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== Tedious Theorem == |
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== Tedious Theorem == |
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# <math>a + b = c + d \Rightarrow a = c </math> "cancellation property" |
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# <math>a + b = c + d \Rightarrow a = c </math> "cancellation property" |
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#: Proof: |
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#: By F4, <math>\exists d \mbox{ s.t. } b + d = 0</math> |
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#: <math>\,\! (a + b) + d = (c + b) + d</math> |
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#: <math>\Rightarrow a + (b + d) = c + (b + d)</math> by F2 |
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#: <math>\Rightarrow a + 0 = c + 0</math> by choice of ''d'' |
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#: <math>\Rightarrow a = c</math> by F3 |
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# <math> a \cdot b = c \cdot b , (b \ne 0) \Rightarrow a = c </math> |
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# <math> a \cdot b = c \cdot b , (b \ne 0) \Rightarrow a = c </math> |
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# <math>a + O' = a \Rightarrow O' = 0</math> |
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#: <math>a + O' = a</math> |
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... |
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#: <math>\Rightarrow a + O' = a + 0</math> by F3 |
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#: <math>\Rightarrow O' = 0</math> by adding the additive inverse of ''a'' to both sides |
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# <math>a \cdot l' = a, a \ne 0 \Rightarrow l' = 1</math> |
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# <math>a + b = 0 = a + b' \Rightarrow b = b'</math> |
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# <math>a \cdot b = 1 = a \cdot b' \Rightarrow b = b' = a^{-1}</math> |
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#: <math>\,\! \mbox{Aside: } a - b = a + (-b)</math> |
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#: <math>\frac ab = a \cdot b^{-1}</math> |
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# <math>\,\! -(-a) = a, (a^{-1})^{-1}</math> |
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# <math>a \cdot 0 = 0</math> |
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#: Proof: |
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#: <math>a \cdot 0 = a(0 + 0)</math> by F3 |
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#: <math>= a \cdot 0 + a \cdot 0</math> by F5 |
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#: <math>= 0 = a \cdot 0</math> |
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# <math>\forall b, 0 \cdot b \ne 1</math> |
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#: So there is no 0<sup>−1</sup> |
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# <math>(-a) \cdot b = a \cdot (-b) = -(a \cdot b)</math> |
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# <math>(-a) \cdot (-b) = a \cdot b</math> |
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# (Bonus) <math>\,\! (a + b)(a - b) = a^2 - b^2</math> |
Revision as of 22:35, 15 September 2009
The real numbers A set with two binary operators and two special elements s.t.
- Note: or means inclusive or in math.
Definition: A field is a set F with two binary operators : F×F → F, : F×F → F and two elements s.t.
Examples
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- is not a field because not every element has a multiplicative inverse.
- Let
- Then
- Therefore F4 fails; there is no number b in F6 s.t. a · b = 1
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Ex. 5
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Ex. 5
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Theorem: for is a field iff (if and only if) is a prime number
Tedious Theorem
- "cancellation property"
- Proof:
- By F4,
- by F2
- by choice of d
- by F3
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- by F3
- by adding the additive inverse of a to both sides
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- Proof:
- by F3
- by F5
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- So there is no 0−1
- (Bonus)