Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 $A$.

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.

share|cite|improve this answer

$$\forall \epsilon > 0,\space\exists\delta>0:(|x-a|\lt\delta\implies|f(x)-L|\lt\epsilon).$$

Perhaps what you are asked to do is to replace the quantified (internal) implication with it's contrapositive:
$$\forall \epsilon > 0,\space\exists\delta>0:(|x-a|\lt\delta\rightarrow|f(x)-L|\lt\epsilon)$$ $$\iff \forall \epsilon > 0,\space\exists\delta>0:[\lnot(|f(x)-L|\lt\epsilon) \rightarrow \lnot(|x-a|\lt\delta)]$$ $$\iff \forall \epsilon >0, \; \exists \delta >0 :[\lnot\lnot(|f(x)-L|\lt\epsilon) \lor \lnot(|x-a|\lt\delta)]$$ $$\iff \forall \epsilon > 0, \; \exists \delta > 0:\lnot[\lnot (|f(x)-L|\lt\epsilon) \land (|x-a|\lt\delta)]$$ $$\iff \lnot \exists \epsilon > 0,\;\forall\delta>0: [(|f(x) - L| \geq \epsilon) \land (|x - a|\lt\delta)]$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.