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Show that sqrt(n!) is irrational

  1. Oct 13, 2011 #1
    Let n>2. Where n is integer show that sqrt(n!) is irrational.

    I am supposed to use the Chebyshev theorem that for n>2. There is a prime p such that n<p<2n.

    So far I am up to inductive hypothesis. Assume it holds for k then show it holds for k+1.

    Well if k! is irrational==> k!= 2^(e_2)***p^(e_n)

    then there is one power of prime which is odd. Using number theory. But don't know any ideas how to go from there.
     
  2. jcsd
  3. Oct 13, 2011 #2
    Induction is not necessary. Assume that n = either 2k or 2k-1 with respect to the Chebyshev theorem in terms of k.
     
    Last edited: Oct 13, 2011
  4. Oct 13, 2011 #3
    I think ramsey2879 means that the proof may be different for the cases n=odd and n=even. I've not gone through the details myself, but the idea of the proof is simple, if this helps:

    Of all the numbers that make n!, namely 1.2.3...n, by Chebyschev theorem there is a prime p in the upper half (roughly speaking, between n/2 and n). The problem is that all multiples of that prime (2p, 3p, ...) are greater than n (since, if p > n/2, then 2p > n). So here you have a prime that appears only once in the factorization of n!... meaning that n! cannot be a square.
     
  5. Oct 14, 2011 #4
    Darn, I expected the Op to figure this out for himself!
     
  6. Oct 14, 2011 #5
    Thanks I got it. But I did it a bit differently so proof as follows:

    Let n!=(2k)!
    Let's name the set N={1,2,3,...,k,...p,...,2k} basically all the factors of n!
    p is a prime between k and 2k which is true by Chebyshev (we're not ask to show this thank god)

    -First we show that p does not appear twice since p>k ==> 2p>2k so 2p is not in N.
    -But we also show p^2 is not in N as well.
    Basically I used induction to show p^2>2k when k>2. Therefore p only appears once and is only power to the one.


    Hence since there is a power of prime that is not divisible by 2 (since p=p^1) ==> n! is irrational.

    (Similar for n!=(2k+1)! I'll left it out).
     
  7. Oct 15, 2011 #6
    Good work. One thought though; since you showed that 2p > 2k then, as all other multiples of p such as p^2 are greater than 2p for p> 2, all multiples of p are necessarily > n. Also, for a prime p in n! to be squared, only p and 2p need appear in N since p*2p = 2p^2. Therefore, your effort to show that p^2 > n was needless. Overall though a good proof for one relatively new to number theory.

    Edit, here is a new problem for the op that is a little bit tougher:
    If p is prime, show that p^p will never appear in the prime factorization of n!
    Have fun solving this!
     
    Last edited: Oct 15, 2011
  8. Oct 15, 2011 #7

    mathwonk

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    by "appear" do you mean that the valuation (exponent) of n! at p will never be exactly equal to p?
     
  9. Oct 15, 2011 #8
    Yes, The exponent will skip from (p-1) to (p+1) in the valuation of n! at p.
     
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