# Discrete Math Logic Homework

Consider the following statement: $$\forall \epsilon > 0,\space\exists\delta>0:(|x-a|\lt\delta\implies|f(x)-L|\lt\epsilon).$$ (a) Write the converse of the statement.
(b) Write the contrapositive of the statement.

I am stuck on how to complete these problems because I do not understand the notation

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Have you read you course notes ? – Siméon Dec 6 '12 at 18:08
Yes, I still just do not understand where to begin. – CHZ Dec 6 '12 at 18:09
Do you understand the notation $$\forall x,\exists y:(P\implies Q)\;?$$ That’s all the notation that you need to understand in order to do the problem. – Brian M. Scott Dec 6 '12 at 18:25
So for the converse do I want to say: ∀p,∃q: (X>Y)? And I know that the contrapositive is just the negation of both sides. So, For all x there does not exist a Y where P yields Q? or is it: For all x there does not exist a Y where P does not yield Q? – CHZ Dec 6 '12 at 18:29

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 $(\lnot B) \Rightarrow (\lnot A)$, where $\lnot A$ is the negation of $A$.
$$\forall \epsilon > 0,\space\exists\delta>0:(|x-a|\lt\delta\implies|f(x)-L|\lt\epsilon).$$
$$\forall \epsilon > 0,\space\exists\delta>0:(|x-a|\lt\delta\rightarrow|f(x)-L|\lt\epsilon)$$ $$\iff \forall \epsilon > 0,\space\exists\delta>0:[\lnot(|f(x)-L|\lt\epsilon) \rightarrow \lnot(|x-a|\lt\delta)]$$ $$\iff \forall \epsilon >0, \; \exists \delta >0 :[\lnot\lnot(|f(x)-L|\lt\epsilon) \lor \lnot(|x-a|\lt\delta)]$$ $$\iff \forall \epsilon > 0, \; \exists \delta > 0:\lnot[\lnot (|f(x)-L|\lt\epsilon) \land (|x-a|\lt\delta)]$$ $$\iff \lnot \exists \epsilon > 0,\;\forall\delta>0: [(|f(x) - L| \geq \epsilon) \land (|x - a|\lt\delta)]$$