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>; |
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<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>. |
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proof of 9: By F3 , <math>\times b = 0 |
proof of 9: By F3 , <math>\times b = 0 \neq 1</math>. |
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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>. |
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=> : 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; |
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Otherwise , by P8 , <math>a |
Otherwise , by P8 , <math>a \neq 0 </math> and we have <math>a \times b = 0 = a \times 0</math>; |
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by cancellation (P2) , <math>b = 0</math>. |
by cancellation (P2) , <math>b = 0</math>. |
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<math>(a + b) \times (a - b) = a^2 - b^2</math>. |
<math>(a + b) \times (a - b) = a^2 - b^2</math>. |
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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> |
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= 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> |
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= a^2 - b^2</math> |
<math>= a^2 - b^2</math> |
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Theorem : |
Theorem : |
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<math>\exists! \iota : \Z \rightarrow F</math> s.t. |
<math>\exists! \iota : \Z \rightarrow F</math> s.t. |
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1. <math>\iota(0) = 0 , \iota(1) = 1</math>; |
1. <math>\iota(0) = 0 , \iota(1) = 1</math>; |
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2. |
2. <math>\forall m ,n \in \Z, \iota(m+n) = \iota(m) + \iota(n)</math>; |
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3. |
3. <math>\forall m ,n \in \Z, \iota(m\times n) = \iota(m) \times \iota(n)</math>. |
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iota(2) = iota(1+1) = iota(1) + iota(1) = 1 + 1; |
<math>\iota(2) = \iota(1+1) = \iota(1) + \iota(1) = 1 + 1;</math> |
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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> |
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...... |
...... |
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In F2 , <math>27 ----> iota(27) = iota(26 + 1) |
In F2 , <math>27 ----> \iota(27) = \iota(26 + 1)</math> |
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= iota(26) + iota(1) |
<math>= \iota(26) + \iota(1)</math> |
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= iota(26) + 1 |
<math>= \iota(26) + 1</math> |
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= iota(13 \times 2) + 1 |
<math>= \iota(13 \times 2) + 1</math> |
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= iota(2) \times iota(13) + 1 |
<math>= \iota(2) \times \iota(13) + 1</math> |
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= (1 + 1) \times iota(13) + 1 |
<math>= (1 + 1) \times \iota(13) + 1</math> |
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= 0 \times iota(13) + 1 |
<math>= 0 \times \iota(13) + 1</math> |
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= 1</math> |
<math>= 1</math> |
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Revision as of 12:38, 16 September 2014
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 \neq 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 \neq 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)) }[/math] [math]\displaystyle{ = a \times a + a \times (-b) + b \times a + (-b) \times b }[/math] [math]\displaystyle{ = 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. [math]\displaystyle{ \forall m ,n \in \Z, \iota(m+n) = \iota(m) + \iota(n) }[/math]; 3. [math]\displaystyle{ \forall m ,n \in \Z, \iota(m\times n) = \iota(m) \times \iota(n) }[/math].
[math]\displaystyle{ \iota(2) = \iota(1+1) = \iota(1) + \iota(1) = 1 + 1; }[/math] [math]\displaystyle{ \iota(3) = \iota(2+1) = \iota(2) + \iota(1) = \iota(2) + 1; }[/math] ...... In F2 , [math]\displaystyle{ 27 ----\gt \iota(27) = \iota(26 + 1) }[/math] [math]\displaystyle{ = \iota(26) + \iota(1) }[/math] [math]\displaystyle{ = \iota(26) + 1 }[/math] [math]\displaystyle{ = \iota(13 \times 2) + 1 }[/math] [math]\displaystyle{ = \iota(2) \times \iota(13) + 1 }[/math] [math]\displaystyle{ = (1 + 1) \times \iota(13) + 1 }[/math] [math]\displaystyle{ = 0 \times \iota(13) + 1 }[/math] [math]\displaystyle{ = 1 }[/math]
