14-240/Classnotes for Monday September 15

Definition:

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

Theorem:

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

[math]\displaystyle{ (a + b) \times (a - b) = a^2 - b^2 }[/math].

        proof: By F5 , [math]\displaystyle{ (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 }[/math]

Theorem :

        [math]\displaystyle{ \exists! \iota : \Z \rightarrow F }[/math]  s.t.
              1. [math]\displaystyle{ \iota(0) = 0 , \iota(1) = 1 }[/math];
              2. For every [math]\displaystyle{ m ,n \in Z }[/math] , [math]\displaystyle{ \iota(m+n) = \iota(m) + \iota(n) }[/math];
              3. For every [math]\displaystyle{ m ,n \in  }[/math] , [math]\displaystyle{ \iota(m\times n) = \iota(m) \times \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 \times 2) + 1
                                         = iota(2) \times iota(13) + 1
                                         = (1 + 1) \times iota(13) + 1
                                         = 0 \times iota(13) + 1
                                         = 1 }[/math]