14-240/Classnotes for Monday September 8: Difference between revisions
(The first half of Wednesday's Lesson) |
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[[File:1.jpg]]===The Real Numbers=== |
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The real numbers are a set <math>\R</math> with 2 binary operations + and *, defined as follows: |
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<math>+: \R \times \R \rightarrow \R</math> |
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<math>*: \R \times \R \rightarrow \R</math> |
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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: |
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[[File:2.jpg]]====The Commutative Law==== |
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<math>a + b = b + a</math> (commutative law for addition) |
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====The Associative Law==== |
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The real numbers a set '''R''' |
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with 2 binary operations +, * |
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+:'''R'''*'''R'''→'''R |
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''' |
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*:'''R'''*'''R'''→'''R |
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''' |
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in addition 2 special element 0,1∈ '''R''' |
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s.t. 0≠1 & furthermore: |
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<math>(a + b) + c = a + (b + c)</math> |
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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)". |
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<math>(PL)G \ne P(LG)</math> |
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a+b=b+a & ab=ba |
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So the associative property does not hold for the English language. |
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====Existence of Units==== |
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R3: For every <math>a \in \R</math>: |
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R2: The associative law |
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<math>a + 0 = a</math> (additive unit) |
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| ⚫ | |||
<math>a * 1 = a</math> (multiplicative unit) |
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(a+b)+c=a+(b+c) |
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(ab)c=a(bc) |
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in our lives |
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pretty little girls |
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(PL)G≠P(LG) |
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R3: a+0=a & a*1=a |
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such that |
such that |
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R1: For every a, b that are elements of |R , a + b = b + a |
R1: For every <math>a, b</math> that are elements of |R , <math>a + b = b + a</math> and <math>ab = ba </math> |
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& ab = ba |
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R2: For every a, b, c that are elements of |R, ( a + b ) + c = a + ( b + c ) |
R2: For every <math>a, b, c</math> that are elements of |R, <math>( a + b ) + c = a + ( b + c )</math> and <math> (ab)c = a(bc)</math> |
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| ⚫ | |||
R3: For every a that is an element of |R, a + 0 = a |
R3: For every <math>a</math> that is an element of |R, <math>a + 0 = a</math> and <math> a * 1 = a</math> |
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& a * 1 = a |
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R4: For every a that is an element of |R there exists b that is an element of |R such that a + b = 0 |
R4: For every a that is an element of |R there exists b that is an element of |R such that <math>a + b = 0</math> |
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& 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 |
& for every a that is an element of |R and <math>a =/= 0</math> there exists b that is an element |R such that <math>a * b = 1</math> |
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R5: For every a, b, c that are elements of |R, ( a + b ) c = ac + bc |
R5: For every <math>a, b, c</math> that are elements of |R, <math>( a + b ) c = ac + bc</math> |
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( a + b ) * ( a - b ) = a^2 - b^2 follows from R1-R5 |
<math>( a + b ) * ( a - b ) = a^2 - b^2</math> follows from R1-R5 |
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The following is true for the Real Numbers but does not follow from R1-R5 |
The following is true for the Real Numbers but does not follow from R1-R5 |
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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 |
For every a that is an element of |R there exists an <math>x</math> that is an element of |R such that <math>a = x^2 or a + x^2 = 0</math> |
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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. |
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. |
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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 |
An example of this is the Rational Numbers |Q. In |Q take <math>a = 2</math> and there does not exist <math>x</math> such that <math>2 = x^2</math> or <math>2 + x^2 = 0</math> |
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The Definition Of A Field: |
The Definition Of A Field: |
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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 |
A "Field" is a set F along with a pair of binary operations +,x : FxF -> F and along with a pair <math>0, 1</math> that are elements of F such that <math>0 =/= 1</math> and such that R1-R5 hold. |
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R1: For every a, b that are elements of F , a + b = b + a |
R1: For every <math>a, b</math> that are elements of F , <math>a + b = b + a</math> and <math> ab = ba</math> |
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& ab = ba |
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R2: For every a, b, c that are elements of F, ( a + b ) + c = a + ( b + c ) |
R2: For every <math>a, b, c</math> that are elements of F, <math>( a + b ) + c = a + ( b + c )</math> and <math>(ab)c = a(bc)</math> |
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& (ab)c = a(bc) |
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R3: For every a that is an element of F, a + 0 = a |
R3: For every <math>a</math> that is an element of F, <math>a + 0 = a</math> and <math>a * 1 = a</math> |
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& a * 1 = a |
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R4: For every a that is an element of F there exists b that is an element of F such that a + b = 0 |
R4: For every <math>a</math> that is an element of F there exists <math>b</math> that is an element of F such that <math>a + b = 0</math> |
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& 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 |
& for every <math>a</math> that is an element of F and <math>a =/= 0</math> there exists <math>b</math> that is an element F such that <math>a * b = 1</math> |
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R5: For every a, b, c that are elements of F, ( a + b ) c = ac + bc |
R5: For every <math>a, b, c</math> that are elements of F, <math>( a + b ) c = ac + bc</math> |
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Example |
Example |
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Proposition: F is a Field |
Proposition: F is a Field |
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checking F5 |
checking F5 |
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a , b , c | ( a + b ) c | ab + bc | good? |
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0 , 0 , 0 | 0 | 0 | yes |
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etc... |
etc... |
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For every prime number P, FP = {0 , 1 , 2 , ... , p-1 } |
For every prime number P, FP = {0 , 1 , 2 , ... , p-1 } |
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along with + & x defined as above |
along with + & x defined as above |
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( a , b ) -> a + b mod p |
<math>( a , b ) -> a + b mod p</math> |
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is a field. |
is a field. |
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Let F be a Field, and let a , b , c denote elements of F |
Let F be a Field, and let a , b , c denote elements of F |
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Then: |
Then: |
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1. a + b = c + b -> a = c |
1. <math>a + b = c + b -> a = c </math> |
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"Cancellation" still holds |
"Cancellation" still holds |
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2. b =/= 0 , ab = cb -> a = c |
2. <math>b =/= 0 , ab = cb -> a = c</math> |
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3. If 0' is an element of F and satisfies for every a , a + 0' = a , then 0' = 0 |
3. If <math>0'</math> is an element of F and satisfies for every <math>a , a + 0' = a</math> , then <math>0' = 0 </math> |
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4. If 1' is "like 1" then 1' = 1 |
4. If <math>1'</math> is "like 1" then <math>1' = 1</math> |
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... to be continued... |
... to be continued... |
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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...

