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Great Proofs

Irrationality Theorem of Pythagoreans
Its other proofs & Generalizations

We saw the proof by the Pythgoreans of the irrationality of √2. Let us call it the first proof. Here is a second proof[1].
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[1]. 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[2] 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.

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.
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.

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[1], 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.

[1] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth Edition, Oxford University Press, 1978
[2] George R. Thomas, Executive Director, MathPath, 2003





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Last updated - Dec 15, 2003