14-240/Tutorial-Sep30: Difference between revisions
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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. |
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(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>. |
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(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: |
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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. |
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:(b) The result is neither proved in class nor in a previous homework. Then |
:(b) The result is neither proved in class nor in a previous homework. Then reference it in the textbook or prove it yourself. |
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==Nikita== |
==Nikita== |
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Latest revision as of 23:13, 4 October 2014
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Boris
Problem
Find a set [math]\displaystyle{ S }[/math] of two elements that satisfies the following:
(1) [math]\displaystyle{ S }[/math] satisfies all the properties of the field except distributivity.
(2) [math]\displaystyle{ \exists x \in S, 0x \neq 0 }[/math].
Solution:
Let [math]\displaystyle{ S = \{ a, b \} }[/math] where [math]\displaystyle{ a }[/math] is the additive identity and [math]\displaystyle{ b }[/math] is the multiplicative identity and [math]\displaystyle{ a \neq b }[/math]. After trial and error, we have the following addition and multiplication tables:
| [math]\displaystyle{ + }[/math] | [math]\displaystyle{ a }[/math] | [math]\displaystyle{ b }[/math] |
|---|---|---|
| [math]\displaystyle{ a }[/math] | [math]\displaystyle{ a }[/math] | [math]\displaystyle{ b }[/math] |
| [math]\displaystyle{ b }[/math] | [math]\displaystyle{ b }[/math] | [math]\displaystyle{ a }[/math] |
| [math]\displaystyle{ \times }[/math] | [math]\displaystyle{ b }[/math] | [math]\displaystyle{ a }[/math] |
|---|---|---|
| [math]\displaystyle{ b }[/math] | [math]\displaystyle{ b }[/math] | [math]\displaystyle{ a }[/math] |
| [math]\displaystyle{ a }[/math] | [math]\displaystyle{ a }[/math] | [math]\displaystyle{ b }[/math] |
We verify that [math]\displaystyle{ S }[/math] satisfies (1). By the addition and multiplication tables, then [math]\displaystyle{ S }[/math] satisfies closure, commutativity, associativity and existence of identities and inverses. Since [math]\displaystyle{ a(b + b) = a(a) = b \neq a = a + a = ab + ab }[/math], then [math]\displaystyle{ S }[/math] does not satisfy distributivity. Then [math]\displaystyle{ S }[/math] satisfies (1).
We verify that [math]\displaystyle{ S }[/math] satisfies (2). Since [math]\displaystyle{ aa = b \neq a }[/math], then [math]\displaystyle{ S }[/math] satisfies (2).
Elementary Errors in Homework
(1) Prove [math]\displaystyle{ A \implies B }[/math]. Assume [math]\displaystyle{ A }[/math] and derive [math]\displaystyle{ B }[/math]. It is not the other way around.
(2) Prove [math]\displaystyle{ A \iff B }[/math]. Show that [math]\displaystyle{ A \implies B }[/math] and [math]\displaystyle{ 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 reference it in the textbook or prove it yourself.
