09-240/Classnotes for Tuesday September 22

2009-09-23T02:44:42Z

Physics: /* Food for thought */

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==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, 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 15

2009-09-17T21:46:45Z

Physics: /* Examples */

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{{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
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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 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.

You may also think of it as 27-n=23 23*23 + 23*n = 27*23.
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 ==
......

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09-240/Classnotes for Tuesday September 22

09-240/Classnotes for Tuesday September 15