Let $$x$$ and $$a$$ be real numbers. Suppose that $$x OpenStudy (zarkon): assume x>a let d=x-a d>0 ... OpenStudy (zarkon): I assume x<a+e is true for all e>0 OpenStudy (anonymous): Yes, Zarkon, that's correct. Thank you for pointing that out. I'll try what you proposed. OpenStudy (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$$