# 14-240/Classnotes for Monday September 15

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## Definition of Subtraction and Division

• Subtraction: if ${\displaystyle a,b\in F,a-b=a+(-b)}$.
• Division: if ${\displaystyle a,b\in F,a/b=a\times b^{-1}}$.

## Basic Properties of a Field (cont'd)

8. ${\displaystyle \forall a\in F}$, ${\displaystyle a\times 0=0}$.

Proof of 8
By F3 , ${\displaystyle a\times 0=a\times (0+0)}$
By F5 , ${\displaystyle a\times (0+0)=a\times 0+a\times 0}$;
By F3 , ${\displaystyle a\times 0=0+a\times 0}$;
By Thm P1, ${\displaystyle 0=a\times 0}$.

9. ${\displaystyle \nexists b\in F}$ s.t. ${\displaystyle 0\times b=1}$;

${\displaystyle \forall b\in F}$ s.t. ${\displaystyle 0\times b\neq 1}$.
Proof of 9
By F3 , ${\displaystyle \times b=0\neq 1}$.

10. ${\displaystyle (-a)\times b=a\times (-b)=-(a\times b)}$.

11. ${\displaystyle (-a)\times (-b)=a\times b}$.

12. ${\displaystyle a\times b=0\iff a=0}$ or ${\displaystyle b=0}$.

Proof of 12
<= :
By P8 , if ${\displaystyle a=0}$ , then ${\displaystyle a\times b=0\times b=0}$;
By P8 , if ${\displaystyle b=0}$ , then ${\displaystyle a\times b=a\times 0=0}$.
=> : Assume ${\displaystyle a\times b=0}$ , if a = 0 we are done;
Otherwise , by P8 , ${\displaystyle a\neq 0}$ and we have ${\displaystyle a\times b=0=a\times 0}$;
by cancellation (P2) , ${\displaystyle b=0}$.

${\displaystyle (a+b)\times (a-b)=a^{2}-b^{2}}$.

Proof
By F5 , ${\displaystyle (a+b)\times (a-b)=a\times (a+(-b))+b\times (a+(-b))}$
${\displaystyle =a^{2}-b^{2}}$

## Theorem

${\displaystyle \exists !\iota :\mathbb {Z} \rightarrow F}$ s.t.

1. ${\displaystyle \iota (0)=0,\iota (1)=1}$;
2. ${\displaystyle \forall m,n\in \mathbb {Z} ,\iota (m+n)=\iota (m)+\iota (n)}$;
3. ${\displaystyle \forall m,n\in \mathbb {Z} ,\iota (m\times n)=\iota (m)\times \iota (n)}$.
Examples

${\displaystyle \iota (2)=\iota (1+1)=\iota (1)+\iota (1)=1+1;}$ ${\displaystyle \iota (3)=\iota (2+1)=\iota (2)+\iota (1)=\iota (2)+1;}$

......

In F2: {\displaystyle {\begin{aligned}27---->\iota (27)&=\iota (26+1)\\&=\iota (26)+\iota (1)\\&=\iota (26)+1\\&=\iota (13\times 2)+1\\&=\iota (2)\times \iota (13)+1\\&=(1+1)\times \iota (13)+1\\&=0\times \iota (13)+1\\&=1\end{aligned}}}