Drorbn - User contributions [en]

09-240/Class Photo

2009-10-22T03:13:39Z

C9db: /* Who We Are... */

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=Bloemendal|first=Daniel|userid=c9db|email=d.bloemendal@ utoronto.ca|location=In between "use coordinates" on the top right blackboard.|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=Danks|first=Andrew|userid=Sknad|email=a.danks at utoronto.ca|location=Towards middle; maroon 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=Muthurajah|first=Michael Praveen|userid=Michael.muthurajah|email=michael.muthurajah@ utoronto.ca|location=farthest right but in the middle of the picture,the brown guy with a black cap wearing blue jeans carrying a backpack on my right shoulder|comments=It's a pleasure to be in this class,math answers everything}}
{{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=Popescu|first=Radu|userid=RPopescu|email=radu.popescu@ utoronto.ca|location=Top left-hand box, second from the right|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=Roias|first=Stephen|userid=Sven|email=stephen.roias@ utoronto.ca|location=top right corner, furthest right|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/Classnotes for Tuesday September 15

2009-09-19T22:04:55Z

C9db:

{{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/Classnotes for Tuesday September 15

2009-09-19T22:04:29Z

C9db:

{{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 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/Classnotes for Tuesday September 15

2009-09-16T19:47:28Z

C9db:

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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 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

2009-09-16T19:46:22Z

C9db:

{{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 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''' gcd(a,m) = 1

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 gcd(a, m) = 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.

== 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

2009-09-16T19:30:47Z

C9db:

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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 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''' gcd(a,m) = 1

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>(ax + my) \bmod{m} = 1\bmod{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 gcd(a, m) = 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.

== 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

2009-09-16T19:18:24Z

C9db: Added a proof

{{09-240/Navigation}}

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Image:09-240 Classnotes for Tuesday September 15 2009 page 1.jpg|[[User:Yangjiay|Yangjiay]] - Page 1
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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 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''' gcd(a,m) = 1

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

<math>\exists x, y s.t. ax + my = 1</math>
<math>(ax + my) \mod m = 1 \mod m</math>
<math>ax = 1 \therefore 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 gcd(a, m) = 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.

== 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 ==
......