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

Proof 2

A standard proof (e.g., [Rademacher and Toeplitz, Ch. 4]) of this result does not differ from the proof that 5–√ is irrational. Still, the argument can be modified a little.

The premise p2=2q2 tells us that p is even. Assuming p and q mutually prime, q is bound to be odd. However, the square of an even number is divisible by 4, which leads us to conclude that q is even. A contradiction.

Proof 3

Following is another intuitive one [Laczkovich, p. 4, Davis & Hersh, p. 299]. 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.

Proof 3'

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.

Proof 4

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/q2≠2.

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."

More than half a century earlier (1831), Augustus De Morgan explained that "... 7 is not made by the multiplication either of any whole number or fraction by itself. The first is evident; the second cannot be readily proved to the beginner, but he may, by taking a number of instances, satisfy himself of this, that no fraction which is really such, that is whose numerator is not measured by its denominator, will give a whole number when multiplied by itself, thus 4/3×4/3=16/9 is not a whole number, and so on."

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, 1991), Richard Beigel set out to check how much of Number Theory is actually needed. He showed that for any integer k and t, k1/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 and Gauss' Lemma. The latter has a proof that does not even mention the divisibility.)

Proof 5

Let x=k1/2 and assume that x is rational but not integer. Then there exists a minimal positive integer n such that xn is an integer. Consider m=n(x−⌊x⌋). Since the fractional part of x, 0≤x−⌊x⌋<1, 0≤m<n. Note that m is an integer for m=nx−n⌊x⌋ which is an integer. Furthermore, mx=nx2−(nx)⌊x⌋=nk−(nx)⌊x⌋ which is also an integer. Due to the minimality of n, m=0. In other words, x=⌊x⌋ and is, thus, an integer in contradiction to our assumption.

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–√ 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! As it does not, the algorithm leads to a continued fraction representation of 2–√. The geometric form of the process is known as anthyphairesis which means (in Greek) continually subtract the smaller from the larger.

Proof 7

And here's another geometric proof I came across in an article by 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:

Essentially the same diagram 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]

[Rademacher and Toeplitz, Ch. 4] gave two proofs of the irrationality of 2–√ of which the first was illustrated by practically the same diagram, without mention of the paperfolding background. As the starting point of the argument, they simply laid a side of a square on the diagonal and then proved the emergence of the smaller right isosceles triangle.

Proof 8

Assume [Laczkovich, 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

(2q−p)/(p−q)=(2−p/q)/(p/q−1)=(2−2–√)/(2–√−1)=(2−2–√)(2–√+1)=2–√.

But this contradicts the minimality of q.

(Prof. Claus I. Doering, Instituto de Matemática - UFRGS, Brazil, has pointed to a much earlier reference. The proof appeared in a footnote of the classic A Course of Modern Analysis by E. T. Whittaker and G. N. Watson, 4th Edition, Cambridge University Press, 1927. The footnote had also been included in the 3rd edition (1920).)

Proof 8'

Obviously, the proof can be restated as (2p−q)2=2(p−q)2, which, too, is easily shown to hold, provided p2=2q2 holds.

(p−q)2(2q−p)2=p2−2pq+q2=3q2−2pq,=4q2−4pq+p2=3p2−4pq=2(3q2−2pq)=2(p−q)2.

(Gary Davis has suggested an alternative formulation: If there is an integer n>0 with n2–√ an integer then m=n(2–√−1) has the same property, but is smaller than n.)

Proof 8''

A kindred proof has been published by Edwin Halfar (Am Math Monthly, Vol. 62, No. 6 (Jun-Jul 1955), p. 437). Suppose 2–√=m/n, where n and m are positive integers. Then n>m, and there is an integer p>0 such that n=m+p, and 2m2=m2+2mp+p2. This implies m>p. Consequently, for some integer a>0, m=p+a, n=2p+a and 2(p+a)2=2(p+a)2. The last equality implies a2=2p2 so that the entire process may be repeated indefinitely giving n>m>a>p…, but since every non-null set of positive integers has a smallest element, this is a contradiction and 2–√ is irrational.

In plain English this asserts that given two squares with integer sides and one having twice the area of the other, there exist a pair of smaller squares with the same properties.

Proof 8'''

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.)

Proof 8IV

Another illustration by Grant Cairns appeared in Math. Mag. 85 (2012), p. 123:

The first two steps of the infinite discent

are suggestive of the continuation

There are extensions of the geometric arguments to regular polygons other than the square that illustrate the irrationality of 3–√, 5–√, 6–√, and - in principle - of other square roots.

Proof 8V

Here's a variant from Am Math Monthly (120 (August/September) 2013, p 674) by Samuel G. Moreno and Esther M. García-Caballero:

If 2–√ is a rational number, then x0=2–√+1 is also rational, so x0=p/q, for some positive integers p and q with q the smallest possible. Since 1<2–√<2, then 2<x0<3, which implies 2q<p<3q. Clearly, x0(x0−2)=(2–√+1)(2–√−1)=1, and thus

1x0=x0−2=pq−2=p−2qq,

so that x0=q/(p−2q). The bound p−2q<q contradicts the minimality of q.

Proof 9

This proof is due to Alex Healy and was once available online at The Braden Files site. I owe the deepest thanks to Rick Mabry (Software Developer turned Professor) for pointing me in the right direction.)

Consider the set W=a+b2–√, a,b integers. Clearly W is closed under multiplication and addition. Define α=(2–√−1), an element of W. Obviously, 0<α<1, so that

(1)

αk→0 as k→∞.

Assume 2–√=p/q. Since W is closed, αk=e+f2–√=(eq+fp)/q≥1/q contradicting (1).

(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 polynomials with integer coefficients, and [Laczkovich, pp. 4-5])

Proof 10

This proof too is by D. Kalman et al (Variations on an Irrational Theme-Geometry, Dynamics, Algebra, Mathematics Magazine, Vol. 70, No. 2 (Apr., 1997), pp. 93-104).

Let A be the matrix

A=(−1 1 2−1)

By the definition,

(−1 1 2−1)(ab)=(2b−aa−b)

Two facts are worth noting: (a) matrix A maps integer lattice onto itself, (b) the line with the equation a=2–√b is an eigenspace L, say, corresponding to the eigenvalue 2–√−1:

(−1 1 2−1)(2–√1)=(2−2–√2–√−1)=(2–√−1)(2–√1).

Since 0<2–√−1<1, the effect of A on L is that of a contraction operator. So that the repeated applications of matrix A to a point on L remain on L and approach the origin. On the other hand, if the starting point was on the lattice, the successive iteration would all remain on the lattice, meaning that there are no lattice points on L.

Proof 11

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

B/A=NA/B.

Assuming that B/A in lowest terms, we observe that 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:

a/A=b/B

so that

b/a=B/A=N−−√.

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

Proof 11'

Geoffrey C. Berresford (Am Math Monthly, Vol. 115, No. 6 (June-July 2008), p. 524) offered a different route from the assumption

N−−√=B/A=NA/B,

with B/A in lowest terms.

If two fractions are equal, with one in lowest terms, the numerator and denominator of the other are a common integer (say c) multiples of the numerator and denominator of the first. Therefore, B=Ac, so that B/A=c, and hence N−−√ is an integer, so N is a perfect square.

(I am grateful to Prof. Claus I. Doering from Instituto de Matemática - UFRGS, Brazil for bringing this proof to my attention.)

Proof 12

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.

Proof 13

Following Nick Lord (Math Gazette, v 91, n 521, July 2007, p. 256) we shall show that, for an integer N>1, N−−√ is irrational. The proof is based on the assertion that, for integer a,m,n, n>1, m and n coprime, the expression

m/n+an/m

is never an integer. Indeed, if say m/n+an/m=k, for an integer k, then

m2=n⋅(km−an),

which would imply that n divides m - in contradiction with either n>1 or coprimality of n and m.

Assume then that N−−√ is rational but not an integer. Then, since N−−√+1 is not an integer, we must have

N−−√+1=m/n,

for coprime m and n, with n>1. But the observation above with a=1−N then leads to a contradiction that

2=N−−√+1+(1−N)/(N−−√+1)

is not an integer.

Proof 14

Gustave Robson (Am Math Monthly, Vol. 63, No. 4 (Apr., 1956), p. 247) published a short proof preceded by a remark: The following proof was invented by Robert James Gauntt, in 1952, while he was a freshman at Purdue. I was unable to induce him to write up his proof.)

a2=2b2 cannot have a non-zero solution in integers because the last non-zero digit of a square, written in the base three, must be 1, whereas the last non-zero digit of twice a square is 2.

Proof 14'

This proof is by Stuart Anderson.

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 \mathbb{Z}_{3}, 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 n1/m is irrational if there exists at least one prime p such that n is not a perfect mth power in \mathbb{Z}_{p}.

Proof 14''

This proof was found by Sergey Markelov when yet in high school. In the decimal system, a square of an interger may only end in one of the following digits: 0,1,4,5,6,9 whereas twice a square may only end with 0,2,8. Thus assuming a2=2b2, both a and b may only end with 0. This triggers an infinite descent which proves that this is also impossible.

Proof 15

2-proofs-in-1 from The American Mathematical Monthly 114 (May 2007), p. 416. The proof is by Xinyun Zhu of Central Michigan University.

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

B/A=NA/B.

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

x/y=(x+tw)/(y+tz),

for any t.

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

0<B−qA<A.

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

N−−√=NA/B=(N⋅A−qB)/(B−qA)

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

N−−√=(sN⋅A+rB)/(sB+rA)=sN⋅A+rB,

which is an integer. A contradiction.

Proof 16

(N. C. Ferreño, Yet Another Proof of the Irrationality of 2–√, Am Math Monthly, v 116, n 1 (Jan 2009), pp. 68-69.)

Consider a linear mapping f: R→R given by f(x)=(2–√−1)x as a dynamical system. For each point x0∈R let us define its orbit O(x0) as the sequence of iterates starting at x0, namely

O(x0)={(2–√−1)nx: n≥0}.

Since 0<2–√−1<1, it is clear that for each x0∈R the orbit O(x0) converges to zero. Now suppose 2–√=p/q is a rational number with p,q∈N. Then, for all n∈N, it holds that

(2–√)nq={2n/2q,2(n−1)/2p,if n is even if n is odd .

In either case, (2–√)nq is a natural number, and from binomial theorem it follows that

0<(2–√−1)nq=∑k[(nk)2–√k(−1)n−kq]∈Z,

k=1,…,n. Therefore O(x0)⊂N which contradicts the fact that O(x0) tends to zero.

Proof 17

Yoram Sagher gave a modification of Dedekind's argument (Am Math Monthly, Vol. 95, No. 2. (Feb., 1988), p. 117):

Suppose k−−√=m/n, where m and n are integers with n>0. If k is not a square, there is an integer q so that q<m/n<q+1. Now m2=kn2 implies m(m−qn)=n(kn−qm) and, hence, m/n=(kn−qm)/(m−qn). From q<m/n<q+1 we get 0<m−qn<n. Therefore we have:

k−−√=(kn−qm)/(m−qn),

where the denominator is positive and smaller than the one in the original fraction. Continuing, we get an infinite decreasing sequence of positive integers, an impossibility.

This proof does not use any properties of primes and would thus be fully accessible to Pythagoras and to Theodorus.

Proof 17'

Following up on Y. Sagher proof, Robert W. Floyd published (Am Math Monthly, Vol. 96, No. 1 (Jan., 1989), p. 67) an extension:

Assuming Pythagoras understood Euclid's algorithm, the following proofs show how he could have demonstrated that any integer root of an integer is an irrational or an integer, and even that the cube root of an integer either is not the root of a quadratic (i.e., not in the form (a+bN−−√)/c or is an integer.

I placed the proof into a separate file.

Proof 18

This is a proof by D. Kalman et al (Variations on an Irrational Theme-Geometry, Dynamics, Algebra, Mathematics Magazine, Vol. 70, No. 2 (Apr., 1997), pp. 93-104).

ABC is an isosceles right triangle. D on AB is such that BD=BC. DE⊥BC; F and G are such that CEGF is a square.

$      $

Observe that CG=DG. Indeed, by the construction of D, ΔBCD is isosceles so that ∠BDC=∠BCD. Also, ∠ECG=∠BDE. Subtracting equals from equals, we see that ∠DCG=∠CDG, implying CG=CD.

Let H on AC be such as to form a parallelogram ADGH. Then ∠FHG=∠CAB=45∘. Hence, ΔCGH is isosceles and CG=GH. But then in the parallelogram ADGH all sides are equal.

The following diagram only shows three equal segments that are important for the proof.

Assume BC and AB are commensurable, i.e. assume that the two have a common unit. Their difference AD is then also wholly measured by the same unit. The unit therefore measures two legs (GH and CG) of the isosceles right ΔCGH. The unit measures AC and AH and so also CH. We obtain an isosceles right ΔCGH all of whose sides are measured by the same unit. But ΔCGH is smaller than ΔABC. Since the process could be continued, we obtain a contradiction.

This proof too does not use any properties of primes and would thus be fully accessible to Pythagoras and to Theodorus.

The proof admits an algebraic equivalent. Suppose e is the unit common to BC and AC: BC=ne and AC=me. We then produce successively AD=(n−m)e, AH=(n−m)e, CH=m−(n−m)e=(2m−n)e; and since triangles ABC and CHG are similar, n/m=(2m−n)/(n−m). The latter identity is equivalent to n2=2m2, meaning that 2–√=n/m. But then also 2–√=(2m−n)/(n−m) which again leads to an infinite descent. The proof appears to be a geometric equivalent of the short Proof 8.

Proof 19

I am grateful to Aharon Meyerowitz from Florida Atlantic University for bringing to my attention the geometric arguments from The Elements of Dynamic Symmetry by Jay Hambidge (the book is available online.)

Cut off of a 2–√×1 rectangle a square of side 1. The rectangle left over will have dimensions (2–√−1)×1. Removing from the latter a square as shown in the diagram leaves a rectangle (2–√−1)×(2−2–√). This rectangle is similar to the one we started with: