14-240/Classnotes for Monday September 8
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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...
