- #1
billy2908
- 12
- 0
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.
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.