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!
On
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 $a=b$.
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.