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**jusstjoe** #2: Let $\displaystyle n = ab$ and $\displaystyle 1 < a \leq b < n$. Suppose $\displaystyle a > \sqrt{n}$ (so $\displaystyle b > \sqrt{n}$ as well). What can you say about $\displaystyle ab$ then? How would you then bring the prime $\displaystyle p$ into this?

OK.

So I think given this situation, ab>n? I believe this is so because if a and b are both bigger than sqrt(n), then it must be that a*b>n,

but then this is a contradiction to our original statement, n=ab...but I don't know if I know how to bring prime numbers into this mix.

Thanks for your help.