Prove that if $n=ab$, $a$ and $b$ are positive integers, then $a \leq \sqrt{n}$ or $b \leq \sqrt{n}$ Prove that if $n=ab$, $a$ and $b$ are positive integers, then $a \leq \sqrt{n}$ or $b \leq \sqrt{n}$


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*Method chosen: Proof by contradiction.

*suppose $n=ab$, $a$ and $b$ are positive integers, then $a>\sqrt{n}$ and $b>\sqrt{n}$

*Let $n=30$ and $a=5$, $b=6$.

*$5 >\sqrt{30}$ and $6>\sqrt{30}$.

*Contradiction, $5 > \sqrt{30}$ is false.

*Therefore it is not the case $a>\sqrt{n}$ and $b>\sqrt{n}$, so it is true that $a \leq \sqrt{n}$ or $b \leq \sqrt{n}$
Did I do this right? Any help would be appreciated.
 A: No, you've only disproved one very specific possible counterexample. A bona fide proof by contra-diction disproves every possible counterexample. I would first establish some machinery:


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*If $u>0$ and $v>0$ then $uv>0$.

*If $u>v>0$ and $w>0$ then $(u-v)>0,w>0\implies (u-v)w>0$, hence $uw>vw$.

*If $u>v>0$ and $w>x>0$ then $uw>vw>0$ and $wv>xv>0$ hence $uw>xv>0$.


This establishes that inequalities can be multiplied together, so long as we pay attention to signs and the symbols. In particular, $a>\sqrt{n}$ and $b>\sqrt{n}$ imply $ab>n$, contradicting our hypothesis $n=ab$.
Hence it cannot be the case that both of $a,b$ are strictly greater than $\sqrt{n}$. The logical negation of this proposition is that at least one of $a,b$ are less than or equal to $\sqrt{n}$.
A: You need to show it more generally than just for n=30. 
from 2 on:
3) Then 
$$n = ab > \left( \sqrt{n}\right)^2$$
4) This implies that $$n>n$$
which is a contradiction. Thus the assumption that $a > \sqrt{n}$ AND $b>\sqrt{n}$ is false. 
Alternatively you could go about it by saying that a and b are the smallest possible values greater than n since they are both integers each would be n+1. 
$$n \neq (n+1)^2$$
Small Changes
2) instead of "then" -> "and" (your not coming to a conclusion your stating an assumption/condition)
A: I think we can solve that by using only the same logic that the author used, it's contradiction of course. 
Suppose n=ab and a,b>0 but a>√n  and b>√n then ab>n
Since ab=n so statement above is disproved 
Hence, the question. 
