posader86

Answered

2022-07-15

Discrete Math Logic Homework

Consider the following statement:

$\mathrm{\forall}\u03f5>0,\text{}\mathrm{\exists}\delta 0:(|x-a|\delta \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f(x)-L|\u03f5).$

(a) Write the converse of the statement.

(b) Write the contrapositive of the statement.

Consider the following statement:

$\mathrm{\forall}\u03f5>0,\text{}\mathrm{\exists}\delta 0:(|x-a|\delta \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f(x)-L|\u03f5).$

(a) Write the converse of the statement.

(b) Write the contrapositive of the statement.

Answer & Explanation

Makenna Lin

Expert

2022-07-16Added 16 answers

Step 1

The usual definitions of "converse" and "contrapositive" used in logic only apply to implications, which are statements of the form $A\Rightarrow B$. The converse of $A\Rightarrow B$ is $B\Rightarrow A$, and the contrapositive is $(\mathrm{\neg}B)\Rightarrow (\mathrm{\neg}A)$, where $\mathrm{\neg}A$ is the negation of A.

Step 2

Because the statement you wrote has two quantifiers at the front, it is not an implication, and the usual definitions do not apply to it. Therefore, you should ask your instructor, or consult your notes, to learn what the instructor wants you to do. It will be difficult to find much help in the usual textbooks or reference sources because these do not give any definition for the "converse" or "contrapositive" of statements that are not implications.

The usual definitions of "converse" and "contrapositive" used in logic only apply to implications, which are statements of the form $A\Rightarrow B$. The converse of $A\Rightarrow B$ is $B\Rightarrow A$, and the contrapositive is $(\mathrm{\neg}B)\Rightarrow (\mathrm{\neg}A)$, where $\mathrm{\neg}A$ is the negation of A.

Step 2

Because the statement you wrote has two quantifiers at the front, it is not an implication, and the usual definitions do not apply to it. Therefore, you should ask your instructor, or consult your notes, to learn what the instructor wants you to do. It will be difficult to find much help in the usual textbooks or reference sources because these do not give any definition for the "converse" or "contrapositive" of statements that are not implications.

Kyle Liu

Expert

2022-07-17Added 4 answers

Step 1

$\mathrm{\forall}\u03f5>0,\text{}\mathrm{\exists}\delta 0:(|x-a|\delta \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f(x)-L|\u03f5).$

Step 2

Perhaps what you are asked to do is to replace the quantified (internal) implication with it's contrapositive:

$\mathrm{\forall}\u03f5>0,\text{}\mathrm{\exists}\delta 0:(|x-a|\delta \to |f(x)-L|\u03f5)$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}\u03f5>0,\text{}\mathrm{\exists}\delta 0:[\mathrm{\neg}(|f(x)-L|\u03f5)\to \mathrm{\neg}(|x-a|\delta )]$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}\u03f5>0,\phantom{\rule{thickmathspace}{0ex}}\mathrm{\exists}\delta >0:[\mathrm{\neg}\mathrm{\neg}(|f(x)-L|<\u03f5)\vee \mathrm{\neg}(|x-a|<\delta )]$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}\u03f5>0,\phantom{\rule{thickmathspace}{0ex}}\mathrm{\exists}\delta >0:\mathrm{\neg}[\mathrm{\neg}(|f(x)-L|<\u03f5)\wedge (|x-a|<\delta )]$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{\neg}\mathrm{\exists}\u03f5>0,\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}\delta >0:[(|f(x)-L|\ge \u03f5)\wedge (|x-a|<\delta )]$

$\mathrm{\forall}\u03f5>0,\text{}\mathrm{\exists}\delta 0:(|x-a|\delta \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f(x)-L|\u03f5).$

Step 2

Perhaps what you are asked to do is to replace the quantified (internal) implication with it's contrapositive:

$\mathrm{\forall}\u03f5>0,\text{}\mathrm{\exists}\delta 0:(|x-a|\delta \to |f(x)-L|\u03f5)$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}\u03f5>0,\text{}\mathrm{\exists}\delta 0:[\mathrm{\neg}(|f(x)-L|\u03f5)\to \mathrm{\neg}(|x-a|\delta )]$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}\u03f5>0,\phantom{\rule{thickmathspace}{0ex}}\mathrm{\exists}\delta >0:[\mathrm{\neg}\mathrm{\neg}(|f(x)-L|<\u03f5)\vee \mathrm{\neg}(|x-a|<\delta )]$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}\u03f5>0,\phantom{\rule{thickmathspace}{0ex}}\mathrm{\exists}\delta >0:\mathrm{\neg}[\mathrm{\neg}(|f(x)-L|<\u03f5)\wedge (|x-a|<\delta )]$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{\neg}\mathrm{\exists}\u03f5>0,\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall}\delta >0:[(|f(x)-L|\ge \u03f5)\wedge (|x-a|<\delta )]$

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