Now if all the ís were even, n would be a perfect square. Since it isnít, at least one of the ís must be odd. Hence, we can write where m is odd and p does not divide k.
Suppose is rational: , where , and .
as p is prime as .
Hence the factor of p occurs an odd number of times in . This is a contradiction because is a perfect square and so each prime factor of has to occur an even number of times.
Thus must be irrational.