Square root of 2 is irrational

The statement we are going to discuss and prove is known as Theorem of Theaetetus due to its appearance in Plato's Theaetetus dialog:

Below we shall concentrate on the one root -- that of 2 -- that Theaetetus has not mentioned and at times suggest an extension to a more general result. But first let's see how Richard Dedekind, one of the people most responsible for the current definition and understanding of irrational numbers, treated Theorem of Theaetetus:

Proof 1

A standard proof of this result does not differ from the proof that √5 is irrational. Still, the argument can be modified a little.

Following is another intuitive one [Lasckovich, p. 4]. This one is based on the assertion that every number is uniquely (up to the order of factors) representable as a product of primes. (A prime is a number divisible only by itself and 1.) The fact which is known as The Fundamental Theorem of Arithmetic. Assuming it as an axiom (or a given fact), let (p/q)² = 2 for some integers p and q. Then p² = 2q². Factor both p and q into a product of primes. p² is factored into a product of the very same primes as p each taken twice. Therefore, p² has an even number of prime factors. So does q². Therefore, 2q² has an odd number of prime factors. Contradiction.

Here's a modification that only counts the factors equal to 2 [Automatic Sequences, p. 40]. p² has an even number of such factors, while 2q² is bound to have an odd number of them. This is equivalent to saying that in the binary expansion of p² the number of unit digits is even, whereas on the right, in 2q², the number of units is odd.

Following is yet another illuminating proof. As in the standard proof, assume p and q are mutually prime (numbers with no common factors). Their squares are still mutually prime for they are built from the same factors. Therefore, the fraction p²/q² cannot cancel out. In particular, p²/q² cannot cancel down to equal 2. Therefore, p²/q²2.

J. L. Lagrange in his Lectures on Elementary Mathematics (1898) argues that "... it's impossible to find a whole number which multiplied by itself will give 2. It cannot be found in fractions, for if you take a fraction reduced to its lowest terms, the square of this fraction will again be a fraction reduced to its lowest terms, and consequently cannot be equal to the whole number 2."

The latter proof makes it entirely obvious that unless a square root of an integer is an integer itself, it is bound to be irrational. Furthermore, the same argument applies to roots other than square. Unless it's an integer itself, a fifth root of an integer is an irrational number!

The proofs above, directly or indirectly, appeal to the Fundamental Theorem of Arithmetic. In an article "Irrationality Without Number Theory" (Am. Math. Monthly, 1990), Richard Beigel set out to check how much of Number Theory is actually needed. He showed that for any integer k and t, k^1/t is either integer or irrational using only the divisibility algorithm and the floor (whole part) function [n]. Following is his proof for t = 2. (A more general result is shown to follow from the Rational Root Theorem.)

Richard's argument can be modified to invoke an infinite regression which is impossible for positive integers. Assuming x to be rational, there exists an integer n such that nx is also an integer. As before, we can find an integer m less than n with the same property. mx = 0 gives an immediate contradiction. Otherwise, applying the same reasoning to m and so on, we potentially have an infinite set of integers with no minimal element which is impossible. This is akin to the following geometric proof [Barbin, Gardner].

Proof 6

If x = 2^1/2 were rational, there would exist a quantity s commensurable both with 1 and x: 1 = sn and x = sm. (It's the same as assuming that x = m/n and taking s = 1/n.) The same will be true of their difference x - 1, which is smaller than x. And the process could continue indefinitely in contradiction with the existence of a minimal element. The game Euclid might have played always ends!

And here's another geometric proof due to Tom Apostol (The American Mathematical Monthly, v 107, N 9, pp 841-842.) This one is so simple it may pass as a proof without words. But this is what his author had to say:

This note presents a remarkably simple proof of the irrationality of √2 that is a variation of the classical Greek geometric proof. By the Pythagorean theorem, an isosceles right triangle of edge-length 1 has hypotenuse of length √2. If √2 is rational, some positive integer multiple of this triangle must have three sides with integer lengths, and hence there must be a smallest isosceles right triangle with this property. But inside any isosceles right triangle whose three sides have integer lengths we can always construct a smaller one with the same property, as shown below. Therefore √2 cannot be rational. Construction. A circular arc with center at the uppermost vertex and radius equal to the vertical leg of the triangle intersects the hypotenuse at a point, from which a perpendicular to the hypotenuse is drawn to the horizontal leg. Each line segment in the diagram has integer length, and the three segments with double tick marks have equal lengths. (Two of them are tangents to the circle from the same point.) Therefore the smaller isosceles right triangle with hypotenuse on the horizontal base also has integer sides. The reader can verify that similar arguments establish the irrationality of √n² + 1 and √n² - 1 for any integer n > 1. For √n² + 1 use a right triangle with legs of lengths 1 and n. For √n² - 1 use a right triangle with hypotenuse n and one leg of length 1.

Essentially the same argument has been used in a Russian geometry textbook by A. P. Kiselev, p. 121. The book, first published in 1892, has been in a systematic use up to the late 1950s with practically no competition, and frequently in the ensuing years. A proof to the same effect but with a paper folding interpretation is due to [Conway and Guy, pp. 183-185]

Assume [Lasckovich, Gardner] that √2 = p/q where p and q are the positive integers and q is the smallest possible. Then p > q and also q > p - q, so that we have

But this contradicts the minimality of q.

A superb graphic illustration for the latter has been popularized by J. Conway around 1990, see [Hahn, ex. 37 for Ch. 1]. Conway discussed the proof at a Darwin Lecture at Cambridge. The lecture appears alongside other Darwin lectures in the book Power published by Cambridge University Press. Conway’s contribution is included as the chapter titled "The Power of Mathematics". The text can be found online. Conway attributes the proof to the Princeton mathematician Stanley Tennenbaum (1927 - 2006) who made the discovery in the early 1950s while a student at the University of Chicago.

Given two squares with integer sides, one twice the other

move the smaller squares

into opposite corners of the bigger square

The intersection of the two forms a square at the center of the diagram. Their union leaves two squares at the free corners of the diagram. By the Carpets Theorem, the two areas are equal:

(Obviously, these squares also have integer sides.)

(This proof has also appeared in an article Irrationality of Square Roots by P. Ungar, Math Magazine, v. 79, n. 2, April 2006, pp. 147-148, with an extension to roots of more general polinomials with integer coefficients, and [Lasckovich, pp. 4-5])

Consider all rational numbers in the interval from 0 to 1, excluding 0. Each number can be written in a unique way as a fraction a/b where a and b have no divisors in common. Imagine now a/b as a centre of an interval of length 1/2b²; in other words, cover a/b by the interval with endpoints a/b - 1/4b² and a/b + 1/4b². Since the rational numbers form a dense set (i.e., in every interval no matter how small there are always rational numbers) and, since the sum of lengths of all covering intervals is found to be infinite, it would seem that, having so generously covered all rational numbers, we have automatically covered all numbers. However, we shall show that 2/2 remains uncovered! In fact the number |b² - 2a²| being an integer must be at least 1; it cannot be 0 since 2 is irrational.

Hence

and the assertion that √2/2 is not covered follows.

Conway and Guy (pp. 184-185) argue that if √N is not an integer but a rational number B/A, then

The fractional parts of B/A and NA/B have the form a/A and b/B, where a, b are positive integers smaller than A, B. But if two numbers are equal, their fractional parts are also equal:

so that

This gives a simpler form for √N, contrary to our assumption.

Alan Cooper found what I would call a common-sense proof of the fact at hand. He based his proof on an observation that squaring a finite decimal fraction, say m.n, never removes the decimal part. This becomes clear on considering the very last (non-zero) digit. This is the only digit responsible for the last digit in the decimal expansion of the square. Thus the latter is never zero.

In case where the fraction is not a finite decimal, we may switch to another number system in which the fraction is finite. The same argument now applies.

Alan notes that the above is a way of interpreting Proof 4.

In Z₃, the field of residues modulo 3, 0² = 0, 1² = 1 and 2² = 1, so there is no element whose square is 2. Now suppose √2 is a rational number a = p/q. Then a maps to (p mod 3) / (q mod 3), which is either p or 2p (mod 3). It follows that a² = p² ≠ 2 (mod 3). But since reduction mod 3 respects all the arithmetic operations, a² = 2 implies a² = 2 (mod 3) in Z₃, a contradiction.

(It can be shown that the equation x² = 2 (mod p), where p is an odd prime, is solvable if p = 1, 7 (mod 8) and is unsolvable if p = 3, 5 (mod 8), see [Stark, pp. 311-313].)

In general, it is easy to see that n^1/m is irrational if there exists at least one prime p such that n is not a perfect m^th power in Z_p.

Starting as in the proof from Conway and Guy, let √N is not an integer but a rational number B/A, then

Assume B/A is in lowest terms so that A is minimal. Recollect that x/y = w/z implies

for any t.

Since B is not divisible by A, there is a q > 0 satisfying

From B/A = NA/B it then follows that

which contradicts B - qA < A. Alternatively, we may use the fact that for mutually prime A and B there are integer r and s such that rA + sB = 1. Then

which is an integer. A contradiction.

It's edifying to recall an estimate of approximation of irrational numbers with rational ones.

The irrationality of √2/2 leads to an interesting investigation.

In case you are curious, √2 is the length of the diagonal of the unit square.

References

1. J-P Allouche & J. Shallit, Automatic Sequences, Cambridge University Press, 2003
2. E. Barbin, The Meanings of Mathematical Proof, in In Eves' Circles, J.M.Anthony, ed., MAA, 1994
3. J. H. Conway, R. K. Guy, The Book of Numbers, Copernicus, 1996
4. P. J. Davis and R. Hersh, The Mathematical Experience, Houghton Mifflin Company, Boston, 1981
5. J. W. R. Dedekind, On Irrational Numbers, in A Source Book in Mathematics by D. E. Smith, Dover, 1959, pp. 38-40
6. M. Gardner, Gardner's Workout, A K Peters, 2001
7. A. Hahn, Basic Calculus: From Archimedes to Newton to its Role in Science, Springer Verlag & Key College, 1998 (Also available online)
8. M. Kac and S. M. Ulam, Mathematics and Logic, Dover Publications, NY, 1968.
9. M. Lasckovitch, Conjecture and Proof, MAA, 2001.
10. S. K. Stein, Mathematics: The Man-Made Universe, 3rd edition, Dover, 2000.
11. H. M. Stark, An Introduction to Number Theory, MIT Press, 1970
12. I. Thomas, Greek Mathematical Works, v1, Harvard University Press, 2006
13. D. Wells, You are a Mathematician, John Wiley & Sons, 1995

Copyright © 1996-2008 Alexander Bogomolny