Show that sqrt(n) is irrational

[billy2908]

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.

 

[ramsey2879]

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.

Induction is not necessary. Assume that n = either 2k or 2k-1 with respect to the Chebyshev theorem in terms of k.

 

[dodo]

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.

 

[ramsey2879]

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.

Darn, I expected the Op to figure this out for himself!

 

[billy2908]

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).

 

[ramsey2879]

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).

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!

 

[mathwonk]

by "appear" do you mean that the valuation (exponent) of n! at p will never be exactly equal to p?

 

[ramsey2879]

by "appear" do you mean that the valuation (exponent) of n! at p will never be exactly equal to p?

Yes, The exponent will skip from (p-1) to (p+1) in the valuation of n! at p.