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

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Definition:  
 
Definition:  
 
             Subtraction: if <math>a, b \in F, a - b = a + (-b)</math>.
 
             Subtraction: if <math>a, b \in F, a - b = a + (-b)</math>.
             Division: if <math>a, b \in F, a / b = a * b^{-1}</math>.
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             Division: if <math>a, b \in F, a / b = a \times b^{-1}</math>.
  
 
Theorem:
 
Theorem:
  
         8. For every <math>a</math> belongs to F , <math>a * 0 = 0</math>.
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         8. <math>\forall a \in F</math>, <math>a \times 0 = 0</math>.
                     proof of 8: By F3 , <math>a * 0 = a * (0 + 0)</math>;
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                     proof of 8: By F3 , <math>a \times 0 = a \times (0 + 0)</math>;
                                 By F5 , <math>a * (0 + 0) = a * 0 + a * 0</math>;
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                                 By F5 , <math>a \times (0 + 0) = a \times 0 + a \times 0</math>;
                                 By F3 , <math>a * 0 = 0 + a * 0</math>;
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                                 By F3 , <math>a \times 0 = 0 + a \times 0</math>;
                                 By Thm P1 ,<math>0 = a * 0</math>.
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                                 By Thm P1,<math>0 = a \times 0</math>.
 
          
 
          
         9. There not exists <math>b</math> belongs to F s.t. <math>0 * b = 1</math>;
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         9. <math>\nexists b \in F</math> s.t. <math>0 \times b = 1</math>;
             For every <math>b</math> belongs to F s.t. <math>0 * b </math>is not equal to <math>1</math>.
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             <math>\forall b \in F</math> s.t. <math>0 \times b \neq 1</math>.
                     proof of 9: By F3 , <math>0 * b = 0 </math>is not equal to <math>1</math>.
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                     proof of 9: By F3 , <math>\times b = 0 </math>is not equal to <math>1</math>.
 
          
 
          
         10. <math>(-a) * b = a * (-b) = -(a * b)</math>.
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         10. <math>(-a) \times b = a \times (-b) = -(a \times b)</math>.
 
        
 
        
         11. <math>(-a) * (-b) = a * b</math>.
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         11. <math>(-a) \times (-b) = a \times b</math>.
 
        
 
        
         12. <math>a * b = 0 iff a = 0 or b = 0</math>.
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         12. <math>a \times b = 0 \iff a = 0 or b = 0</math>.
                     proof of 12: <= : By P8 , if <math>a = 0</math> , then <math>a * b = 0 * b = 0</math>;
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                     proof of 12: <= : By P8 , if <math>a = 0</math> , then <math>a \times b = 0 \times b = 0</math>;
                                       By P8 , if <math>b = 0</math> , then <math>a * b = a * 0 = 0</math>.
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                                       By P8 , if <math>b = 0</math> , then <math>a \times b = a \times 0 = 0</math>.
                                 => : Assume <math>a * b = 0 </math> , if a = 0 we have done;
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                                 => : 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 * b = 0 = a * 0</math>;   
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                                       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>;   
 
                                                   by cancellation (P2) , <math>b = 0</math>.
 
                                                   by cancellation (P2) , <math>b = 0</math>.
 
          
 
          
<math>(a + b) * (a - b) = a^2 - b^2</math>.
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<math>(a + b) \times (a - b) = a^2 - b^2</math>.
         proof: By F5 , <math>(a + b) * (a - b) = a * (a + (-b)) + b * (a + (-b))
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         proof: By F5 , <math>(a + b) \times (a - b) = a \times (a + (-b)) + b \times (a + (-b))
                                                 = a * a + a * (-b) + b * a + (-b) * b
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                                                 = a \times a + a \times (-b) + b \times a + (-b) \times b
 
                                                 = a^2 - b^2</math>
 
                                                 = a^2 - b^2</math>
 
Theorem :  
 
Theorem :  
         There exists !(unique) iota <math>\iota : \Z \rightarrow F</math>  s.t.
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         <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</math> belong to <math>Z</math> , <math>\iota(m+n) = \iota(m) + \iota(n)</math>;
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               2. For every <math>m ,n \in Z</math> , <math>\iota(m+n) = \iota(m) + \iota(n)</math>;
               3. For every <math>m ,n</math> belong to <math>Z</math> , <math>\iota(m*n) = \iota(m) * \iota(n)</math>.
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               3. For every <math>m ,n \in </math> , <math>\iota(m\times n) = \iota(m) \times \iota(n)</math>.
  
 
         iota(2) = iota(1+1) = iota(1) + iota(1) = 1 + 1;
 
         iota(2) = iota(1+1) = iota(1) + iota(1) = 1 + 1;
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                                         = iota(26) + iota(1)
 
                                         = iota(26) + iota(1)
 
                                         = iota(26) + 1
 
                                         = iota(26) + 1
                                         = iota(13 * 2) + 1
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                                         = iota(13 \times 2) + 1
                                         = iota(2) * iota(13) + 1
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                                         = iota(2) \times iota(13) + 1
                                         = (1 + 1) * iota(13) + 1
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                                         = (1 + 1) \times iota(13) + 1
                                         = 0 * iota(13) + 1
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                                         = 0 \times iota(13) + 1
 
                                         = 1</math>
 
                                         = 1</math>

Revision as of 13:31, 16 September 2014

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 = 0 or b = 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 : \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