14-240/Classnotes for Monday September 15

Definition:

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

Theorem:

        8. $\forall a\in F$, $a\times 0=0$.
                   proof of 8: By F3 , $a\times 0=a\times (0+0)$;
                               By F5 , $a\times (0+0)=a\times 0+a\times 0$;
                               By F3 , $a\times 0=0+a\times 0$;
                               By Thm P1,$0=a\times 0$.
       
        9. $\nexists b\in F$ s.t. $0\times b=1$;
           $\forall b\in F$ s.t. $0\times b\neq 1$.
                   proof of 9: By F3 , $\times b=0$is not equal to $1$.
       
       10. $(-a)\times b=a\times (-b)=-(a\times b)$.
     
       11. $(-a)\times (-b)=a\times b$.
      
       12. $a\times b=0\iff a=0orb=0$.
                   proof of 12: <= : By P8 , if $a=0$ , then $a\times b=0\times b=0$;
                                     By P8 , if $b=0$ , then $a\times b=a\times 0=0$.
                                => : Assume $a\times b=0$ , if a = 0 we are done;
                                     Otherwise , by P8 , $a$ is not equal to $0$and we have $a\times b=0=a\times 0$;  
                                                 by cancellation (P2) , $b=0$.
       

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

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

Theorem :

        $\exists !\iota :\mathbb {Z} \rightarrow F$  s.t.
              1. $\iota (0)=0,\iota (1)=1$;
              2. For every $m,n\in Z$ , $\iota (m+n)=\iota (m)+\iota (n)$;
              3. For every $m,n\in$ , $\iota (m\times n)=\iota (m)\times \iota (n)$.

        iota(2) = iota(1+1) = iota(1) + iota(1) = 1 + 1;
        iota(3) = iota(2+1) = iota(2) + iota(1) = iota(2) + 1; 
        ......                                                                          
     
        In F2 , $27---->iota(27)=iota(26+1)=iota(26)+iota(1)=iota(26)+1=iota(13\times 2)+1=iota(2)\times iota(13)+1=(1+1)\times iota(13)+1=0\times iota(13)+1=1$