Proof by infinite descent One proof of the number's irrationality is the following
proof by infinite descent. It is also a
proof of a negation by refutation: it proves the statement "\sqrt{2} is not rational" by assuming that it is rational and then deriving a falsehood. • Assume that \sqrt{2} is a rational number, meaning that there exists a pair of integers whose ratio is exactly \sqrt{2}. • If the two integers have a common
factor, it can be eliminated using the
Euclidean algorithm. • Then \sqrt{2} can be written as an
irreducible fraction \frac{a}{b} such that and are
coprime integers (having no common factor) which additionally means that at least one of or must be
odd. • It follows that \frac{a^2}{b^2}=2 and a^2=2b^2. ( ) ( are integers) • Therefore, is
even because it is equal to . ( is necessarily even because it is 2 times another whole number.) • It follows that must be even (as squares of odd integers are never even). • Because is even, there exists an integer that fulfills a = 2k. • Substituting from step 7 for in the second equation of step 4: 2b^2 = a^2 = (2k)^2 = 4k^2, which is equivalent to b^2=2k^2. • Because is divisible by two and therefore even, and because 2k^2=b^2, it follows that is also even which means that is even. • By steps 5 and 8, and are both even, which contradicts step 3 (that \frac{a}{b} is irreducible). Since a falsehood has been derived, the assumption (1) that \sqrt{2} is a rational number must be false. This means that \sqrt{2} is not a rational number; that is to say, \sqrt{2} is irrational. This proof was hinted at by
Aristotle, in his
Analytica Priora, §I.23. It appeared first as a full proof in
Euclid's
Elements, as proposition 117 of Book X. However, since the early 19th century, historians have agreed that this proof is an
interpolation and not attributable to Euclid.
Proof using reciprocals Assume by way of contradiction that \sqrt 2 were rational. Then we may write \sqrt 2 + 1 = \frac{q}{p} as an irreducible fraction in lowest terms, with coprime positive integers q>p. Since (\sqrt 2-1)(\sqrt 2+1)=2-1^2=1, it follows that \sqrt 2-1 can be expressed as the irreducible fraction \frac{p}{q}. However, since \sqrt 2-1 and \sqrt 2+1 differ by an integer, it follows that the denominators of their irreducible fraction representations must be the same, i.e. q=p. This gives the desired contradiction.
Proof by unique factorization As with the proof by infinite descent, we obtain a^2 = 2b^2. Being the same quantity, each side has the same
prime factorization by the
fundamental theorem of arithmetic, and in particular, would have to have the factor 2 occur the same number of times. However, the factor 2 appears an odd number of times on the right, but an even number of times on the left—a contradiction.
Application of the rational root theorem The irrationality of \sqrt{2} also follows from the
rational root theorem, which states that a rational
root of a
polynomial, if it exists, must be the
quotient of a factor of the constant term and a factor of the
leading coefficient. In the case of p(x) = x^2 - 2, the only possible rational roots are \pm 1 and \pm 2. As \sqrt{2} is not equal to \pm 1 or \pm 2, it follows that \sqrt{2} is irrational. This application also invokes the integer root theorem, a stronger version of the rational root theorem for the case when p(x) is a
monic polynomial with integer
coefficients; for such a polynomial, all roots are necessarily integers (which \sqrt{2} is not, as 2 is not a perfect square) or irrational. The rational root theorem (or integer root theorem) may be used to show that any square root of any
natural number that is not a perfect square is irrational. For other proofs that the square root of any non-square natural number is irrational, see
Quadratic irrational number or
Infinite descent.
Geometric proofs Tennenbaum's proof of A simple proof is attributed to
Stanley Tennenbaum when he was a student in the early 1950s. Assume that \sqrt{2} = a/b, where a and b are coprime positive integers. Then a and b are the smallest positive integers for which a^2 = 2b^2. Geometrically, this implies that a square with side length a will have an area equal to two squares of (lesser) side length b. Call these squares A and B. We can draw these squares and compare their areas - the simplest way to do so is to fit the two B squares into the A squares. When we try to do so, we end up with the arrangement in Figure 1., in which the two B squares overlap in the middle and two uncovered areas are present in the top left and bottom right. In order to assert a^2 = 2b^2, we would need to show that the area of the overlap is equal to the area of the two missing areas, i.e. (2b-a)^2 = 2(a-b)^2. In other terms, we may refer to the side lengths of the overlap and missing areas as p = 2b-a and q = a-b, respectively, and thus we have p^2 = 2q^2. But since we can see from the diagram that p and q , and we know that p and q are integers from their definitions in terms of a and b, this means that we are in violation of the original assumption that a and b are the smallest positive integers for which a^2 = 2b^2. Hence, even in assuming that a and b are the smallest positive integers for which a^2 = 2b^2, we may prove that there exists a smaller pair of integers p and q which satisfy the relation. This contradiction within the definition of a and b implies that they cannot exist, and thus \sqrt{2} must be irrational.
Apostol's proof Tom M. Apostol made another geometric
reductio ad absurdum argument showing that \sqrt{2} is irrational. It is also an example of proof by infinite descent. It makes use of
compass and straightedge construction, proving the theorem by a method similar to that employed by ancient Greek geometers. It is essentially the same algebraic proof as Tennebaum's proof, viewed geometrically in another way. Let be a right isosceles triangle with hypotenuse length and legs as shown in Figure 2. By the
Pythagorean theorem, \frac{m}{n}=\sqrt{2}. Suppose and are integers. Let be a
ratio given in its
lowest terms. Draw the arcs and with centre . Join . It follows that , and and coincide. Therefore, the
triangles and are
congruent by
SAS. Because is a right angle and is half a right angle, is also a right isosceles triangle. Hence implies . By symmetry, , and is also a right isosceles triangle. It also follows that . Hence, there is an even smaller right isosceles triangle, with hypotenuse length and legs . These values are integers even smaller than and and in the same ratio, contradicting the hypothesis that is in lowest terms. Therefore, and cannot be both integers; hence, \sqrt{2} is irrational.
Constructive proof While the proofs by infinite descent are constructively valid when "irrational" is defined to mean "not rational", we can obtain a constructively stronger statement by using a positive definition of "irrational" as "quantifiably apart from every rational". Let and be positive integers such that (as satisfies these bounds). Now and cannot be equal, since the first has an odd number of factors 2 whereas the second has an even number of factors 2. Thus . Multiplying the absolute difference by in the numerator and denominator, we get :\left|\sqrt2 - \frac{a}{b}\right| = \frac{b^2\!\left(\sqrt{2}+\frac{a}{b}\right)} \ge \frac{1}{b^2\!\left(\sqrt2 + \frac{a}{b}\right)} \ge \frac{1}{3b^2}, the latter
inequality being true because it is assumed that , giving (otherwise the quantitative apartness can be trivially established). This gives a lower bound of for the difference , yielding a direct proof of irrationality in its constructively stronger form, not relying on the
law of excluded middle. This proof constructively exhibits an explicit discrepancy between \sqrt{2} and any rational.
Proof by Pythagorean triples This proof uses the following property of primitive
Pythagorean triples: : If , , and are coprime positive integers such that , then is never even. This lemma can be used to show that two identical perfect squares can never be added to produce another perfect square. Suppose the contrary that \sqrt2 is rational. Therefore, :\sqrt2 = {a \over b} :where a,b \in \mathbb{Z} and \gcd(a,b) = 1 :Squaring both sides, :2 = {a^2 \over b^2} :2b^2 = a^2 :b^2+b^2 = a^2 Here, is a primitive Pythagorean triple, and from the lemma is never even. However, this contradicts the equation which implies that must be even. ==Multiplicative inverse==