We saw the proof by the Pythgoreans of the irrationality of √2. Let us call it the first proof. Here is a second proof.
If √2 were rational, let √2 = a/b. Then a2 = 2b2, where a and b have no common factors (other than 1). Then, b divides a2 and therefore p divides a2 for any prime factor p of b. It follows that p divides a. Since a and b have no common factors, this is impossible. Hence b has no prime factors and so b = 1. But this means that √2 = a which is false.
Here is G.H.Hardy's comment on these two proofs:
"The first proof and the second proof are very similar but there is an important difference. In the first proof we consider divisibility by 2, a given number. Clearly, if 2 divides a2, then 2 divides a, since the square of an odd number is certainly odd. In the second proof, on the other hand, we consider divisibility by the unknown prime p and, in fact, we assume" the powerful theorem known as Euclid's First Theorem, which is that if p is a prime, and p divides ab, then p divides a or p divides b. "Thus the first proof is the logically simpler proof, while, as we shall see in a moment, the second proof lends itself more readily to generalization."
Hardy's comment above precedes Theorem 44 in his book. Theorem 44 is the following.
N1/m is irrational, unless N is the m-th power of an integer.
Suppose that N1/m is a rational.
Then N1/m = a/b, where a and b have no common factors. So am = Nbm.
Then b divides am, and p divides am for every prime factor p of b. Hence p divides a, and from this follows - as in the second proof - that b = 1 and N1/m = a, which contradicts that N is not the m-th power of an integer. It follows that N1/m is not rational.
Another proof of the above result is given below. This proof also assumes Euclid's First Theorem, but contains an additional interesting proposition in the form of a claim that immediately achieves the result. Further, it is instructive in working in any integer base.
Let N be a positive integer that is not the m-th power of a positive integer.
If N1/m were rational, let it equal X + p/q, where X, p, and q are positive integers, p and q have no common factors, and p < q.
Let X, p, and q be positive integers, p and q have no common factors, and p < q.
Then (X + p/q)m has exactly m digits to the right of the radix point in the representation in base q.
PROOF OF CLAIM:
By virtue of the claim, N = (X + p/q)m is not an integer in base q, and hence in any base. This contradicts that N is an integer. Therefore N1/m is not rational. Q.E.D.
Since X + p/q has exactly one non-zero digit (p) to the right of the radix point in base q, the claim is true for m = 1. Let this representation of (X + p/q) be Y.p in base q.
Suppose the claim is true for n ≥ 1. That is, (X + p/q)n = Z.r1 r2…rn, 0 < rn < q.
Then (X + p/q)n+1 = (X + p/q)(X + p/q)n = (Y.p)(Z.r1 r2…rn)
= YZ +Y(0.r1r2 …rn) + (0.p)Z + (0.p)(0.r1r2…rn), where the last summand
is the only term that can possibly contribute an (n+1)-th digit to the right of the radix point.
But (0.p)(0.r1r2…rn) = (p/q)(r1/q + r2/q2 +…+ rn/qn).
Now, since p and q have no common factors, prn = tq + rn+1, where t and rn+1are integers and 0 < rn+1, t < q.
Therefore, (p/q)(rn/qn) = (t/qn) + (rn+1/qn+1) = 0.0…0trn+1), with n-1 0's between the radix point and t. It follows that the non-integer part of (X + p/q)n+1 is 0.s1s2…snrn+1, where the si are non-negative integers less than q.
So the claim is true for n+1, and, hence, by induction, for all m. This proves the claim.
Note that the last proof above is a technical proof that does not qualify as a great proof - this does not mean that there are no technical proofs that are great proofs. We close this discussion by presenting a generalization, known as Gauss's Lemma, of the above result. We have noted elsewhere that generalization is a phenomenon that is a hallmark of mathematics, that it is the generous vein that mathematicians often fruitfully pursue.
If x is a root of an equation
xm + c1xm-1 + …+ cm = 0,
with integer coefficients of which the first is unity, then x is either integer or irrational.
(The particular case in which the equation is xm - N = 0 was proved twice above.)
Without loss of generality, we may assume that cm ≠ 0.
(For if it were zero, there must be a cr that is the first non-zero coefficient before cm. Otherwise x = 0 is the only solution. If 0 is not the solution, divide the equation by xm-r. This yields an equation that has the same form as the original equation.)
If x = a/b, where a and b have no common factors, then
am + c1am-1b + …+ cmbm = 0.
Hence b divides am (why?), and from this follows as before that b = 1 which is clearly false. Therefore x≠a/b. Q.E.D.
There are many proofs known of the irrationality of √2. We give elsewhere a few proofs that also serve to illustrate various proof techniques.
 G. H. Hardy and E. M. Wright, An Introduction to
the Theory of Numbers, Fifth Edition, Oxford University Press, 1978
 George R. Thomas, Executive Director, MathPath, 2003