14-240/Classnotes for Monday September 15

Definition:

           Subtract: if [math]\displaystyle{ a , b  }[/math]belong to [math]\displaystyle{ F , a - b = a + (-b) }[/math].
           Divition: if [math]\displaystyle{ a , b  }[/math]belong to F , [math]\displaystyle{ a / b = a * (b to the power (-1) }[/math].

Theorem:

        8. For every [math]\displaystyle{ a belongs to F , a * 0 = 0 }[/math].
                   proof of 8: By F3 , [math]\displaystyle{ a * 0 = a * (0 + 0) }[/math];
                               By F5 , [math]\displaystyle{ a * (0 + 0) = a * 0 + a * 0 }[/math];
                               By F3 , [math]\displaystyle{ a * 0 = 0 + a * 0 }[/math];
                               By Thm P1 ,[math]\displaystyle{ 0 = a * 0 }[/math].
       
        9. There not exists [math]\displaystyle{ b }[/math] belongs to F s.t. [math]\displaystyle{ 0 * b = 1 }[/math];
           For every [math]\displaystyle{ b }[/math] belongs to F s.t. [math]\displaystyle{ 0 * b  }[/math]is not equal to [math]\displaystyle{ 1 }[/math].
                   proof of 9: By F3 , [math]\displaystyle{ 0 * b = 0  }[/math]is not equal to [math]\displaystyle{ 1 }[/math].
       
       10. [math]\displaystyle{ (-a) * b = a * (-b) = -(a * b) }[/math].
     
       11. [math]\displaystyle{ (-a) * (-b) = a * b }[/math].
      
       12. [math]\displaystyle{ a * b = 0 iff a = 0 or b = 0 }[/math].
                   proof of 12: <= : By P8 , [math]\displaystyle{ if a = 0 , then a * b = 0 * b = 0 }[/math];
                                     By P8 , [math]\displaystyle{ if b = 0 , then a * b = a * 0 = 0 }[/math].
                                => : Assume [math]\displaystyle{ a * b = 0 }[/math] , if a = 0 we have done;
                                     Otherwise , by P8 , [math]\displaystyle{ a  }[/math]is not equal to [math]\displaystyle{ 0  }[/math]and we have [math]\displaystyle{ a * b = 0 = a * 0 }[/math];  
                                                 by cancellation (P2) , [math]\displaystyle{ b = 0 }[/math].
       

[math]\displaystyle{ (a + b) * (a - b) = a square - b square }[/math].

        proof: By F5 , [math]\displaystyle{ (a + b) * (a - b) = a * (a + (-b)) + b * (a + (-b))
                                                = a * a + a * (-b) + b * a + (-b) * b
                                                = a square - b square }[/math]

Theorem :

        There exists !(unique) [math]\displaystyle{ iota : Z ---\gt  F }[/math]  s.t.
              1. [math]\displaystyle{ iota(0) = 0 , iota(1) = 1 }[/math];
              2. For every [math]\displaystyle{ m ,n }[/math] belong to Z , [math]\displaystyle{ iota(m+n) = iota(m) + iota(n) }[/math];
              3. >For every [math]\displaystyle{ m ,n }[/math] belong to Z , [math]\displaystyle{ iota(m*n) = iota(m) * iota(n) }[/math].

        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 , [math]\displaystyle{ 27 ----\gt  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 }[/math]