(n-1)! is divisible by n

  • Thread starter xax
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xax
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Main Question or Discussion Point

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

Answers and Replies

695
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Think a bit about the prime factors of n... are they smaller than n? Then think of what (n-1)! means.
 
HallsofIvy
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Dodo's hint is excellent- but the crucial point is whether the factors of n are less than n-1, not jus n itself!
 
mathman
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If n is not prime, then n=j*k, where j and k are >1, therefore both <n. As a result both are factors in (n-1)!, so n divides (n-1)!.
 
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Almost there, one last thing to consider is the squares of prime numbers. What happens when n=9=3*3 for example? This is where the n>=5 comes in.
 
xax
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You all have good points and I thought of all of them and the reason I've posted this is because of the square numbers as rodigee said. Well I think if n = k*k then k is one of the numbers between 1 and n-1. since n=k*k then k divides (n-k) and this is smaller the n-1 which means n divides (n-1)!.
 
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Sorry, xax but I don't really understand your idea heh. This is how I see it:

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

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