
iff proof
Please Help!
Prove that n is a square iff the exponent of every prime number occuring in the factorisation of n is even.
Hence prove that the square root of any natural number is either an integer or an irrational number.
Any help would be greatly appreicated I don't know anyone else who can.

Hello,
Write the decomposition into prime factors of n.
$\displaystyle n = \prod_{i=1}^k p_i ^{\alpha_i}$
So $\displaystyle n^2 = \prod_{i=1}^k p_i ^{\alpha_i} \prod_{i=1}^k p_i^{\alpha_i} = \prod_{i=1}^k [p_i^{\alpha_i} \times p_i^{\alpha_i}]$
As $\displaystyle a^b a^c = a^{b+c}$, the previous expression equals to :
$\displaystyle \prod_{i=1}^k p_i^{\alpha_i + \alpha_i} = \prod_{i=1}^k p_i^{2 \alpha_i}$
Hence the exponent of every prime number occuring in the factorisation of n is even because multiple of 2.
And as it's equalities, there is equivalence.