14-240/Classnotes for Wednesday September 10: Difference between revisions
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Definition: |
Definition: |
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A "field" is a set F along with a pair of binary operations : |
A "field" is a set F along with a pair of binary operations : |
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+ : F + F -> F; |
+ : F + F -> F; |
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* : F * F -> F; |
* : F * F -> F; |
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and along with a pair 0 , 1 belong to F , 0 is not equal to 1 & s.t. F1 to F5 hold. |
and along with a pair 0 , 1 belong to F , 0 is not equal to 1 & s.t. F1 to F5 hold. |
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F1 ~ F5 : |
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R1 : for every a , b belong to R , a + b = b + a & a * b = b * a; |
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R2 : for every a , b , c belong to R , (a + b) + c = a + (b + c) & (a * b) * c = a * (b * c); |
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R3 : "existence of units" : for every a belongs to R , a + 0 = a ("additive unit") & a * 1 = a ("multiplicative unit"); |
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R4 : "the existence of inverses" : for every a belongs to R , there exists b belongs to R , s.t. a + b = 0; |
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for every a belongs to R , a is not equal to 0 , there exists b , s.t a * b = 1; |
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R5 : "the distributive law" : for every a , b , c belong to R , (a + b) * c = a * c + b * c; |
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Examples: |
Examples: |
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1. a + b = c + b => a = c "concellation law" |
1. a + b = c + b => a = c "concellation law" |
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2. b is not equal to 0 & a * b = c * b => a = c; |
2. b is not equal to 0 & a * b = c * b => a = c; |
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Proof of 1 : |
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1. By F4 , there is some b' belongs to F , s.t. b + b' = 0; |
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we have a + b = c + b; |
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=> (a + b) + b' = (c + b) + b'; |
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2. By F2 , a + (b + b') = c + (b + b'); |
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so by the choice of b' , we can know that a + 0 = c + 0; |
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3. By F3 , a = c. |
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^_^ |
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Proof of 2 : more or less the same; |
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3. If 0' belongs to F is "like 0" , then it is 0; |
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If 0' belongs to F satisfies that for every a belongs to F , a + 0' = a , then 0' = 0; |
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4. If 1' belongs to F is "like 1" , then it is 1; |
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If 1' belongs to F satisfies that for every a belongs to F , a * 1' = a , then 1' = 1; |
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Proof of 3 : |
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1. By F3 , 0' = 0' + 0 |
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2. By F1 , 0' + 0 = 0 + 0' |
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3. By assuption on 0' , 0 + 0' = 0. |
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^_^ |
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5. If b & b' satisfy a + b = 0 & a + b' = 0 => b = b'; |
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=> in any field "-a" makes sense ----- unambigous meaning |
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(-a) : the b s.t. a + b = 0; |
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6. If a is not equal to 0 & a * b = 1 = a * b' |
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=> b = b'; |
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=> if a is not equal to 0 , "-a" makes sense. |
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Proof of 5 : |
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1. a + b = 0 = a + b'; |
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2. By F1 , b + a = b'+ a; |
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3. By concellation , b = b'; |
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^_^ |
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7. -(-a) = a and when a is not equal to 0 , ( a to the power (-1) ) to the power (-1) = a; |
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Proof of 7 : |
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1. By definition , a + (-a) = 0; (*) |
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2. By definition , (-a) + (-(-a) = 0; |
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3. By (*) and F1 , (-a) + a = 0; |
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4. By property 5 , -(-a) & a must be the same. |
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^_^ |
Revision as of 15:39, 11 September 2014
Knowledge about Fields :
During this lecture, at the beginning, we talked about the properties of the real numbers. Then we applied these properties to the "Field". At the end of the lecture, we learned how to prove basic properties of field.
The real numbers: A set R with
+ : R + R -> R; * : R * R -> R;
s.t.
R1 : for every a , b belong to R , a + b = b + a & a * b = b * a; R2 : for every a , b , c belong to R , (a + b) + c = a + (b + c) & (a * b) * c = a * (b * c); R3 : "existence of units" : for every a belongs to R , a + 0 = a ("additive unit") & a * 1 = a ("multiplicative unit"); R4 : "the existence of inverses" : for every a belongs to R , there exists b belongs to R , s.t. a + b = 0; for every a belongs to R , a is not equal to 0 , there exists b , s.t a * b = 1; R5 : "the distributive law" : for every a , b , c belong to R , (a + b) * c = a * c + b * c;
True for R , yet does not follow from R1 to R5. For every a belongs to R , there exists x belongs to R , s.t. a = x square or -a = x square;
PS: This "or" here means "inclusive or" !!!
in Q which means rational numbers , let us take a = 2 , there not exists x belongs to Q , s.t. 2 = x square;
Field :
Definition:
A "field" is a set F along with a pair of binary operations : + : F + F -> F; * : F * F -> F; and along with a pair 0 , 1 belong to F , 0 is not equal to 1 & s.t. F1 to F5 hold.
F1 ~ F5 : R1 : for every a , b belong to R , a + b = b + a & a * b = b * a; R2 : for every a , b , c belong to R , (a + b) + c = a + (b + c) & (a * b) * c = a * (b * c); R3 : "existence of units" : for every a belongs to R , a + 0 = a ("additive unit") & a * 1 = a ("multiplicative unit"); R4 : "the existence of inverses" : for every a belongs to R , there exists b belongs to R , s.t. a + b = 0; for every a belongs to R , a is not equal to 0 , there exists b , s.t a * b = 1; R5 : "the distributive law" : for every a , b , c belong to R , (a + b) * c = a * c + b * c;
Examples:
1. R is a field; 2. Q which means rational numbers is a field; 3. C which means complex numbers is a field; 4. F = {0 , 1} is a field: + 0 1 * 0 1 0 0 1 0 0 0 1 1 0 1 0 1 F = {0 , 1} = F2 = Z/2 5. For every prime number P : Fp = {0 , 1 , 2 , 3 , ... , p-1} is a field; along with + & * defined as above; (a , b) -> a + b mod p foe example : F7 = {1 , 2 , 3 , 4 , 5 , 6}; + like remainder when you divided by 7; like remainders mod 7.
Thm (basic properties of fields) : Let F be a field, and let a , b , c denote elements of F' , then
1. a + b = c + b => a = c "concellation law" 2. b is not equal to 0 & a * b = c * b => a = c;
Proof of 1 : 1. By F4 , there is some b' belongs to F , s.t. b + b' = 0; we have a + b = c + b; => (a + b) + b' = (c + b) + b'; 2. By F2 , a + (b + b') = c + (b + b'); so by the choice of b' , we can know that a + 0 = c + 0; 3. By F3 , a = c. ^_^ Proof of 2 : more or less the same; 3. If 0' belongs to F is "like 0" , then it is 0; If 0' belongs to F satisfies that for every a belongs to F , a + 0' = a , then 0' = 0; 4. If 1' belongs to F is "like 1" , then it is 1; If 1' belongs to F satisfies that for every a belongs to F , a * 1' = a , then 1' = 1;
Proof of 3 : 1. By F3 , 0' = 0' + 0 2. By F1 , 0' + 0 = 0 + 0' 3. By assuption on 0' , 0 + 0' = 0. ^_^
5. If b & b' satisfy a + b = 0 & a + b' = 0 => b = b'; => in any field "-a" makes sense ----- unambigous meaning (-a) : the b s.t. a + b = 0; 6. If a is not equal to 0 & a * b = 1 = a * b' => b = b'; => if a is not equal to 0 , "-a" makes sense. Proof of 5 : 1. a + b = 0 = a + b'; 2. By F1 , b + a = b'+ a; 3. By concellation , b = b'; ^_^
7. -(-a) = a and when a is not equal to 0 , ( a to the power (-1) ) to the power (-1) = a; Proof of 7 : 1. By definition , a + (-a) = 0; (*) 2. By definition , (-a) + (-(-a) = 0; 3. By (*) and F1 , (-a) + a = 0; 4. By property 5 , -(-a) & a must be the same. ^_^