14-240/Classnotes for Monday September 8 - Revision history

Boyang.wu: /* The Commutative Law */

2014-12-08T04:48:07Z

The Commutative Law

← Older revision		Revision as of 00:48, 8 December 2014
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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:

	====The Commutative Law====		[[File:2.jpg]]====The Commutative Law====

	R1: For every <math>a, b \in \R</math>, we have:		R1: For every <math>a, b \in \R</math>, we have:

Boyang.wu: /* The Real Numbers */

2014-12-08T04:46:50Z

The Real Numbers

← Older revision		Revision as of 00:46, 8 December 2014
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	===The Real Numbers===		[[File:1.jpg]]===The Real Numbers===

	The real numbers are a set <math>\R</math> with 2 binary operations + and *, defined as follows:		The real numbers are a set <math>\R</math> with 2 binary operations + and *, defined as follows:

Hameeral: Cleaned up and typeset Monday's notes.

2014-09-14T16:02:49Z

Cleaned up and typeset Monday's notes.

← Older revision		Revision as of 12:02, 14 September 2014
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			===The Real Numbers===

			The real numbers are a set <math>\R</math> with 2 binary operations + and *, defined as follows:

			<math>+: \R \times \R \rightarrow \R</math>

			<math>*: \R \times \R \rightarrow \R</math>

			in addition to 2 special elements <math>0, 1 \in \R</math> such that <math>0 \ne 1</math>, with the following properties:

			====The Commutative Law====

		⚫	R1: For every <math>a, b \in \R</math>, we have:
			<math>a + b = b + a</math> (commutative law for addition)
		⚫	<math>ab = ba</math> (commutative law for multiplication)

			====The Associative Law====
	The real numbers a set '''R'''
	with 2 binary operations +, *

		⚫	R2: For every <math>a, b, c \in \R</math>, we have:
	+:'''R'''*'''R'''→'''R
	'''
	:'''R''''''R'''→'''R
	'''
	in addition 2 special element 0,1∈ '''R'''
	s.t. 0≠1 & furthermore:

			<math>(a + b) + c = a + (b + c)</math>
⚫	~~R1:~~ ~~The~~ commutative law (for ~~both addition and~~ multiplication)
		⚫	<math>(ab)c = a(bc)</math>

			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)".
⚫	For every a,~~b∈'''~~R~~'''~~, we have
			<math>(PL)G \ne P(LG)</math>
	a+b=b+a & ab=ba

			So the associative property does not hold for the English language.

			====Existence of Units====

			R3: For every <math>a \in \R</math>:
	R2: The associative law

			<math>a + 0 = a</math> (additive unit)
⚫	For every a,b~~,c∈'''~~R~~'''~~, we have
			<math>a * 1 = a</math> (multiplicative unit)
	(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 ==		== Wednesday September 10th 2014 - Fields ==

R.mcclure at 19:32, 11 September 2014

2014-09-11T19:32:05Z

← Older revision		Revision as of 15:32, 11 September 2014
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	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 & <math>0=/=1</math> are elements of \|R
⚫	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		such that

	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>
	& ab = ba

	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>
	& (ab)c = a(bc)

	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>
	& 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		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>
	& 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>

	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>

	( 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

	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
	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>

	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.
	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>

	The Definition Of A Field:		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.		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.

	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>
	& ab = ba

	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>
	& (ab)c = a(bc)

	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>
	& 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		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>
	& 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>

	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>

	Example		Example
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	Proposition: F is a Field		Proposition: F is a Field
	checking F5		checking F5

	a , b , c \| ( a + b ) c \| ab + bc \| good?
	0 , 0 , 0 \| 0 \| 0 \| yes

	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 }
	along with + & x defined as above		along with + & x defined as above
	( a , b ) -> a + b mod p		<math>( a , b ) -> a + b mod p</math>
	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
	Then:		Then:
	1. a + b = c + b -> a = c		1. <math>a + b = c + b -> a = c </math>
	"Cancellation" still holds		"Cancellation" still holds
	2. b =/= 0 , ab = cb -> a = c		2. <math>b =/= 0 , ab = cb -> a = c</math>
	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>
	4. If 1' is "like 1" then 1' = 1		4. If <math>1'</math> is "like 1" then <math>1' = 1</math>

	... to be continued...		... to be continued...

R.mcclure: The first half of Wednesday's Lesson

2014-09-11T16:19:28Z

The first half of Wednesday's Lesson

← Older revision		Revision as of 12:19, 11 September 2014
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	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...

Boyang.wu at 19:31, 9 September 2014

2014-09-09T19:31:45Z

← Older revision		Revision as of 15:31, 9 September 2014
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	For every a,b∈'''R''', we have		For every a,b∈'''R''', we have
	a+b=b+a & ab=ba		a+b=b+a & ab=ba



	R2: The associative law		R2: The associative law

Boyang.wu at 19:31, 9 September 2014

2014-09-09T19:31:15Z

← Older revision		Revision as of 15:31, 9 September 2014
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	{{14-240:Dror/Students Divider}}		{{14-240:Dror/Students Divider}}










			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

Drorbn: Created page with "{{14-240/Navigation}} We went over "What is this class about?" ({{Pensieve link|Classes/14-240/one/What_is_This_Class_AboutQ.pdf|PDF}}, {{Pensieve link|Classes/14-240/What_is_..."

2014-09-08T14:44:00Z

Created page with "{{14-240/Navigation}} We went over "What is this class about?" ({{Pensieve link|Classes/14-240/one/What_is_This_Class_AboutQ.pdf|PDF}}, {{Pensieve link|Classes/14-240/What_is_..."

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{{14-240/Navigation}}
We went over "What is this class about?" ({{Pensieve link|Classes/14-240/one/What_is_This_Class_AboutQ.pdf|PDF}}, {{Pensieve link|Classes/14-240/What_is_This_Class_AboutQ.html|HTML}}), then over "[[14-240/About This Class|About This Class]]", and then over the first few properties of real numbers that we will care about.

{{14-240:Dror/Students Divider}}