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 & [math]\displaystyle{ 0=/=1 }[/math] are elements of |R such that
R1: For every [math]\displaystyle{ a, b }[/math] that are elements of |R , [math]\displaystyle{ a + b = b + a }[/math] and [math]\displaystyle{ ab = ba }[/math]
R2: For every [math]\displaystyle{ a, b, c }[/math] that are elements of |R, [math]\displaystyle{ ( a + b ) + c = a + ( b + c ) }[/math] and [math]\displaystyle{ (ab)c = a(bc) }[/math]
R3: For every [math]\displaystyle{ a }[/math] that is an element of |R, [math]\displaystyle{ a + 0 = a }[/math] and [math]\displaystyle{ a * 1 = a }[/math]
R4: For every a that is an element of |R there exists b that is an element of |R such that [math]\displaystyle{ a + b = 0 }[/math] & for every a that is an element of |R and [math]\displaystyle{ a =/= 0 }[/math] there exists b that is an element |R such that [math]\displaystyle{ a * b = 1 }[/math]
R5: For every [math]\displaystyle{ a, b, c }[/math] that are elements of |R, [math]\displaystyle{ ( a + b ) c = ac + bc }[/math]
[math]\displaystyle{ ( a + b ) * ( a - b ) = a^2 - b^2 }[/math] 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 [math]\displaystyle{ x }[/math] that is an element of |R such that [math]\displaystyle{ a = x^2 or a + x^2 = 0 }[/math]
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 [math]\displaystyle{ a = 2 }[/math] and there does not exist [math]\displaystyle{ x }[/math] such that [math]\displaystyle{ 2 = x^2 }[/math] or [math]\displaystyle{ 2 + x^2 = 0 }[/math]
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 [math]\displaystyle{ 0, 1 }[/math] that are elements of F such that [math]\displaystyle{ 0 =/= 1 }[/math] and such that R1-R5 hold.
R1: For every [math]\displaystyle{ a, b }[/math] that are elements of F , [math]\displaystyle{ a + b = b + a }[/math] and [math]\displaystyle{ ab = ba }[/math]
R2: For every [math]\displaystyle{ a, b, c }[/math] that are elements of F, [math]\displaystyle{ ( a + b ) + c = a + ( b + c ) }[/math] and [math]\displaystyle{ (ab)c = a(bc) }[/math]
R3: For every [math]\displaystyle{ a }[/math] that is an element of F, [math]\displaystyle{ a + 0 = a }[/math] and [math]\displaystyle{ a * 1 = a }[/math]
R4: For every [math]\displaystyle{ a }[/math] that is an element of F there exists [math]\displaystyle{ b }[/math] that is an element of F such that [math]\displaystyle{ a + b = 0 }[/math] & for every [math]\displaystyle{ a }[/math] that is an element of F and [math]\displaystyle{ a =/= 0 }[/math] there exists [math]\displaystyle{ b }[/math] that is an element F such that [math]\displaystyle{ a * b = 1 }[/math]
R5: For every [math]\displaystyle{ a, b, c }[/math] that are elements of F, [math]\displaystyle{ ( a + b ) c = ac + bc }[/math]
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
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 [math]\displaystyle{ ( a , b ) -\gt a + b mod p }[/math] is a field.
Theorem: (basic properties of Fields)
Let F be a Field, and let a , b , c denote elements of F Then: 1. [math]\displaystyle{ a + b = c + b -\gt a = c }[/math] "Cancellation" still holds 2. [math]\displaystyle{ b =/= 0 , ab = cb -\gt a = c }[/math] 3. If [math]\displaystyle{ 0' }[/math] is an element of F and satisfies for every [math]\displaystyle{ a , a + 0' = a }[/math] , then [math]\displaystyle{ 0' = 0 }[/math] 4. If [math]\displaystyle{ 1' }[/math] is "like 1" then [math]\displaystyle{ 1' = 1 }[/math]
... to be continued...
