In Stephen Abbott's book, Understanding Analysis, theorem 1.2.6 is stated as

Two real numbers $a$ and $b$ are equal if and only if for every real number $\epsilon$ it follows that $\vert{a-b}\vert<\epsilon$

For proving $(\Rightarrow)$ we must show that:

if $a=b$, then $\forall{\epsilon>0}:\bigl(\vert{a-b}\vert<\epsilon\bigr)$

which is fairly straightforward. If $a=b$, then $\vert{a-b}\vert=0$ which is smaller than every $\epsilon$.

And for proving $(\Leftarrow)$ we must show that:

if $\forall{\epsilon>0}:\bigl(\vert{a-b}\vert<\epsilon\bigr)$, then $a=b$

which in the book is proved by contradiction.

I am trying to prove $(\Leftarrow)$ through a contrapositive instead but I seem to have confused myself with what the contrapositive statement to $(\Leftarrow)$ should be.

Will it be the negation of the quantifiers and statements, and then switching the if and then around so that instead of:

if $\forall{\epsilon>0}:\bigl(\vert{a-b}\vert<\epsilon\bigr)$, then $a=b$

we get:

if $a\neq{b}$, then $\exists{\epsilon>0}:\bigl(\vert{a-b}\vert\geq\epsilon\bigr)$

and now to prove by contrapositive, we must show that if $a-b\neq0$ then there exists some $\epsilon>0$ that is smaller than or equal to $\vert{a-b}\vert$?

  • $\begingroup$ Correct. In otehr words, if $a \ne B$ we have that the "distance" between them is some not-null quantitiy. $\endgroup$ May 27, 2021 at 14:11
  • 2
    $\begingroup$ A natural choice might be $\epsilon = \frac{|a-b|}{2}$. You might need to show it is positive rather than zero $\endgroup$
    – Henry
    May 27, 2021 at 14:14
  • $\begingroup$ You're right that it's preferable to use contrapositive over contradiction. See here. $\endgroup$
    – Joe
    May 27, 2021 at 14:30
  • $\begingroup$ Does this answer your question? Intuition about : $a = b \iff | a − b| &lt; \epsilon$, for every $\epsilon &gt; 0$ $\endgroup$
    – user1211588
    Sep 22, 2023 at 7:46

2 Answers 2


The statement is

If for every $\DeclareMathOperator{\epsilon}{\varepsilon}\epsilon>0$ we have that $\lvert a-b\rvert<\epsilon$ then $a=b$.

The contrapositive is

If $a \neq b$ then it is not true that for every $\epsilon>0$ we have $\lvert a-b\rvert<\epsilon$.

In other words,

If $a\neq b$ then there exists an $\epsilon>0$ such that $\lvert a-b \rvert\geq\epsilon$.

This is certainly true. Take, for instance, $\epsilon=\dfrac{\lvert a-b\rvert}{2}$; this works for any values of $a$ and $b$ such that $a\neq b$.

  • $\begingroup$ This is essentially my answer ... $\endgroup$ May 28, 2021 at 15:18
  • $\begingroup$ @VivaanDaga: I don't agree with that. For instance, my answer does not explain what the term 'contrapositive' means in general; your answer does not write down what the contrapositive of the specific statement at hand is. $\endgroup$
    – Joe
    May 28, 2021 at 15:22
  • $\begingroup$ I would rather know the contra positive of general statements than for a single statemnet $\endgroup$ May 28, 2021 at 15:24
  • $\begingroup$ @VivaanDaga: Well, I thought that given that OP uses the term 'contrapositive' in his question, he would be familiar with what it means in general. Sometimes it is a little tricky to work out what $\neg P$ means in the specific context. That is what I tried to clear up. $\endgroup$
    – Joe
    May 28, 2021 at 15:27
  • $\begingroup$ But this answer is still saying things a little better (+1) $\endgroup$ May 28, 2021 at 15:28

You are correct if P implies Q then the contra postive stamemnet is that the negation of Q implies the negation of P you can prove the stammer by contra postive by noting the fact that $|a-b|>0$ so you can let $\epsilon$ be $|a-b|$


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