Drorbn - User contributions [en]

09-240/Classnotes for Thursday September 24

2009-10-22T02:47:23Z

Raymct:

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{{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, +, \cdot</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 = z</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

2009-10-22T02:44:55Z

Raymct:

{{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, +, \cdot</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 Tuesday September 15

2009-10-22T02:42:15Z

Raymct:

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

<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
Image:09-240 Classnotes for Tuesday September 15 2009 page 3.jpg|Page 3
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 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:

<math>a \in \mathbb Z</math> has a multiplicative inverse modulo <math>m</math> if and only if a and m are relatively prime.

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 modulo m if <math>a</math> and <math>m</math> are relatively prime. It is therefore a natural conclusion that if <math>m</math> is a prime number all elements in the set will be relatively prime to m.

----

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 + b \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/Class Photo

2009-10-06T17:00:23Z

Raymct:

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=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=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=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=Sinn|first=Daniel|userid=c8sd|email=|location=second-last row, fourth from left|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=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=}}