# Difference between revisions of "09-240/Classnotes for Tuesday September 15"

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

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

Definition: A field is a set F with two binary operators $\,\!+$: F×FF, $\times\,\!$: F×FF and two elements $0, 1 \in \mathbb R$ s.t.

$F1\quad \mbox{Commutativity } a + b = b + a \mbox{ and } a \cdot b = b \cdot a \forall a, b \in F$
$F2\quad \mbox{Associativity } (a + b) + c = a + (b + c) \mbox{ and } (a \cdot b) \cdot c = a \cdot (b \cdot c)$
$F3\quad a + 0 = a, a \cdot 1 = a$
$F4\quad \forall a, \exists b, a + b = 0 \mbox{ and } \forall a \ne 0, \exists b, a \cdot b = 1$
$F5\quad \mbox{Distributivity } (a + b) \cdot c = a \cdot c + b \cdot c$

## Examples

1. $F = \mathbb R$
2. $F = \mathbb Q$
3. $\mathbb C = \{ a + bi : a, b \in \mathbb R \}$
$i = \sqrt{-1}$
$\,\!(a + bi) + (c + di) = (a + c) + (b + d)i$
$\,\!0 = 0 + 0i, 1 = 1 + 0i$
4. $\,\!F_2 = \{ 0, 1 \}$
5. $\,\!F_7 = \{ 0, 1,2,3,4,5,6 \}$
6. $\,\!F_6 = \{ 0, 1,2,3,4,5 \}$ is not a field because not every element has a multiplicative inverse.
Let $a = 2.$
Then $a \cdot 0 = 0, a \cdot 1 = 2, a \cdot 3 = 0, a \cdot 4 = 2, a \cdot 5 = 4$
Therefore F4 fails; there is no number b in F6 s.t. a · b = 1
Ex. 4
+ 0 1
0 0 1
1 1 0
Ex. 4
× 0 1
0 0 0
1 0 1
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
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: F2 is a field.

In order to prove that the associative property holds, make a table (similar to a truth table) for a, b and c.

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: $\,\! F_p$ for $p > 1$ is a field iff (if and only if) $p$ is a prime number

Proof:

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

This can be shown using Bézout's identity:

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

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

$23 \times 27 = \begin{matrix} 27 \\ \overbrace{23 + 23 + 23 + \cdots + 23} \end{matrix} = 621$

$27 \times 23 = \begin{matrix} 23 \\ \overbrace{27 + 27 + 27 + \cdots + 27} \end{matrix} = 621$

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

1. $a + b = c + d \Rightarrow a = c$ "cancellation property"
Proof:
By F4, $\exists d \mbox{ s.t. } b + d = 0$
$\,\! (a + b) + d = (c + b) + d$
$\Rightarrow a + (b + d) = c + (b + d)$ by F2
$\Rightarrow a + 0 = c + 0$ by choice of d
$\Rightarrow a = c$ by F3
2. $a \cdot b = c \cdot b , (b \ne 0) \Rightarrow a = c$
3. $a + O' = a \Rightarrow O' = 0$
Proof:
$\,\! a + O' = a$
$\Rightarrow a + O' = a + 0$ by F3
$\Rightarrow O' = 0$ by adding the additive inverse of a to both sides
4. $a \cdot l' = a, a \ne 0 \Rightarrow l' = 1$
5. $a + b = 0 = a + b' \Rightarrow b = b'$
6. $a \cdot b = 1 = a \cdot b' \Rightarrow b = b' = a^{-1}$
$\,\! \mbox{Aside: } a - b = a + (-b)$
$\frac ab = a \cdot b^{-1}$
7. $\,\! -(-a) = a, (a^{-1})^{-1}$
8. $a \cdot 0 = 0$
Proof:
$a \cdot 0 = a(0 + 0)$ by F3
$= a \cdot 0 + a \cdot 0$ by F5
$= 0 = a \cdot 0$
9. $\forall b, 0 \cdot b \ne 1$
So there is no 0−1
10. $(-a) \cdot b = a \cdot (-b) = -(a \cdot b)$
11. $(-a) \cdot (-b) = a \cdot b$
12. (Bonus) $\,\! (a + b)(a - b) = a^2 - b^2$

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