14-240/Classnotes for Monday September 8: Difference between revisions
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in addition to 2 special elements <math>0, 1 \in \R</math> such that <math>0 \ne 1</math>, with the following properties: |
in addition to 2 special elements <math>0, 1 \in \R</math> such that <math>0 \ne 1</math>, with the following properties: |
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====The Commutative Law==== |
[[File:2.jpg]]====The Commutative Law==== |
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R1: For every <math>a, b \in \R</math>, we have: |
R1: For every <math>a, b \in \R</math>, we have: |
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Latest revision as of 23:48, 7 December 2014
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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 are a set with 2 binary operations + and *, defined as follows:
in addition to 2 special elements such that , with the following properties:
R1: For every , we have:
(commutative law for addition) (commutative law for multiplication)
The Associative Law
R2: For every , we have:
This is not true for a number of other sets in our lives! For example, the associative law does not hold for the English language. Consider the phrase "pretty little girls": "(pretty little) girls" does not mean the same thing as "pretty (little girls)".
So the associative property does not hold for the English language.
Existence of Units
R3: For every :
(additive unit) (multiplicative unit)
Wednesday September 10th 2014 - Fields
The real numbers: A set |R with +,x : |R x |R -> |R & are elements of |R such that
R1: For every that are elements of |R , and
R2: For every that are elements of |R, and
R3: For every that is an element of |R, and
R4: For every a that is an element of |R there exists b that is an element of |R such that & for every a that is an element of |R and there exists b that is an element |R such that
R5: For every that are elements of |R,
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 that is an element of |R such that
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 and there does not exist such that or
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 that are elements of F such that and such that R1-R5 hold.
R1: For every that are elements of F , and
R2: For every that are elements of F, and
R3: For every that is an element of F, and
R4: For every that is an element of F there exists that is an element of F such that & for every that is an element of F and there exists that is an element F such that
R5: For every that are elements of F,
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 is a field.
Theorem: (basic properties of Fields)
Let F be a Field, and let a , b , c denote elements of F Then: 1. "Cancellation" still holds 2. 3. If is an element of F and satisfies for every , then 4. If is "like 1" then
... to be continued...

