'''Theorem''': ''F''2 is a field.

In order to prove that the associative property holds, make a table (similar to a [http://en.wikipedia.org/wiki/Truth_table truth table]) for ''a'', ''b'' and ''c''.

{| border="1" cellspacing="0" style="text-align: center;"
! a !! b !! c !!  
|-
| 0 || 0 || 0 ||  
|-
| 0 || 0 || 1 ||  
|-
| 0 || 1 || 0 ||  
|-
| 0 || 1 || 1 || (0 + 1) + 1 =? 0 + (1 + 1) 1 + 1 =? 0 + 0 0 = 0
|-
| 1 || 0 || 0 ||  
|-
| 1 || 0 || 1 ||  
|-
| 1 || 1 || 0 ||  
|-
| 1 || 1 || 1 ||  
|}

'''Theorem''': <math>\,\! F_p </math> for <math>p > 1</math> is a field ''iff'' ([http://en.wikipedia.org/wiki/If_and_only_if if and only if]) <math>p</math> is a prime number

----

Multiplication is repeated addition.

<math>23 \times 27 = \begin{matrix} 27 \\ \overbrace{23 + 23 + 23 + \cdots + 23} \end{matrix} = 621</math>

<math>27 \times 23 = \begin{matrix} 23 \\ \overbrace{27 + 27 + 27 + \cdots + 27} \end{matrix} = 621</math>

One may interpret this as counting the units in a 23×27 rectangle; one may choose to count along either 23 rows or 27 columns, but both ways lead to the same answer.

Exponentiation is repeated multiplication, but it does not have the same properties as multiplication; 23 = 8, but 32 = 9.

== Tedious Theorem ==
# <math>a + b = c + d \Rightarrow a = c </math> "cancellation property"
#: Proof:
#: By F4, <math>\exists d \mbox{ s.t. } b + d = 0</math>
#: <math>\,\! (a + b) + d = (c + b) + d</math>
#: <math>\Rightarrow a + (b + d) = c + (b + d)</math> by F2
#: <math>\Rightarrow a + 0 = c + 0</math> by choice of ''d''
#: <math>\Rightarrow a = c</math> by F3
# <math> a \cdot b = c \cdot b , (b \ne 0) \Rightarrow a = c </math>
# <math>a + O' = a \Rightarrow O' = 0</math>
#: Proof:
#: <math>\,\! a + O' = a</math>
#: <math>\Rightarrow a + O' = a + 0</math> by F3
#: <math>\Rightarrow O' = 0</math> by adding the additive inverse of ''a'' to both sides
# <math>a \cdot l' = a, a \ne 0 \Rightarrow l' = 1</math>
# <math>a + b = 0 = a + b' \Rightarrow b = b'</math>
# <math>a \cdot b = 1 = a \cdot b' \Rightarrow b = b' = a^{-1}</math>
#: <math>\,\! \mbox{Aside: } a - b = a + (-b)</math>
#: <math>\frac ab = a \cdot b^{-1}</math>
# <math>\,\! -(-a) = a, (a^{-1})^{-1}</math>
# <math>a \cdot 0 = 0</math>
#: Proof:
#: <math>a \cdot 0 = a(0 + 0)</math> by F3
#: <math>= a \cdot 0 + a \cdot 0</math> by F5
#: <math>= 0 = a \cdot 0</math>
# <math>\forall b, 0 \cdot b \ne 1</math>
#: So there is no 0−1
# <math>(-a) \cdot b = a \cdot (-b) = -(a \cdot b)</math>
# <math>(-a) \cdot (-b) = a \cdot b</math>
# (Bonus) <math>\,\! (a + b)(a - b) = a^2 - b^2</math>

== Quotation of the Day ==
......

09-240/Classnotes for Tuesday September 15

2009-09-16T05:46:47Z

C8sd: /* Examples */ <hr>

{{09-240/Navigation}}

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The real numbers A set <math>\mathbb R</math> with two binary operators and two special elements <math>0, 1 \in \mathbb R</math> s.t.

: <math>F1.\quad \forall a, b \in \mathbb R, a + b = b + a \mbox{ and } a \cdot b = b \cdot a</math>
: <math>F2.\quad \forall a, b, c, (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)</math>
: <math>\mbox{(So for any real numbers } a_1, a_2, ..., a_n, \mbox{ one can sum them in any order and achieve the same result.}</math>
: <math>F3.\quad \forall a, a + 0 = a \mbox{ and } a \cdot 0 = 0 \mbox{ and } a \cdot 1 = a</math>
: <math>F4.\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1</math>
: <math>\mbox{So } a + (-a) = 0 \mbox{ and } a \cdot a^{-1} = 1</math>
: <math>\mbox{(So } (a + b) \cdot (a - b) = a^2 - b^2)</math>
: <math>\forall a, \exists x, x \cdot x = a \mbox{ or } a + x \cdot x = 0</math>
: Note: '''or''' means '''inclusive or''' in math.
: <math>F5.\quad (a + b) \cdot c = a \cdot c + b \cdot c</math>

Definition: A '''field''' is a set ''F'' with two binary operators <math>\,\!+</math>: ''F''×''F'' → ''F'', <math>\times\,\!</math>: ''F''×''F'' → ''F'' and two elements <math>0, 1 \in \mathbb R</math> s.t.
: <math>F1\quad \mbox{Commutativity } a + b = b + a \mbox{ and } a \cdot b = b \cdot a \forall a, b \in F</math>
: <math>F2\quad \mbox{Associativity } (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)</math>
: <math>F3\quad a + 0 = a, a \cdot 1 = a</math>
: <math>F4\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1</math>
: <math>F5\quad \mbox{Distributivity } (a + b) \cdot c = a \cdot c + b \cdot c</math>

== Examples ==

# <math>F = \mathbb R</math>
# <math>F = \mathbb Q</math>
# <math>\mathbb C = \{ a + bi : a, b \in \mathbb R \}</math>
#: <math>i = \sqrt{-1}</math>
#: <math>\,\!(a + bi) + (c + di) = (a + c) + (b + d)i</math>
#: <math>\,\!0 = 0 + 0i, 1 = 1 + 0i</math>
# <math>\,\!F_2 = \{ 0, 1 \}</math>
# <math>\,\!F_7 = \{ 0, 1,2,3,4,5,6 \}</math>
# <math>\,\!F_6 = \{ 0, 1,2,3,4,5 \}</math> is not a field because not every element has a multiplicative inverse.
#: Let <math>a = 2.</math>
#: Then <math>a \cdot 0 = 0, a \cdot 1 = 2, a \cdot 3 = 0, a \cdot 4 = 2, a \cdot 5 = 4</math>
#: Therefore F4 fails; there is '''no''' number ''b'' in ''F''6 s.t. ''a · b'' = 1

{|
|
{| border="1" cellspacing="0"
|+ Ex. 4
|-
! + !! 0 !! 1
|-
! 0
| 0 || 1
|-
! 1
| 1 || 0
|-
|}

|
{| border="1" cellspacing="0"
|+ Ex. 4
|-
! × !! 0 !! 1
|-
! 0
| 0 || 0
|-
! 1
| 0 || 1
|}

|-
|
{| border="1" cellspacing="0"
|+ Ex. 5
|-
! + !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
|-
! 0
| 0 || 1 || 2 || 3 || 4 || 5 || 6
|-
! 1
| 1 || 2 || 3 || 4 || 5 || 6 || 0
|-
! 2
| 2 || 3 || 4 || 5 || 6 || 0 || 1
|-
! 3
| 3 || 4 || 5 || 6 || 0 || 1 || 2
|-
! 4
| 4 || 5 || 6 || 0 || 1 || 2 || 3
|-
! 5
| 5 || 6 || 0 || 1 || 2 || 3 || 4
|-
! 6
| 6 || 0 || 1 || 2 || 3 || 4 || 5
|}

|
{|border="1" cellspacing="0"
|+ Ex. 5
|-
! × !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
|-
! 0
| 0 || 0 || 0 || 0 || 0 || 0 || 0
|-
! 1
| 0 || 1 || 2 || 3 || 4 || 5 || 6
|-
! 2
| 0 || 2 || 4 || 6 || 1 || 3 || 5
|-
! 3
| 0 || 3 || 6 || 2 || 5 || 1 || 4
|-
! 4
| 0 || 4 || 1 || 5 || 2 || 6 || 3
|-
! 5
| 0 || 5 || 3 || 1 || 6 || 4 || 2
|-
! 6
| 0 || 6 || 5 || 4 || 3 || 2 || 1
|}
|}

'''Theorem''': ''F''2 is a field.

In order to prove that the associative property holds, make a [http://en.wikipedia.org/wiki/Truth_table truth table] for ''a'', ''b'' and ''c''.

{| border="1" cellspacing="0" style="text-align: center;"
! a !! b !! c !!  
|-
| 0 || 0 || 0 ||  
|-
| 0 || 0 || 1 ||  
|-
| 0 || 1 || 0 ||  
|-
| 0 || 1 || 1 || (0 + 1) + 1 =? 0 + (1 + 1) 1 + 1 =? 0 + 0 0 = 0
|-
| 1 || 0 || 0 ||  
|-
| 1 || 0 || 1 ||  
|-
| 1 || 1 || 0 ||  
|-
| 1 || 1 || 1 ||  
|}

'''Theorem''': <math>\,\! F_p </math> for <math>p > 1</math> is a field ''iff'' ([http://en.wikipedia.org/wiki/If_and_only_if if and only if]) <math>p</math> is a prime number

----

Multiplication is repeated addition.

<math>23 \times 27 = \begin{matrix} 27 \\ \overbrace{23 + 23 + 23 + \cdots + 23} \end{matrix} = 621</math>

<math>27 \times 23 = \begin{matrix} 23 \\ \overbrace{27 + 27 + 27 + \cdots + 27} \end{matrix} = 621</math>

One may interpret this as counting the units in a 23×27 rectangle; one may choose to count along either 23 rows or 27 columns, but both ways lead to the same answer.

Exponentiation is repeated multiplication, but it does not have the same properties as multiplication; 23 = 8, but 32 = 9.

== Tedious Theorem ==
# <math>a + b = c + d \Rightarrow a = c </math> "cancellation property"
#: Proof:
#: By F4, <math>\exists d \mbox{ s.t. } b + d = 0</math>
#: <math>\,\! (a + b) + d = (c + b) + d</math>
#: <math>\Rightarrow a + (b + d) = c + (b + d)</math> by F2
#: <math>\Rightarrow a + 0 = c + 0</math> by choice of ''d''
#: <math>\Rightarrow a = c</math> by F3
# <math> a \cdot b = c \cdot b , (b \ne 0) \Rightarrow a = c </math>
# <math>a + O' = a \Rightarrow O' = 0</math>
#: Proof:
#: <math>\,\! a + O' = a</math>
#: <math>\Rightarrow a + O' = a + 0</math> by F3
#: <math>\Rightarrow O' = 0</math> by adding the additive inverse of ''a'' to both sides
# <math>a \cdot l' = a, a \ne 0 \Rightarrow l' = 1</math>
# <math>a + b = 0 = a + b' \Rightarrow b = b'</math>
# <math>a \cdot b = 1 = a \cdot b' \Rightarrow b = b' = a^{-1}</math>
#: <math>\,\! \mbox{Aside: } a - b = a + (-b)</math>
#: <math>\frac ab = a \cdot b^{-1}</math>
# <math>\,\! -(-a) = a, (a^{-1})^{-1}</math>
# <math>a \cdot 0 = 0</math>
#: Proof:
#: <math>a \cdot 0 = a(0 + 0)</math> by F3
#: <math>= a \cdot 0 + a \cdot 0</math> by F5
#: <math>= 0 = a \cdot 0</math>
# <math>\forall b, 0 \cdot b \ne 1</math>
#: So there is no 0−1
# <math>(-a) \cdot b = a \cdot (-b) = -(a \cdot b)</math>
# <math>(-a) \cdot (-b) = a \cdot b</math>
# (Bonus) <math>\,\! (a + b)(a - b) = a^2 - b^2</math>

== Quotation of the Day ==
......

09-240/Classnotes for Tuesday September 15

2009-09-16T05:46:00Z

C8sd: /* Examples */ truth table

{{09-240/Navigation}}

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The real numbers A set <math>\mathbb R</math> with two binary operators and two special elements <math>0, 1 \in \mathbb R</math> s.t.

: <math>F1.\quad \forall a, b \in \mathbb R, a + b = b + a \mbox{ and } a \cdot b = b \cdot a</math>
: <math>F2.\quad \forall a, b, c, (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)</math>
: <math>\mbox{(So for any real numbers } a_1, a_2, ..., a_n, \mbox{ one can sum them in any order and achieve the same result.}</math>
: <math>F3.\quad \forall a, a + 0 = a \mbox{ and } a \cdot 0 = 0 \mbox{ and } a \cdot 1 = a</math>
: <math>F4.\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1</math>
: <math>\mbox{So } a + (-a) = 0 \mbox{ and } a \cdot a^{-1} = 1</math>
: <math>\mbox{(So } (a + b) \cdot (a - b) = a^2 - b^2)</math>
: <math>\forall a, \exists x, x \cdot x = a \mbox{ or } a + x \cdot x = 0</math>
: Note: '''or''' means '''inclusive or''' in math.
: <math>F5.\quad (a + b) \cdot c = a \cdot c + b \cdot c</math>

Definition: A '''field''' is a set ''F'' with two binary operators <math>\,\!+</math>: ''F''×''F'' → ''F'', <math>\times\,\!</math>: ''F''×''F'' → ''F'' and two elements <math>0, 1 \in \mathbb R</math> s.t.
: <math>F1\quad \mbox{Commutativity } a + b = b + a \mbox{ and } a \cdot b = b \cdot a \forall a, b \in F</math>
: <math>F2\quad \mbox{Associativity } (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)</math>
: <math>F3\quad a + 0 = a, a \cdot 1 = a</math>
: <math>F4\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1</math>
: <math>F5\quad \mbox{Distributivity } (a + b) \cdot c = a \cdot c + b \cdot c</math>

== Examples ==

# <math>F = \mathbb R</math>
# <math>F = \mathbb Q</math>
# <math>\mathbb C = \{ a + bi : a, b \in \mathbb R \}</math>
#: <math>i = \sqrt{-1}</math>
#: <math>\,\!(a + bi) + (c + di) = (a + c) + (b + d)i</math>
#: <math>\,\!0 = 0 + 0i, 1 = 1 + 0i</math>
# <math>\,\!F_2 = \{ 0, 1 \}</math>
# <math>\,\!F_7 = \{ 0, 1,2,3,4,5,6 \}</math>
# <math>\,\!F_6 = \{ 0, 1,2,3,4,5 \}</math> is not a field because not every element has a multiplicative inverse.
#: Let <math>a = 2.</math>
#: Then <math>a \cdot 0 = 0, a \cdot 1 = 2, a \cdot 3 = 0, a \cdot 4 = 2, a \cdot 5 = 4</math>
#: Therefore F4 fails; there is '''no''' number ''b'' in ''F''6 s.t. ''a · b'' = 1

{|
|
{| border="1" cellspacing="0"
|+ Ex. 4
|-
! + !! 0 !! 1
|-
! 0
| 0 || 1
|-
! 1
| 1 || 0
|-
|}

|
{| border="1" cellspacing="0"
|+ Ex. 4
|-
! × !! 0 !! 1
|-
! 0
| 0 || 0
|-
! 1
| 0 || 1
|}

|-
|
{| border="1" cellspacing="0"
|+ Ex. 5
|-
! + !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
|-
! 0
| 0 || 1 || 2 || 3 || 4 || 5 || 6
|-
! 1
| 1 || 2 || 3 || 4 || 5 || 6 || 0
|-
! 2
| 2 || 3 || 4 || 5 || 6 || 0 || 1
|-
! 3
| 3 || 4 || 5 || 6 || 0 || 1 || 2
|-
! 4
| 4 || 5 || 6 || 0 || 1 || 2 || 3
|-
! 5
| 5 || 6 || 0 || 1 || 2 || 3 || 4
|-
! 6
| 6 || 0 || 1 || 2 || 3 || 4 || 5
|}

|
{|border="1" cellspacing="0"
|+ Ex. 5
|-
! × !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
|-
! 0
| 0 || 0 || 0 || 0 || 0 || 0 || 0
|-
! 1
| 0 || 1 || 2 || 3 || 4 || 5 || 6
|-
! 2
| 0 || 2 || 4 || 6 || 1 || 3 || 5
|-
! 3
| 0 || 3 || 6 || 2 || 5 || 1 || 4
|-
! 4
| 0 || 4 || 1 || 5 || 2 || 6 || 3
|-
! 5
| 0 || 5 || 3 || 1 || 6 || 4 || 2
|-
! 6
| 0 || 6 || 5 || 4 || 3 || 2 || 1
|}
|}

'''Theorem''': ''F''2 is a field.

In order to prove that the associative property holds, make a [http://en.wikipedia.org/wiki/Truth_table truth table] for ''a'', ''b'' and ''c''.

{| border="1" cellspacing="0" style="text-align: center;"
! a !! b !! c !!  
|-
| 0 || 0 || 0 ||  
|-
| 0 || 0 || 1 ||  
|-
| 0 || 1 || 0 ||  
|-
| 0 || 1 || 1 || (0 + 1) + 1 =? 0 + (1 + 1) 1 + 1 =? 0 + 0 0 = 0
|-
| 1 || 0 || 0 ||  
|-
| 1 || 0 || 1 ||  
|-
| 1 || 1 || 0 ||  
|-
| 1 || 1 || 1 ||  
|}

'''Theorem''': <math>\,\! F_p </math> for <math>p > 1</math> is a field ''iff'' ([http://en.wikipedia.org/wiki/If_and_only_if if and only if]) <math>p</math> is a prime number

Multiplication is repeated addition.

<math>23 \times 27 = \begin{matrix} 27 \\ \overbrace{23 + 23 + 23 + \cdots + 23} \end{matrix} = 621</math>

<math>27 \times 23 = \begin{matrix} 23 \\ \overbrace{27 + 27 + 27 + \cdots + 27} \end{matrix} = 621</math>

One may interpret this as counting the units in a 23×27 rectangle; one may choose to count along either 23 rows or 27 columns, but both ways lead to the same answer.

Exponentiation is repeated multiplication, but it does not have the same properties as multiplication; 23 = 8, but 32 = 9.

== Tedious Theorem ==
# <math>a + b = c + d \Rightarrow a = c </math> "cancellation property"
#: Proof:
#: By F4, <math>\exists d \mbox{ s.t. } b + d = 0</math>
#: <math>\,\! (a + b) + d = (c + b) + d</math>
#: <math>\Rightarrow a + (b + d) = c + (b + d)</math> by F2
#: <math>\Rightarrow a + 0 = c + 0</math> by choice of ''d''
#: <math>\Rightarrow a = c</math> by F3
# <math> a \cdot b = c \cdot b , (b \ne 0) \Rightarrow a = c </math>
# <math>a + O' = a \Rightarrow O' = 0</math>
#: Proof:
#: <math>\,\! a + O' = a</math>
#: <math>\Rightarrow a + O' = a + 0</math> by F3
#: <math>\Rightarrow O' = 0</math> by adding the additive inverse of ''a'' to both sides
# <math>a \cdot l' = a, a \ne 0 \Rightarrow l' = 1</math>
# <math>a + b = 0 = a + b' \Rightarrow b = b'</math>
# <math>a \cdot b = 1 = a \cdot b' \Rightarrow b = b' = a^{-1}</math>
#: <math>\,\! \mbox{Aside: } a - b = a + (-b)</math>
#: <math>\frac ab = a \cdot b^{-1}</math>
# <math>\,\! -(-a) = a, (a^{-1})^{-1}</math>
# <math>a \cdot 0 = 0</math>
#: Proof:
#: <math>a \cdot 0 = a(0 + 0)</math> by F3
#: <math>= a \cdot 0 + a \cdot 0</math> by F5
#: <math>= 0 = a \cdot 0</math>
# <math>\forall b, 0 \cdot b \ne 1</math>
#: So there is no 0−1
# <math>(-a) \cdot b = a \cdot (-b) = -(a \cdot b)</math>
# <math>(-a) \cdot (-b) = a \cdot b</math>
# (Bonus) <math>\,\! (a + b)(a - b) = a^2 - b^2</math>

== Quotation of the Day ==
......

09-240/Classnotes for Tuesday September 15

2009-09-16T05:36:44Z

C8sd: Attribute scans.

{{09-240/Navigation}}

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The real numbers A set <math>\mathbb R</math> with two binary operators and two special elements <math>0, 1 \in \mathbb R</math> s.t.

: <math>F1.\quad \forall a, b \in \mathbb R, a + b = b + a \mbox{ and } a \cdot b = b \cdot a</math>
: <math>F2.\quad \forall a, b, c, (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)</math>
: <math>\mbox{(So for any real numbers } a_1, a_2, ..., a_n, \mbox{ one can sum them in any order and achieve the same result.}</math>
: <math>F3.\quad \forall a, a + 0 = a \mbox{ and } a \cdot 0 = 0 \mbox{ and } a \cdot 1 = a</math>
: <math>F4.\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1</math>
: <math>\mbox{So } a + (-a) = 0 \mbox{ and } a \cdot a^{-1} = 1</math>
: <math>\mbox{(So } (a + b) \cdot (a - b) = a^2 - b^2)</math>
: <math>\forall a, \exists x, x \cdot x = a \mbox{ or } a + x \cdot x = 0</math>
: Note: '''or''' means '''inclusive or''' in math.
: <math>F5.\quad (a + b) \cdot c = a \cdot c + b \cdot c</math>

Definition: A '''field''' is a set ''F'' with two binary operators <math>\,\!+</math>: ''F''×''F'' → ''F'', <math>\times\,\!</math>: ''F''×''F'' → ''F'' and two elements <math>0, 1 \in \mathbb R</math> s.t.
: <math>F1\quad \mbox{Commutativity } a + b = b + a \mbox{ and } a \cdot b = b \cdot a \forall a, b \in F</math>
: <math>F2\quad \mbox{Associativity } (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)</math>
: <math>F3\quad a + 0 = a, a \cdot 1 = a</math>
: <math>F4\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1</math>
: <math>F5\quad \mbox{Distributivity } (a + b) \cdot c = a \cdot c + b \cdot c</math>

== Examples ==

# <math>F = \mathbb R</math>
# <math>F = \mathbb Q</math>
# <math>\mathbb C = \{ a + bi : a, b \in \mathbb R \}</math>
#: <math>i = \sqrt{-1}</math>
#: <math>\,\!(a + bi) + (c + di) = (a + c) + (b + d)i</math>
#: <math>\,\!0 = 0 + 0i, 1 = 1 + 0i</math>
# <math>\,\!F_2 = \{ 0, 1 \}</math>
# <math>\,\!F_7 = \{ 0, 1,2,3,4,5,6 \}</math>
# <math>\,\!F_6 = \{ 0, 1,2,3,4,5 \}</math> is not a field because not every element has a multiplicative inverse.
#: Let <math>a = 2.</math>
#: Then <math>a \cdot 0 = 0, a \cdot 1 = 2, a \cdot 3 = 0, a \cdot 4 = 2, a \cdot 5 = 4</math>
#: Therefore F4 fails; there is '''no''' number ''b'' in ''F''6 s.t. ''a · b'' = 1

{|
|
{| border="1" cellspacing="0"
|+ Ex. 4
|-
! + !! 0 !! 1
|-
! 0
| 0 || 1
|-
! 1
| 1 || 0
|-
|}

|
{| border="1" cellspacing="0"
|+ Ex. 4
|-
! × !! 0 !! 1
|-
! 0
| 0 || 0
|-
! 1
| 0 || 1
|}

|-
|
{| border="1" cellspacing="0"
|+ Ex. 5
|-
! + !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
|-
! 0
| 0 || 1 || 2 || 3 || 4 || 5 || 6
|-
! 1
| 1 || 2 || 3 || 4 || 5 || 6 || 0
|-
! 2
| 2 || 3 || 4 || 5 || 6 || 0 || 1
|-
! 3
| 3 || 4 || 5 || 6 || 0 || 1 || 2
|-
! 4
| 4 || 5 || 6 || 0 || 1 || 2 || 3
|-
! 5
| 5 || 6 || 0 || 1 || 2 || 3 || 4
|-
! 6
| 6 || 0 || 1 || 2 || 3 || 4 || 5
|}

|
{|border="1" cellspacing="0"
|+ Ex. 5
|-
! × !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6
|-
! 0
| 0 || 0 || 0 || 0 || 0 || 0 || 0
|-
! 1
| 0 || 1 || 2 || 3 || 4 || 5 || 6
|-
! 2
| 0 || 2 || 4 || 6 || 1 || 3 || 5
|-
! 3
| 0 || 3 || 6 || 2 || 5 || 1 || 4
|-
! 4
| 0 || 4 || 1 || 5 || 2 || 6 || 3
|-
! 5
| 0 || 5 || 3 || 1 || 6 || 4 || 2
|-
! 6
| 0 || 6 || 5 || 4 || 3 || 2 || 1
|}
|}

Multiplication is repeated addition.

<math>23 \times 27 = \begin{matrix} 27 \\ \overbrace{23 + 23 + 23 + \cdots + 23} \end{matrix} = 621</math>

<math>27 \times 23 = \begin{matrix} 23 \\ \overbrace{27 + 27 + 27 + \cdots + 27} \end{matrix} = 621</math>

One may interpret this as counting the units in a 23×27 rectangle; one may choose to count along either 23 rows or 27 columns, but both ways lead to the same answer.

Exponentiation is repeated multiplication, but it does not have the same properties as multiplication; 23 = 8, but 32 = 9.

'''Theorem''': <math>\,\! F_p </math> for <math>p > 1</math> is a field ''iff'' ([http://en.wikipedia.org/wiki/If_and_only_if if and only if]) <math>p</math> is a prime number

== Tedious Theorem ==
# <math>a + b = c + d \Rightarrow a = c </math> "cancellation property"
#: Proof:
#: By F4, <math>\exists d \mbox{ s.t. } b + d = 0</math>
#: <math>\,\! (a + b) + d = (c + b) + d</math>
#: <math>\Rightarrow a + (b + d) = c + (b + d)</math> by F2
#: <math>\Rightarrow a + 0 = c + 0</math> by choice of ''d''
#: <math>\Rightarrow a = c</math> by F3
# <math> a \cdot b = c \cdot b , (b \ne 0) \Rightarrow a = c </math>
# <math>a + O' = a \Rightarrow O' = 0</math>
#: Proof:
#: <math>\,\! a + O' = a</math>
#: <math>\Rightarrow a + O' = a + 0</math> by F3
#: <math>\Rightarrow O' = 0</math> by adding the additive inverse of ''a'' to both sides
# <math>a \cdot l' = a, a \ne 0 \Rightarrow l' = 1</math>
# <math>a + b = 0 = a + b' \Rightarrow b = b'</math>
# <math>a \cdot b = 1 = a \cdot b' \Rightarrow b = b' = a^{-1}</math>
#: <math>\,\! \mbox{Aside: } a - b = a + (-b)</math>
#: <math>\frac ab = a \cdot b^{-1}</math>
# <math>\,\! -(-a) = a, (a^{-1})^{-1}</math>
# <math>a \cdot 0 = 0</math>
#: Proof:
#: <math>a \cdot 0 = a(0 + 0)</math> by F3
#: <math>= a \cdot 0 + a \cdot 0</math> by F5
#: <math>= 0 = a \cdot 0</math>
# <math>\forall b, 0 \cdot b \ne 1</math>
#: So there is no 0−1
# <math>(-a) \cdot b = a \cdot (-b) = -(a \cdot b)</math>
# <math>(-a) \cdot (-b) = a \cdot b</math>
# (Bonus) <math>\,\! (a + b)(a - b) = a^2 - b^2</math>

== Quotation of the Day ==
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