1. ## Approximation proof

Consider two numbers with this property:

(e stands for epsilon)

for any e > 0, |a - b| < e

Prove a = b by contraposition.

Thanks.

2. Suppose $a\ne b$.
Let $\varepsilon = \left| {a - b} \right| > 0$.

3. That is the full solution?

4. Originally Posted by cgiulz
That is the full solution?
Well can this be true: $0 < \left| {a - b} \right| < \left| {a - b} \right|?$

BTW: We are not here to give you complete solutions.
You are to use our help to complete the problem.
After all, it is your problem.

5. I understand that. From previous posts I've found that posters have provided me with very helpful tips that eventually lead me to the answer. This is my first analysis class ever, so at first glance I thought that was just what I needed to start with. Then I realized that those two steps implied the answer, so thanks. One last thing if you would be so kind: I thought epsilon was "lurking in the problem," and you couldn't just let it be equal to some answer, or maybe I'm getting ahead of myself?

Thanks again.

6. Originally Posted by cgiulz
Consider two numbers with this property:

(e stands for epsilon)

for any e > 0, |a - b| < e

Prove a = b by contraposition.

Thanks.
The 'contrapositive" is "If $a\ne b$ then there exist e> 0 such that $|a- b|\ge e$.

If $a\ne b$, then |a- b|> 0. Take $\epsilon= \frac{|a-b|}{2}$, say. What remains to be shown?