14-240/Classnotes for Monday September 15: Difference between revisions

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9. <math>\nexists b \in F</math> s.t. <math>0 \times b = 1</math>;
9. <math>\nexists b \in F</math> s.t. <math>0 \times b = 1</math>;
<math>\forall b \in F</math> s.t. <math>0 \times b \neq 1</math>.
<math>\forall b \in F</math> s.t. <math>0 \times b \neq 1</math>.
proof of 9: By F3 , <math>\times b = 0 </math>is not equal to <math>1</math>.
proof of 9: By F3 , <math>\times b = 0 \neq 1</math>.
10. <math>(-a) \times b = a \times (-b) = -(a \times b)</math>.
10. <math>(-a) \times b = a \times (-b) = -(a \times b)</math>.
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By P8 , if <math>b = 0</math> , then <math>a \times b = a \times 0 = 0</math>.
By P8 , if <math>b = 0</math> , then <math>a \times b = a \times 0 = 0</math>.
=> : Assume <math>a \times b = 0 </math> , if a = 0 we are done;
=> : Assume <math>a \times b = 0 </math> , if a = 0 we are done;
Otherwise , by P8 , <math>a </math> is not equal to <math>0 </math>and we have <math>a \times b = 0 = a \times 0</math>;
Otherwise , by P8 , <math>a \neq 0 </math> and we have <math>a \times b = 0 = a \times 0</math>;
by cancellation (P2) , <math>b = 0</math>.
by cancellation (P2) , <math>b = 0</math>.
<math>(a + b) \times (a - b) = a^2 - b^2</math>.
<math>(a + b) \times (a - b) = a^2 - b^2</math>.
proof: By F5 , <math>(a + b) \times (a - b) = a \times (a + (-b)) + b \times (a + (-b))
proof: By F5 , <math>(a + b) \times (a - b) = a \times (a + (-b)) + b \times (a + (-b))</math>
= a \times a + a \times (-b) + b \times a + (-b) \times b
<math>= a \times a + a \times (-b) + b \times a + (-b) \times b</math>
= a^2 - b^2</math>
<math>= a^2 - b^2</math>
Theorem :
Theorem :
<math>\exists! \iota : \Z \rightarrow F</math> s.t.
<math>\exists! \iota : \Z \rightarrow F</math> s.t.
1. <math>\iota(0) = 0 , \iota(1) = 1</math>;
1. <math>\iota(0) = 0 , \iota(1) = 1</math>;
2. For every <math>m ,n \in Z</math> , <math>\iota(m+n) = \iota(m) + \iota(n)</math>;
2. <math>\forall m ,n \in \Z, \iota(m+n) = \iota(m) + \iota(n)</math>;
3. For every <math>m ,n \in </math> , <math>\iota(m\times n) = \iota(m) \times \iota(n)</math>.
3. <math>\forall m ,n \in \Z, \iota(m\times n) = \iota(m) \times \iota(n)</math>.


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

Revision as of 12:38, 16 September 2014

Definition: 

           Subtraction: if .
           Division: if .

Theorem: 

        8. , .
                   proof of 8: By F3 , ;
                               By F5 , ;
                               By F3 , ;
                               By Thm P1,.
       
        9.  s.t. ;
            s.t. .
                   proof of 9: By F3 , .
       
       10. .
     
       11. .
      
       12. .
                   proof of 12: <= : By P8 , if  , then ;
                                     By P8 , if  , then .
                                => : Assume  , if a = 0 we are done;
                                     Otherwise , by P8 ,  and we have ;  
                                                 by cancellation (P2) , .

. 

        proof: By F5 ,

Theorem : 

          s.t.
              1. ;
              2. ;
              3. .

        
         
        ......                                                                          
     
        In F2 ,
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