Proving the Triangle Inequality: ##|a-b| < \epsilon##

[Mr Davis 97]

Homework Statement

If $\forall \epsilon > 0$ it follows that $|a-b| < \epsilon$, then $a=b$.

Homework Equations

The Attempt at a Solution

Proof by contraposition. Suppose that $a \neq b$. We need to show that $\exists \epsilon > 0$ such that $|a-b| \ge \epsilon$. Well, let $\epsilon_0 = |a-b| > 0$. Since $|a-b| \ge \epsilon_0$, we are done.

Is this proof okay? It doesn't seem very enlightening as to why the theorem is true...

 

[andrewkirk]

Since $|a-b| \ge \epsilon_0$, we are done.

Not quite. It remains to prove that
$$|a-b|>0\rightarrow a\neq b$$

 

[Mr Davis 97]

Not quite. It remains to prove that
$$|a-b|>0\rightarrow a\neq b$$

Why do I have to show that?

 

[andrewkirk]

As you were. I misread the question.