14-240/Classnotes for Monday September 8: Difference between revisions
No edit summary |
(The first half of Wednesday's Lesson) |
||
Line 42: | Line 42: | ||
R3: a+0=a & a*1=a |
R3: a+0=a & a*1=a |
||
Wednesday September 10th 2014 - Fields |
|||
The real numbers: A set |R with +,x : |R x |R -> |R & 0=/=1 are elements of |R |
|||
such that |
|||
R1: For every a, b that are elements of |R , a + b = b + a |
|||
& ab = ba |
|||
R2: For every a, b, c that are elements of |R, ( a + b ) + c = a + ( b + c ) |
|||
& (ab)c = a(bc) |
|||
R3: For every a that is an element of |R, a + 0 = a |
|||
& a * 1 = a |
|||
R4: For every a that is an element of |R there exists b that is an element of |R such that a + b = 0 |
|||
& for every a that is an element of |R and a =/= 0 there exists b that is an element |R such that a * b = 1 |
|||
R5: For every a, b, c that are elements of |R, ( a + b ) c = ac + bc |
|||
( a + b ) * ( a - b ) = a^2 - b^2 follows from R1-R5 |
|||
The following is true for the Real Numbers but does not follow from R1-R5 |
|||
For every a that is an element of |R there exists an x that is an element of |R such that a = x^2 or a + x^2 = 0 |
|||
However we can see that it does not follow from R1-R5 because we can find a field that obeys R1-R5 yet does not follow the above rule. |
|||
An example of this is the Rational Numbers |Q. In |Q take a = 2 and there does not exist x such that 2 = x^2 or 2 + x^2 = 0 |
|||
The Definition Of A Field: |
|||
A "Field" is a set F along with a pair of binary operations +,x : FxF -> F and along with a pair 0, 1 that are elements of F such that 0 =/= 1 & such that R1-R5 hold. |
|||
R1: For every a, b that are elements of F , a + b = b + a |
|||
& ab = ba |
|||
R2: For every a, b, c that are elements of F, ( a + b ) + c = a + ( b + c ) |
|||
& (ab)c = a(bc) |
|||
R3: For every a that is an element of F, a + 0 = a |
|||
& a * 1 = a |
|||
R4: For every a that is an element of F there exists b that is an element of F such that a + b = 0 |
|||
& for every a that is an element of F and a =/= 0 there exists b that is an element F such that a * b = 1 |
|||
R5: For every a, b, c that are elements of F, ( a + b ) c = ac + bc |
|||
Example |
|||
1. |R is a field (real numbers) |
|||
2. |Q is a field (rational numbers) |
|||
3. |C is a field (complex numbers) |
|||
4. F = {0, 1} |
|||
*insert table of addition and multiplication* |
|||
Proposition: F is a Field |
|||
checking F5 |
|||
a , b , c | ( a + b ) c | ab + bc | good? |
|||
0 , 0 , 0 | 0 | 0 | yes |
|||
etc... |
|||
F = {0 , 1} = F2 = Z/2 |
|||
Do the same for F7 |
|||
*insert table of addition and multiplication* |
|||
"Like remainders when you divide by 7" |
|||
"like remainders mod 7' |
|||
Theorem (that shall remain unproved) : |
|||
For every prime number P, FP = {0 , 1 , 2 , ... , p-1 } |
|||
along with + & x defined as above |
|||
( a , b ) -> a + b mod p |
|||
is a field. |
|||
---- |
|||
Theorem: (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 |
|||
"Cancellation" still holds |
|||
2. b =/= 0 , ab = cb -> a = c |
|||
3. If 0' is an element of F and satisfies for every a , a + 0' = a , then 0' = 0 |
|||
4. If 1' is "like 1" then 1' = 1 |
|||
... to be continued... |
Revision as of 11:19, 11 September 2014
|
We went over "What is this class about?" (PDF, HTML), then over "About This Class", and then over the first few properties of real numbers that we will care about.
Dror's notes above / Students' notes below |
The real numbers a set R
with 2 binary operations +, *
+:R*R→R
*:R*R→R
in addition 2 special element 0,1∈ R s.t. 0≠1 & furthermore:
R1: The commutative law (for both addition and multiplication)
For every a,b∈R, we have a+b=b+a & ab=ba
R2: The associative law
For every a,b,c∈R, we have (a+b)+c=a+(b+c) (ab)c=a(bc)
in our lives pretty little girls (PL)G≠P(LG)
R3: a+0=a & a*1=a
Wednesday September 10th 2014 - Fields
The real numbers: A set |R with +,x : |R x |R -> |R & 0=/=1 are elements of |R such that
R1: For every a, b that are elements of |R , a + b = b + a & ab = ba
R2: For every a, b, c that are elements of |R, ( a + b ) + c = a + ( b + c ) & (ab)c = a(bc)
R3: For every a that is an element of |R, a + 0 = a & a * 1 = a
R4: For every a that is an element of |R there exists b that is an element of |R such that a + b = 0 & for every a that is an element of |R and a =/= 0 there exists b that is an element |R such that a * b = 1
R5: For every a, b, c that are elements of |R, ( a + b ) c = ac + bc
( a + b ) * ( a - b ) = a^2 - b^2 follows from R1-R5
The following is true for the Real Numbers but does not follow from R1-R5 For every a that is an element of |R there exists an x that is an element of |R such that a = x^2 or a + x^2 = 0
However we can see that it does not follow from R1-R5 because we can find a field that obeys R1-R5 yet does not follow the above rule. An example of this is the Rational Numbers |Q. In |Q take a = 2 and there does not exist x such that 2 = x^2 or 2 + x^2 = 0
The Definition Of A Field: A "Field" is a set F along with a pair of binary operations +,x : FxF -> F and along with a pair 0, 1 that are elements of F such that 0 =/= 1 & such that R1-R5 hold.
R1: For every a, b that are elements of F , a + b = b + a & ab = ba
R2: For every a, b, c that are elements of F, ( a + b ) + c = a + ( b + c ) & (ab)c = a(bc)
R3: For every a that is an element of F, a + 0 = a & a * 1 = a
R4: For every a that is an element of F there exists b that is an element of F such that a + b = 0 & for every a that is an element of F and a =/= 0 there exists b that is an element F such that a * b = 1
R5: For every a, b, c that are elements of F, ( a + b ) c = ac + bc
Example
1. |R is a field (real numbers) 2. |Q is a field (rational numbers) 3. |C is a field (complex numbers) 4. F = {0, 1}
- insert table of addition and multiplication*
Proposition: F is a Field checking F5
a , b , c | ( a + b ) c | ab + bc | good? 0 , 0 , 0 | 0 | 0 | yes
etc...
F = {0 , 1} = F2 = Z/2
Do the same for F7
- insert table of addition and multiplication*
"Like remainders when you divide by 7" "like remainders mod 7'
Theorem (that shall remain unproved) : For every prime number P, FP = {0 , 1 , 2 , ... , p-1 } along with + & x defined as above ( a , b ) -> a + b mod p is a field.
Theorem: (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 "Cancellation" still holds 2. b =/= 0 , ab = cb -> a = c 3. If 0' is an element of F and satisfies for every a , a + 0' = a , then 0' = 0 4. If 1' is "like 1" then 1' = 1
... to be continued...