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

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 $$\vert{a-b}\vert$$?

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

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 $$a\neq b$$.

• This is essentially my answer ... May 28, 2021 at 15:18
• @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.
– Joe
May 28, 2021 at 15:22
• I would rather know the contra positive of general statements than for a single statemnet May 28, 2021 at 15:24
• @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.
– Joe
May 28, 2021 at 15:27
• But this answer is still saying things a little better (+1) 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 $$|a-b|$$