Question

Let \(x\) and \(a\) be real numbers. Suppose that \(x 10 years ago

Answer 1

assume x>a

let d=x-a

d>0
...

Answer 2

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

Answer 3

Yes, Zarkon, that's correct. Thank you for pointing that out.

I'll try what you proposed.

Answer 4

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, \(x\leq a\). \(\blacksquare\)\[\]