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:

Theaetetus: Theodorus was proving to us a certain thing about square roots, I mean the square roots of three square feet and five square feet, namely, that these roots are not commensurable in length with the foot-length, and he proceeded in this way, taking each case in turn up to the root of seventeen square feet; at this point for some reason he stopped. Now it occurred to us, since the number of square roots appeared to be unlimited, to try to gather them into one class, by which we could henceforth describe all the roots.

Socrates: And did you find such a class?

Theaetetus: I think we did; but see if you agree.

Socrates: Speak on.

Theaetetus: We divided all numbers into two classes. The one, consisting of numbers that can be represented as the product of equal factors, we likened in shape to the square and called them square or equilateral numbers.

Socrates: And properly so.

Theaetetus: The numbers between these, among which are three and five and all that cannot be represented as the product of equal factors, but only as the product of a greater by a less or a less by a greater, and are therefore contained by greater and less sides, we likened to oblong shape and called oblong numbers.

Socrates: Excellent. And what after this?

Theaetetus: Such lines as form the sides of equilateral plane numbers we called lengths, and such as form the oblong numbers we called roots, because they are not commensurable with others in length, but only with the plane areas which they have the power to form. And similarly in the case of solids.

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

Let D be a positive integer but not the square of an integer, then there exists a positive integer λ such that

λ² < D < (λ + 1)².

... If there exists a rational number whose square is D, then there exist two positive integers t, u, that satisfy the equation:

t² - Du² = 0,

and we may assume that u is the least positive integer possessing the property that its square, by multiplication by D, may be converted into the square of an integer t. Since evidently

λu < t < (λ + 1)u,

the number v = t - λu is a positive integer certainly less than u. If further we put

s = Du - λt,

s is likewise a positive integer, and we have

s² - Dv² = (λ² - D)(t² - Du²) = 0,

which is contrary to the assumption respecting u. Hence the square of every rational number is either less than D or greater than D ...

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 [Lasckovich, 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.)

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 = 21/2 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!

Proof 7

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 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 [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

(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.

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

Proof 8''

A similar 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 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 + b √2}, 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 + f √2 = (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 polinomials with integer coefficients, and [Lasckovich, 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 =(
-12
1-1
)

By the definition,

  (
-12
1-1
) (
a
b
) = (
2b - a
a - b
)

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

  (
-12
1-1
) (
√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 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 n1/m is irrational if there exists at least one prime p such that n is not a perfect mth power in Zp.

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, if n is even
2(n-1)/2p, 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 [(
n
k
) (√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 thus is fully accessible to Pythagoras and to Theodorus.

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 + b√n) / c or is an integer.

In the following, all variables but r are restricted to integer values.

The (shorter) proof of Sagher's special case comes first, to motivate the others. If r = √k = m/n (in lowest terms), Euclid's algorithm gives α and β for which αm + beta;n = 1. Then rn = m, rm = r²n = kn, and

r = r(αm + βn) = αrm + βrn = αkn + βm,

an integer.

Now take r = m/n = k1/3, αm + βn = 1. Then

m = rn, rn² = mn, rmn = m², rm² = r³n² = kn².

r = r(αm + βn)² = α²rm² + 2αβrmn + β²rn² = α²kn² + 2αβm² + β²mn,

an integer.

Now make the weaker assumption that r = k1/3, and that r satisfies a proper quadratic equation ar² + br + c = 0, a ≠ 0. then

0 = (ar² + br + c)(ar - b) = a²r³ + (ac - b²)r - bc.

If ac = b², divide the equation by a² to find r³ = (b/a)³ and r = b/a. Otherwise, put k for r³ and find r = (bc - a²k)/(ac - b²). Either way, r is rational, and consequently an integer.

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 is perpendicular to 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 triangle CGH. 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 thus is 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 CH 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.)

two squares in a sqrt(2)x1 rectangle

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 start with:

(2 - √2) / (√2 - 1) = √2 / 1.

which shows that the standard "infinite descent" argument implies. Were √2 rational, it would be possible to select the smallest rectangle with dimensions proportional to (√2 - 1)×1. As the argument shows, the existence of such a smallest rectangle would lead to a contradiction.

This argument may serve as an illustration to Proof 8.

The book contains another approach illustrated by the following diagram:

two squares in a sqrt(2)x1 rectangle

Here two 1×1 squares are drawn at the opposite ends of the rectangle. This splits the original √2×1 rectangle into one big and two small squares and three rectangles of dimensions (√2 - 1)×(2 - √2). The same argument applies again.

Proof 20

The irrationality of √2 is a consequence of the p-adic Local-Global Principle. √2 is irrational because 2 is not a quadratic residue modulo 5!

The irrationality of √2 can be rephrased in a way that appears quite paradoxical [Kac & Ulam, pp. 6-7]:

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.

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.

Ludmila Duchêne and Agnès Leblanc put together an enchanting literary tribute to the question of irrationality of √2 (in French.)

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. A. De Morgan, On the Study and Difficulties of Mathematics, Dover, 2005, p. 130
  7. M. Gardner, Gardner's Workout, A K Peters, 2001
  8. A. Hahn, Basic Calculus: From Archimedes to Newton to its Role in Science, Springer Verlag & Key College, 1998 (Also available online)
  9. M. Kac and S. M. Ulam, Mathematics and Logic, Dover Publications, NY, 1968.
  10. M. Lasckovitch, Conjecture and Proof, MAA, 2001.
  11. H. Rademacher, O. Toeplitz, The Enjoment of Mathematics, Dover, 1990.
  12. S. K. Stein, Mathematics: The Man-Made Universe, 3rd edition, Dover, 2000.
  13. H. M. Stark, An Introduction to Number Theory, MIT Press, 1970
  14. I. Thomas, Greek Mathematical Works, v1, Harvard University Press, 2006
  15. D. Wells, You are a Mathematician, John Wiley & Sons, 1995

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