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Show that if $n$ is composite then there exists a prime $p \leq n^\frac{1}{2}$ such that $p\mid n$.

I would like to use contradiction to prove this claim but I'm not sure about how I should contradict this statement.

Daniel W. Farlow
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Keyang Zhang
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  • What if all $p$ prime divisors of $n$ satisfy $p>n^\frac{1}{2}$? – Daniel Apr 01 '15 at 01:15
  • Your question has been asked before. See, for instance, http://math.stackexchange.com/questions/431930/prove-that-if-a-number-n-1-is-not-prime-then-it-has-a-prime-factor-le-sq – Barry Smith Apr 01 '15 at 01:16

1 Answers1

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Assume no prime $p<\sqrt n$ divides $n$, then there must be a prime $p$ such that $\sqrt n< p\le n$ and $p|n$ Let $m=n/p$. Because $p>\sqrt n$, $m\le\sqrt n$. But then there is a prime $q$ such that $q|m$ and $q\le \sqrt n$.

Tim Raczkowski
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