1.4: Irrational Numbers
The best known of all irrational numbers is \(\sqrt{2}\). We establish \(\sqrt{2} \ne \dfrac{a}{b}\) with a novel proof which does not make use of divisibility arguments.
Suppose \(\sqrt{2} = \dfrac{a}{b}\) (\(a\), \(b\) integers), with \(b\) as small as possible. Then \(b < a < 2b\) so that
\(\dfrac{2ab}{ab} = 2\), \(\dfrac{a^2}{b^2} = 2,\) and \(\dfrac{2ab - a^2}{ab - b^2} = 2 = \dfrac{a(2b - a)}{b(a - b)}\).
Thus
\(\sqrt{2} = \dfrac{2b - a}{a - b}\).
But \(a < 2b\) and \(a - b < b\); hence we have a rational representation of \(\sqrt{2}\) with denominator smaller than the smallest possible!
To convince students of the existence of irrationals one might begin with a proof of the irrationality of \(\log_{10} 2\). If \(\log_{10} 2 = \dfrac{a}{b}\) then \(10^{a/b} = 2\) or \(10^a = 2^b\). But now the left hand side is divisible by 5 while the right hand side is not.
Also not as familiar as it should be is the fact that \(\cos 1^{\circ}\) (and \(\sin 1^{\circ}\)) is irrational. From
\(\cos 45^{\circ} + i \sin 45^{\circ} = (\cos 1^{\circ} + i \sin 1^{\circ})^{45}\)
we deduce that \(45^{\circ}\) can be expressed as a polynomial in integer coefficients in \(\cos 1^{\circ}\). Hence if \(\cos 1^{\circ}\) were rational so would be \(\cos 45^{\circ} = \dfrac{1}{\sqrt{2}}\).
The fact that
\(\cos 1 = 1 - \dfrac{1}{2!} + \dfrac{1}{4!} - \cdot\cdot\cdot\)
is irrational can be proved in the same way as the irrationality of \(e\). In the latter case, assuming \(e\) rational,
\(\dfrac{b}{a} = e = 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \cdot\cdot\cdot + \dfrac{1}{(a + 1)!} + \dfrac{1}{(a + 2)!} + \cdot\cdot\cdot\),
which, after multiplication by \(a!\), would imply that \(\dfrac{1}{a + 1} + \dfrac{1}{(a + 1)(a + 2)} + \cdot\cdot\cdot\) is a positive integer less than 1.
A slightly more complicated argument can be used to show that \(e\) is not of quadratic irrationality, i.e., that if \(a, b, c\) are integers then \(ae^2 + be + c \ne 0\). However, a proof of the transcendentality of \(e\) is still not easy. The earlier editions of Hardy and Wright claimed that there was no easy proof that π is transcendental but this situation was rectified in 1947 by I. Niven whose proof of the irrationality of \(\pi\) we now present.
Let
\(\pi = \dfrac{a}{b}, f(x) = \dfrac{x^n (a - by)^n}{n!}\), and \(F(x) = f(x) - f^{(2)}(x) + f^{(4)}(x) - \cdot\cdot\cdot,\)
the positive integer \(n\) being specified later. Since \(n!f(x)\) has integral coefficients and terms in \(x\) of degree \(\le 2n\), \(f(x)\) and all its derivatives will have integral values at \(x= 0\). Also for \(x = \pi = \dfrac{a}{b}\), since \(f(x) = f(\dfrac{a}{b} - x)\). By elementary calculus we have
\(\dfrac{d}{dx} [F'(x) \sin x - F(x) \cos x] = F'' (x) \sin x + F(x) \sin x = f(x) \sin x.\)
Hence
\(\int_{0}^{\pi} f(x) \sin x dx = [F'(x) - F(x) \cos x]_{0}^{\pi} =\) an integer.
However, for \(0 < x < \pi\),
\(0 < f(x) \sin x < \dfrac{\pi^n a^n}{n!} \to 0\)
for large \(n\). Hence the definite integral is positive but arbitrarily small for large \(n\); this contradiction shows that the assumption \(\pi = \dfrac{a}{b}\) is untenable.
This proof has been extended in various ways. For example, Niven also proved that the cosine of a rational number is irrational. If now \(\pi\) were rational, \(\cos \pi = −1\) would be irrational. Further, the method can also be used to prove the irrationality of certain numbers defined as the roots of the solutions of second order differential equations satisfying special boundary conditions. Recently, a variation of Niven’s proof has been given which, although more complicated, avoids the use of integrals or infinite series. A really simple proof that \(\pi\) is transcendental, i.e., does not satisfy any polynomial equation with integer coefficients is still lacking.
With regard to transcendental numbers there are essentially three types of problems: to prove the existence of such numbers, to construct such numbers, and finally (and this is much more difficult than the first two) to prove that certain numbers which arise naturally in analysis are transcendental. Examples of numbers which have been proved transcendental are \(\pi\), \(e\), \(e^{-\pi}\), and \(\dfrac{\log 3}{\log 2}\). It is interesting to remark here that Euler’s constant \(\gamma\) and
\(\sum_{n = 1}^{\infty} \dfrac{1}{n^{2s + 1}}\) (\(s\) is an integer)
have not even been proved irrational.
Cantor’s proof of the existence of transcendental numbers proceeds by showing that the algebraic numbers are countable while the real numbers are not. Thus every uncountable set of numbers contains transcendental numbers. For example there is a transcendental number of the form \(e^{i\theta}\), \(0 < \theta < \dfrac{\pi}{2}\), say.
Although it is not entirely relevant here we will perform now a little disappearing stunt using such a transcendental number \(e^{i\theta}\) and a construction due to Kuratowski.
Consider the following set of points in the complex plane. Start with the point \(O\) and let \(\tilde{S}\) be the set of all points obtainable from it by a succession of the operations of translating the points 1 unit to the right and rotating them through an angle \(\theta\) about \(O\). If we denote such translations and rotations by \(T\) and \(R\) respectively then a typical point of our set \(\tilde{S}\) may be denoted by \(T^{a}R^{b}T^{c}R^{d} \cdot\cdot\cdot\). We next observe that every point of \(\tilde{S}\) must have a unique representation in this form. Indeed, \(T\) means adding 1 to the complex number corresponding to the point and \(R\) means multiplication by \(e^{i\theta}\). Hence all our points are polynomials in \(e^{i\theta}\) with positive coefficients, say \(z = P (e^{i\theta})\). But now if a point has a double representation, then \(P (e^{i\theta}) = R(e^{i\theta})\) and we would obtain a polynomial in \(e^{i\theta}\) which would negate the transcendental character of \(e^{i\theta}\).
Let \(\tilde{T}\) denote the subset of \(\tilde{S}\) which consists of those points of \(\tilde{S}\) for which the last operation needed to reach them is a \(T\), and let \(\tilde{R}\) denote the subset which consist of those points of \(\tilde{S}\) for which the last operation needed to reach them is an \(R\). Clearly \(\tilde{S} = \tilde{T} \cup \tilde{R}\) and \(\tilde{T} \cap \tilde{R} = \emptyset\). A translation of \(\tilde{S}\) of one unit to the right sends \(\tilde{S}\) into \(\tilde{T}\), i.e., it makes \(\tilde{R}\) vanish! On the other hand, a rotation of the plane through \(\theta\) sends \(\tilde{S}\) into \(\tilde{R}\) making \(\tilde{T}\) vanish!
So far we have discussed only the existence of transcendental numbers. The easiest approach to the actual construction of such numbers is via a theorem due to Liouville.
We say that an algebraic number is of degree \(n\) if it satisfies a polynomial equation of degree \(n\). We say that a real number \(\lambda\) is approximable to order \(n\) provided the inequality
\(|\lambda - \dfrac{a}{b}| < \dfrac{c}{b^n}\)
has an infinity of solutions for some constant \(c\). Liouville’s theorem states that a real algebraic number of degree \(n\) is not approximable to any order greater than \(n\).
Suppose \(\lambda\) is of degree \(n\). Then it satisfies an equation
\(f(\lambda) = a_0 \lambda^n + a_1 \lambda^{n - 1} + \cdot\cdot\cdot + a_n = 0.\)
There is a number \(M = M(\lambda)\) such that \(|f'(x)| < M\) where \(\lambda - 1 < x < \lambda + 1\). Suppose now that \(\dfrac{p}{1} \ne \lambda\) is an approximation to \(\lambda\). We may assume the approximation good enough to ensure that \(\dfrac{p}{q}\) lies in the interval \(\lambda - 1, \lambda + 1)\),
and is nearer to \(\lambda\) than any other root of \(f(x) = 0\), so that \(f(p/q) \ne 0\).
Clearly (see Figure 2),
\(|f(\dfrac{p}{q})| = \dfrac{1}{q^n} |a_0p^n + a_1 p^{n - 1} q + \cdot\cdot\cdot + a_n q^n| \ge \dfrac{1}{q^n}\)
and
\(|dfrac{f(p/q)}{\lambda - p/q}| < M\)
so that
\(|\lambda - \dfrac{p}{q}| > \dfrac{c}{q^n}\)
and the theorem is proved.
Although Liouville’s theorem suffices for the construction of many transcendental numbers, much interest centers on certain refinements. In particular, it is desirable to have a theorem of the following type. If \(\lambda\) is of degree \(n\) then
\(|\lambda - \dfrac{p}{q}| < \dfrac{M}{q^{f(n)}}\)
has at most a finite number of solutions. Here \(f(n)\) may be taken as \(n\) by Liouville’s theorem . Can it be decreased? Thue, about 1909, first showed that one could take \(f(n) = \dfrac{n}{2}\) and Siegel (1921) showed that we can take \(f(n) = 2\sqrt{n}\). This was slightly improved by Dyson and Schneider to \(\sqrt{2n}\). Very recently (1955), F. K. Roth created a sensation by proving that we can take \(f(n) = 2 + \epsilon\). His proof is long and complicated. That we cannot take \(f(n) = 2\) (hence Roth’s result is in a way best possible) can be seen from the following result due to Dirichlet.
For irrational \(\lambda\) there exist infinitely many solutions of
\(|\lambda - \dfrac{p}{q}| < \dfrac{1}{q^2}.\)
The proof is not difficult, Let \(\lambda\) be irrational and consider, for fixed \(n\), the numbers (\(\lambda\)), (\(2 \lambda\)), ..., (\(n \lambda\)), where \((x)\) means "fractional part of \(x\)". These \(n\) points are distinct points on \(0, 1)\); hence there exist two of them say \(i\lambda\) and \(j \lambda\) whose distance apart is \(\le \dfrac{1}{n}\). Thus we have
\((i\lambda) - (j \lambda) < \dfrac{1}{n}\)
or
\(k \lambda - m \le \dfrac{1}{n}\) (\(k\) and \(m\) integers \(\le n\))
and
\(|\lambda - \dfrac{m}{k}| \le \dfrac{1}{nk} \le \dfrac{1}{n^2},\)
as required.
We now return to the application of Liouville’s theorem to the construction of transcendental numbers.
Consider
\(\dfrac{1}{10^{1!}} + \dfrac{1}{10^{2!}} + \cdot\cdot\cdot + \dfrac{1}{10^{p!}} = \lambda_p\)
as well as the real number \(\lambda = \lambda_{\infty}\). It is easily checked that \(|\lambda_{\infty} - \lambda_p| < \dfrac{1}{\lambda^{n + 1}\) for every \(p\). Hence \(\lambda\) is approximable to order \(n\) for any \(n\) and hence is not algebraic.