## anonymous 5 years ago Let $$x$$ and $$a$$ be real numbers. Suppose that $$x<a+\epsilon$$. Prove that $$x\leq a$$. I'm supposed to prove this by contradiction, but I have no idea how to begin; all I need is a hint on how to start.

1. Zarkon

assume x>a let d=x-a d>0 ...

2. Zarkon

I assume x<a+e is true for all e>0

3. anonymous

Yes, Zarkon, that's correct. Thank you for pointing that out. I'll try what you proposed.

4. anonymous

Does this seem legit? Proof: Let $$x$$ and $$a$$ be real numbers and suppose that $$x<a+\epsilon$$ for every positive number $$\epsilon$$. Assume that $$x>a$$, and let $$d=x-a$$. Then $$d>0$$. It follows that $$x<a+d$$, $$x<a+x-a$$, $$x<x$$. The last statement is a contradiction. Therefore, $$x\leq a$$. $$\blacksquare$$