If $\lvert a-b \rvert <\epsilon$, with $a$ and $b$ constants and $\epsilon >0$, then prove that $a=b$

Question

I already know the part when if $a\neq b$, then we can say $\epsilon=a-b$ which leads to contradiction. So I decided to try it considering the cases:

If $\lvert a-b \rvert <\epsilon$, with $a$ and $b$ constants and $\epsilon >0$, then proove that $a=b$. And I try by contradiction, so I have this:

Suppose $a\neq b$. Then:

Case) $a>b$, and $a>0>b$

$\implies a-b>0$

$\implies \lvert a-b\vert=a-b<\epsilon$, so we can say $\epsilon=a-b>0$ and $\lvert a-b\rvert<e=a-b$, then $\lvert a-b\rvert<a-b$ wich is a contradiction.

Case) $a<b$, and $a<0<b$

$\implies a-b<0$

$\implies \lvert a-b\rvert=-(a-b)<\epsilon$, so $a-b>-\epsilon$

$\implies \lvert \epsilon\rvert>\lvert a-b\rvert$, then $\lvert a-b\vert <\epsilon$

I can't find a contradiction here, can you tell me what am I omitting? I'm really struggling and don't know why. I would appreciate your help.

Answer 1

Let $a$ and $b$ be constants such that $|a - b| < \epsilon$ for all real numbers $\epsilon > 0$. Suppose $a \neq b$. Then $a - b$ is a nonzero number and $|a - b|$ is a positive real number.

Let $\epsilon = \dfrac{|a - b|}{2}$. Then $|a - b| > \epsilon$, which is a contradiction.

Thus $a = b$.