# Prove for each non-negative number s, there exists a non-negative t such that s $\geq$ t

Statement to prove (or disprove with counter-example) is:

"For each non-negative number $s$, there exists a non-negative $t$ such that $s \geq t$."

I am pretty sure this statement is true, but I do not really know a way to approach a proof for this statement. I don't know how to prove it for every non-negative number $s$.

Thanks.

• Or you sure you have the question right? Batman's answer shows that that is.... light. Commented Sep 28, 2017 at 5:17

Just take $t=s$. It doesn't even matter what "non-negative" means for this to work! The same proof will apply for any other adjective.
• (+1) It only matters that $s \geq t$ means $s > t$ or $s = t$. We don't even have to know what $>$ means :) Commented Sep 28, 2017 at 9:14
Every non-negative number is greater than or equal to $0$. This is by definition of non-negativity. $0$ itself is non-negative (since $0 \geq 0$).
If $s \ge 0$ take $t=s/2$. Then $t \ge 0$ and $s \ge t$.