Question

Prove: If x ge 0 and x le (alpha) for all (alpha) > 0, then x=0.

Any hints on how to start this or what the proof should include?

Answer 1

If \[x \ge 0\] and \[x < (\alpha)\] for all \[(\alpha) > 0\] then x=0

Answer 2

might want to start by assuming
\(x\neq 0\) and arrive at a contradiction

Answer 3

since \(x\geq 0\) if \(x\neq 0\) then \(x=\beta\) and \(\beta>0\)
now consider \(\frac{\beta}{2}\) and get a contradiction

Answer 4

ahhh i like that, thanks!

Answer 5

Ug, my 2nd line should be \[x \le (\alpha)\] not strictly less than. I think that changes things.

Answer 6

It seems to me that because the inequalities include "or equal to" that x > 0 wouldnt provide the contradiction I am looking for.