14-240/Classnotes for Monday September 15

Definition:

           Subtract: if $a,b$ belong to $F,a-b=a+(-b)$.
           Divition: if $a,b$ belong to $F,a/b=a*(b$to the power Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-1)}
.

Theorem:

        8. For every Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a}
 belongs to F , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a * 0 = 0}
.
                   proof of 8: By F3 , $a*0=a*(0+0)$;
                               By F5 , $a*(0+0)=a*0+a*0$;
                               By F3 , $a*0=0+a*0$;
                               By Thm P1 ,$0=a*0$.
       
        9. There not exists $b$ belongs to F s.t. $0*b=1$;
           For every $b$ belongs to F s.t. $0*b$is not equal to $1$.
                   proof of 9: By F3 , $0*b=0$is not equal to $1$.
       
       10. $(-a)*b=a*(-b)=-(a*b)$.
     
       11. $(-a)*(-b)=a*b$.
      
       12. $a*b=0iffa=0orb=0$.
                   proof of 12: <= : By P8 , if $a=0$ , then $a*b=0*b=0$;
                                     By P8 , if $b=0$ , then $a*b=a*0=0$.
                                => : Assume $a*b=0$ , if a = 0 we have done;
                                     Otherwise , by P8 , $a$ is not equal to $0$and we have $a*b=0=a*0$;  
                                                 by cancellation (P2) , $b=0$.
       

$(a+b)*(a-b)=asquare-bsquare$.

        proof: By F5 , $(a+b)*(a-b)=a*(a+(-b))+b*(a+(-b))=a*a+a*(-b)+b*a+(-b)*b=asquare-bsquare$

Theorem :

        There exists !(unique) $iota:Z--->F$  s.t.
              1. $iota(0)=0,iota(1)=1$;
              2. For every $m,n$ belong to $Z$ , $iota(m+n)=iota(m)+iota(n)$;
              3. >For every $m,n$ belong to $Z$ , $iota(m*n)=iota(m)*iota(n)$.

        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 , $27---->iota(27)=iota(26+1)=iota(26)+iota(1)=iota(26)+1=iota(13*2)+1=iota(2)*iota(13)+1=(1+1)*iota(13)+1=0*iota(13)+1=1$