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I'm having a hard time proving this statement that makes intuitive sense:

Prove if $a,b\in\Bbb Q$, and for all $\epsilon$, $| a - b |<\epsilon$, then $a = b$.

I understand that if $a = b$, then $a - b = 0$ and $0 < \epsilon$. But I don't know how to show that.

Thanks everyone!

John Doe
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Joesteffy
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2 Answers2

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Suppose for contradiction's sake that $a\neq b$. Since the statement is symmetric in $a,b$ you may assume WLOG that $b>a$.

Take $\epsilon = b-a$. You get $b-a = |b-a| < b-a$, a contradiction.

Gabriel Romon
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  • I feel like an idiot. Thank you! – Joesteffy Apr 13 '18 at 14:18
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    @Joesteffy it happens to everybody, don't get discouraged ! – Gabriel Romon Apr 13 '18 at 14:19
  • Question for you @Gabriel-Romon , if you assume the contrapositive, doesn't that change the inequality statement from less than to greater than or equal to? – Joesteffy Apr 13 '18 at 14:22
  • @Joesteffy I haven't used the contrapositive. If you want to proceed like that, you need to suppose that $a\neq b$ and show $\exists \epsilon, |a-b|\geq \epsilon$. Note that $\epsilon = \max(a,b)-\min(a,b)$ fits the bill. – Gabriel Romon Apr 13 '18 at 14:28
  • @Gabriel-Roman what's the difference between supposing the contrapositive and 'supposing for contradiction's sake' like you did in the proof above? Sorry, I'm still relatively new to proof writing. – Joesteffy Apr 13 '18 at 14:37
  • @Joesteffy have a look at this answer https://math.stackexchange.com/a/262831/66096 – Gabriel Romon Apr 13 '18 at 14:41
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Let $a=c/d$ and $b=e/f$ where $c,e\in \Bbb Z$ and $d,f\in \Bbb Z^+.$ Then $$a\ne b\implies a-b\ne 0\implies c/d-e/f\ne 0\implies \frac { cf-de}{df}\ne 0\implies$$ $$\implies cf-de\ne 0\implies |cf-de|\geq 1 \implies$$ $$\implies |a-b|=\frac {|cf-de|}{|df|}\geq \frac {1}{|df|}=\frac {1}{df}.$$ So if $\epsilon =\frac {1}{1+df}$ then $|a-b|>\epsilon.$

So if $a,b\in \Bbb Q$ and $a\ne b$ then we cannot have $|a-b|<\epsilon$ for every $\epsilon \in \Bbb Q^+. $

Note: $cf-de\ne 0\implies |cf-de|\geq 1$ because $cf-de$ is an integer.

More importantly, it follows from the axiomatic definition of $\Bbb R$ that if $a,b$ are any real numbers then $|a-b|$ cannot be less than every member of $\Bbb Q^+$ unless $a=b$.

DanielWainfleet
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