14-240/Tutorial-Sep30: Difference between revisions
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==Boris== |
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====Problem==== |
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Find a set <math>S</math> of two elements that satisfies the following: |
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(1) <math>S</math> satisfies all the properties of the field except distributivity. |
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(2) <math>\exists x \in S, 0x \neq 0</math>. |
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Solution: |
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Let <math>S = \{ a, b \} </math> where <math>a</math> is the additive identity and <math>b</math> is the multiplicative identity and <math>a \neq b</math>. After trial and error, we have the following addition and multiplication tables: |
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{| class="wikitable" |
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|- |
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! <math>+</math> |
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! <math>a</math> |
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! <math>b</math> |
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|- |
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! <math>a</math> |
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| <math>a</math> |
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| <math>b</math> |
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|- |
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! <math>b</math> |
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| <math>b</math> |
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| <math>a</math> |
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|} |
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{| class="wikitable" |
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|- |
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! <math>\times</math> |
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! <math>b</math> |
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! <math>a</math> |
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|- |
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! <math>b</math> |
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| <math>b</math> |
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| <math>a</math> |
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|- |
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! <math>a</math> |
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| <math>a</math> |
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| <math>b</math> |
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|} |
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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). |
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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). |
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====Elementary Errors in Homework==== |
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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. |
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(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: |
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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. |
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:(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== |
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Latest revision as of 23:13, 4 October 2014
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Boris
Problem
Find a set of two elements that satisfies the following:
(1) satisfies all the properties of the field except distributivity.
(2) .
Solution:
Let where is the additive identity and is the multiplicative identity and . After trial and error, we have the following addition and multiplication tables:
We verify that satisfies (1). By the addition and multiplication tables, then satisfies closure, commutativity, associativity and existence of identities and inverses. Since , then does not satisfy distributivity. Then satisfies (1).
We verify that satisfies (2). Since , then satisfies (2).
Elementary Errors in Homework
(1) Prove . Assume and derive . It is not the other way around.
(2) Prove . Show that and .
(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.
