Prove this lemma the square root of a positive integer is ei

Prove this lemma: the square root of a positive integer is either an integer or irrational. (Hint: let n be a positive integer and suppose (x)^(1/2) is rational but not an integer. If x^(1/2) = p/q in lowest terms, show that q has at least one prime factor that does not divide p)

Solution

What about the case of general n? Well, of course n² is not only rational but is an integer, namely n. Moreover, an arbitrary positive integer n can be factored to get one of these two limiting cases: namely, any n can be uniquely decomposed as

n = sN² , where s is squarefree. (Prove it!) Since sN² = N s, we have that n is rational iff s is rational; by the above result, this only occurs if s = 1. Thus: the square root of a positive integer is either an integer or irrational.

or

if n is square, then sqrt(x) is an
integer and if n is nonsquare then n = p^2/q^2 gives
xq^2 = p^2 and n nonsquare implies that some prime P in n
will occur to an odd power. Yet if P is or is not in q,
P will still occur to an odd power in xq^2. Yet P must occur
to an even power in a^2. Therefore nb^2 does not
equal p^2, a contradiction, so sqrt(x) is irrational.
Conclusion: for every natural n, sqrt(x) is either an
integer or is irrational.