14-240/Tutorial-Sep30 - Revision history

Bug: /* Elementary Errors in Homework */

2014-10-05T04:13:43Z

Elementary Errors in Homework

← Older revision		Revision as of 00:13, 5 October 2014
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	:(a) The result is already proved in class or in a previous homework. Then state the result and use it without proof.		:(a) The result is already proved in class or in a previous homework. Then state the result and use it without proof.

	:(b) The result is neither proved in class nor in a previous homework. Then reference ~~the textbook if the proof is~~ in the textbook or prove it yourself.		:(b) The result is neither proved in class nor in a previous homework. Then reference it in the textbook or prove it yourself.

	==Nikita==		==Nikita==

Bug: /* Elementary Errors in Homework */

2014-10-05T04:11:58Z

Elementary Errors in Homework

← Older revision		Revision as of 00:11, 5 October 2014
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	:(a) The result is already proved in class or in a previous homework. Then state the result and use it without proof.		:(a) The result is already proved in class or in a previous homework. Then state the result and use it without proof.

	:(b) The result is neither proved in class nor in a previous homework. Then ~~prove~~ it ~~before~~ ~~using~~ ~~its~~ ~~result~~.		:(b) The result is neither proved in class nor in a previous homework. Then reference the textbook if the proof is in the textbook or prove it yourself.

	==Nikita==		==Nikita==

Bug: /* Elementary Errors in Homework */

2014-10-05T04:06:43Z

Elementary Errors in Homework

← Older revision		Revision as of 00:06, 5 October 2014
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	(1) Prove <math>A \implies B</math>. Assume <math>A</math> and derive <math>B</math>. It is not the other way around.		(1) Prove <math>A \implies B</math>. Assume <math>A</math> and derive <math>B</math>. It is not the other way around.


	(2) Prove <math>A \iff B</math>. Show that <math>A \implies B</math> and <math>B \implies A</math>.		(2) Prove <math>A \iff B</math>. Show that <math>A \implies B</math> and <math>B \implies A</math>.


	(3) '''This is for Boris's section only'''. When a proof requires a previous result, there are two possibilities:		(3) '''This is for Boris's section only'''. When a proof requires a previous result, there are two possibilities:

Bug: /* Elementary Errors in Homework */

2014-10-05T04:06:33Z

Elementary Errors in Homework

← Older revision		Revision as of 00:06, 5 October 2014
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	(1) Prove <math>A \implies B</math>. Assume <math>A</math> and derive <math>B</math>. It is not the other way around.		(1) Prove <math>A \implies B</math>. Assume <math>A</math> and derive <math>B</math>. It is not the other way around.


	(2) Prove <math>A \iff B</math>. Show that <math>A \implies B</math> and <math>B \implies A</math>.		(2) Prove <math>A \iff B</math>. Show that <math>A \implies B</math> and <math>B \implies A</math>.


	(3) '''This is for Boris's section only'''. When a proof requires a previous result, there are two possibilities:		(3) '''This is for Boris's section only'''. When a proof requires a previous result, there are two possibilities:

Bug: /* Problem */

2014-10-05T04:06:02Z

Problem

← Older revision		Revision as of 00:06, 5 October 2014
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	We verify that <math>S</math> satisfies (2). Since <math>aa = b \neq a</math>, then <math>S</math> satisfies (2).		We verify that <math>S</math> satisfies (2). Since <math>aa = b \neq a</math>, then <math>S</math> satisfies (2).

			====Elementary Errors in Homework====

			(1) Prove <math>A \implies B</math>. Assume <math>A</math> and derive <math>B</math>. It is not the other way around.

			(2) Prove <math>A \iff B</math>. Show that <math>A \implies B</math> and <math>B \implies A</math>.

			(3) '''This is for Boris's section only'''. When a proof requires a previous result, there are two possibilities:

			:(a) The result is already proved in class or in a previous homework. Then state the result and use it without proof.

			:(b) The result is neither proved in class nor in a previous homework. Then prove it before using its result.

	==Nikita==		==Nikita==

Bug: /* Problem */

2014-10-05T03:30:00Z

Problem

← Older revision		Revision as of 23:30, 4 October 2014
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	We verify that <math>S</math> satisfies (2). Since <math>aa = b \neq = a</math>, then <math>S</math> satisfies (2).		We verify that <math>S</math> satisfies (2). Since <math>aa = b \neq a</math>, then <math>S</math> satisfies (2).

	==Nikita==		==Nikita==

Bug: /* Problem */

2014-10-05T03:29:12Z

Problem

← Older revision		Revision as of 23:29, 4 October 2014
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	We verify that <math>S</math> satisfies (1). By the addition and multiplication ~~tables, <math>S</math>. By the addition and multplication~~ tables, then <math>S</math> satisfies closure, commutativity, associativity and existence of identities and inverses. Since <math>a(b + b) = a(a) = b \neq a = a + a = ab + ab</math>, then <math>S</math> does not satisfy distributivity. Then <math>S</math> satisfies (1).		We verify that <math>S</math> satisfies (1). By the addition and multiplication tables, then <math>S</math> satisfies closure, commutativity, associativity and existence of identities and inverses. Since <math>a(b + b) = a(a) = b \neq a = a + a = ab + ab</math>, then <math>S</math> does not satisfy distributivity. Then <math>S</math> satisfies (1).

Bug: /* Problem */

2014-10-05T03:28:34Z

Problem

← Older revision		Revision as of 23:28, 4 October 2014
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	\| <math>b</math>		\| <math>b</math>
	\|}		\|}


	We verify that <math>S</math> satisfies (1). By the addition and multiplication tables, <math>S</math>. By the addition and multplication tables, then <math>S</math> satisfies closure, commutativity, associativity and existence of identities and inverses. Since <math>a(b + b) = a(a) = b \neq a = a + a = ab + ab</math>, then <math>S</math> does not satisfy distributivity. Then <math>S</math> satisfies (1).		We verify that <math>S</math> satisfies (1). By the addition and multiplication tables, <math>S</math>. By the addition and multplication tables, then <math>S</math> satisfies closure, commutativity, associativity and existence of identities and inverses. Since <math>a(b + b) = a(a) = b \neq a = a + a = ab + ab</math>, then <math>S</math> does not satisfy distributivity. Then <math>S</math> satisfies (1).


	We verify that <math>S</math> satisfies (2). Since <math>aa = b \neq = a</math>, then <math>S</math> satisfies (2).		We verify that <math>S</math> satisfies (2). Since <math>aa = b \neq = a</math>, then <math>S</math> satisfies (2).

Bug: /* Problem */

2014-10-05T03:28:02Z

Problem

← Older revision		Revision as of 23:28, 4 October 2014
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	\|}		\|}

	We verify that <math>S</math> satisfies (1). By the addition and multiplication tables, <math>S</math> is ~~closed~~ ~~under~~ addition and ~~scalar~~ ~~multiplication. Since <math>a + b = b + a = b</math> and <math>ab = ba = a</math>~~, then <math>S</math> is ~~commutative.~~ It ~~can~~ ~~also~~ be ~~shown that <math>S</math> is associative. Observe that <math>a</math> is the additive~~ and ~~<math>b</math> is the multiplicative identity~~. Since <math>a(b + b) = a(a) = b \neq a = a + a = ab + ab</math>, then ~~<math>S</math> is not distributive. Then~~ <math>S</math> does not satisfy distributivity. Then <math>S</math> satisfies (1).		We verify that <math>S</math> satisfies (1). By the addition and multiplication tables, <math>S</math>. By the addition and multplication tables, then <math>S</math> satisfies closure, commutativity, associativity and existence of identities and inverses. Since <math>a(b + b) = a(a) = b \neq a = a + a = ab + ab</math>, then <math>S</math> does not satisfy distributivity. Then <math>S</math> satisfies (1).

	We verify that <math>S</math> satisfies (2). Since <math>aa = b \neq = a</math>, then <math>S</math> satisfies (2).		We verify that <math>S</math> satisfies (2). Since <math>aa = b \neq = a</math>, then <math>S</math> satisfies (2).

Bug: /* Problem */

2014-10-05T03:18:26Z

Problem

← Older revision		Revision as of 23:18, 4 October 2014
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	\|}		\|}

	We verify that <math>S</math> satisfies (1~~) and (2~~). By the addition and multiplication tables, <math>S</math> is closed under addition and scalar multiplication. Since <math>a + b = b + a = b</math> and <math>ab = ba = a</math>, then <math>S</math> is commutative. By ~~extension~~, <math>S</math> is ~~also~~ ~~associative~~.		We verify that <math>S</math> satisfies (1). By the addition and multiplication tables, <math>S</math> is closed under addition and scalar multiplication. Since <math>a + b = b + a = b</math> and <math>ab = ba = a</math>, then <math>S</math> is commutative. It can also be shown that <math>S</math> is associative. Observe that <math>a</math> is the additive and <math>b</math> is the multiplicative identity. Since <math>a(b + b) = a(a) = b \neq a = a + a = ab + ab</math>, then <math>S</math> is not distributive. Then <math>S</math> does not satisfy distributivity. Then <math>S</math> satisfies (1).

			We verify that <math>S</math> satisfies (2). Since <math>aa = b \neq = a</math>, then <math>S</math> satisfies (2).

	==Nikita==		==Nikita==