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I need to prove this for any n natural, n>= 5, n not prime.

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- Thread starter xax
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- #1

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I need to prove this for any n natural, n>= 5, n not prime.

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HallsofIvy

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mathman

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[tex]n > 4[/tex] implies that

[tex]\sqrt{n} > 2[/tex] so we also have that [tex]\sqrt{n}\sqrt{n} > 2\sqrt{n}[/tex] or simply [tex]n > 2\sqrt{n}[/tex], meaning that [tex]2\sqrt{n}[/tex] and [tex]\sqrt{n}[/tex] both show up in [tex](n-1)![/tex] so we're good to go.

- #8

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You make perfect sense rodigee and this is a nicer demonstation than mine. Thank you.

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