Archive for July, 2009

Niven’s proof that the trigonometric and inverse trigonometric functions are irrational for rational non-zero arguments

July 17, 2009

In the late 1940s, Niven discovered a way to apply Hermite’s method to the cos function.  The following proof that cos x is irrational for non-zero rational x is adapted from that given in Niven’s book on irrational numbers [1].  I have tried to simplify the proof and make the motivation a little clearer.

Once again, we begin with the fact that if x is a real number

x is irrational if there exist integers h and k such that 0<|kx – h|< ε for any given real number 1>ε>0

Now consider the integral ∫sin(ht)Fn(t)dt with h a positive integer and F(t) a polynomial in t with integer coefficients, and whose degree is 4n.  If we integrate repeatedly by parts we obtain

h^{4n+1}\int \sin ht F_n(t)dt=\cos ht\sum_{k/2=0}^{2n}(-1)^{k+1}h^{4n-k}F^{(k)}_n(t)+\\\indent\sin ht\sum_{(k+1)/2=0}^{2n}(-1)^{k+1}h^{4n-k}F^{(k)}_n(t)

where in the first and second sums k is permitted to take only even and odd values respectively.  Now suppose we take

F_n(t)=(1-t)^{2n}(1-(1-t)^2)^n

Since this is an even polynomial in (1-t), its odd derivatives are zero at t=1.  Hence the second sum vanishes when t=1, and since sin ht=0 when t=0, if we integrate from t=0 to t=1 we obtain

h^{4n+1}\int_0^1 \sin ht F_n(t)dt=\cos h\sum_{k/2=0}^{2n}(-1)^{k+1}h^{4n-k}F^{(k)}_n(1)-\\\indent\sum_{k/2=0}^{2n}(-1)^{k+1}h^{4n-k}F^{(k)}_n(0)=(\cos h)k_n-h_n

with hn, kn integers.  F_n(t) has the properties that 1≥ F_n(t)≥0 if 1≥t≥ 0, and (see lemma below) F^{(k)}_n(1) and F^{(k)}_n(0) are divisible by n! so we have

h\frac{(h^4)^n}{n!}\int_0^1\sin htF_n(t)dt=(\cos h)k_n^\prime+h_n^\prime

with k’n(h) = kn(h)/n! and h’n(h) = hn(h)/n! integers.

The absolute value of the integral is less than 1.   (h4)n/n! can be made arbitrarily small by taking n large enough, so for all n greater than some sufficiently large value, nε say, the LHS of the equation can be made less in absolute value than any given positive real number ε.

To finish we need only show that for any given real number r and some n≥nε , rk’n+h’n≠0.  This must be the case if r is irrational so we suppose r = p/q with p, q>0 and g.c.d.(p,q) = 1.  rk’n+h’n cannot be zero if pk’n+qh’n is not zero.

By the Lemma, F^{(k)}_n(1)/n! and hence k’n is divisible by n+1.  We take n even.  By the Lemma, all the terms F^{(k)}_n(0)/n! in the sum

h^\prime_n=\sum_{k/2=0}^{2n}(-1)^kh^{4n-k}F^{(k)}_n(0)

are divisible by n + 1 with the exception of F^{(n)}_n(0).  We take n + 1 greater than nε , and such that it contains a prime factor that is not a prime factor of either h or q.  Then qh’n is not divisible by n + 1.  If n + 1 divides pk’n but not qh’n , their sum cannot be zero.

The fact that cos h is irrational for a positive integer h implies that cos x is irrational for all non-zero rational numbers x.  Let x be positive and rational and equal to a/b, with a, b positive integers.  Then as a consequence of the recursive trigonometric identity

 \cos nt = 2\cos t \cos(n-1)t-\cos (n-2)t

cos nt is a polynomial in cos t, and so cos bx = cos a is a polynomial in cos x.  If cos x is rational, then cos a is rational, so by the counterpositive, since cos a is irrational, cos x is irrational.  Since cos (-x)=cos x, the result is true also for negative rational x.

The following is also due to Niven.

If sin x were rational for rational x ≠ 0, so would be 1- 2\sin^2x=\cos 2x.  Thus sin x is irrational for rational x ≠ 0.

If tan x were rational for rational x ≠ 0, so would be

(1-\tan^2x)/(1+\tan^2x)=\cos 2x

 Thus tan x is irrational for rational x≠0.

As a consequence of the following

Let f(x) be a function whose argument is a set of real numbers and let it have an inverse function f-1(y) where y is the set of numbers f(x).  Let f be such that if x is rational then f(x) is irrational.  Then if y is rational f-1(y) is irrational.

the inverse trigonometric functions can be shown to be irrational for non-zero rational x.  Since \arctan 1 = \pi/4, \pi is necessarily irrational.

Niven provides a proof along similar lines that \pi^2 is irrational.  However Lambert’s continued fraction method can also be used to prove the same result, and I will provide this proof in a later post.

Lemma:  Let

F_n(t)=(1-t)^{2n}(1-(1-t)^2)n=(1-t)^{2n}t^n(2-t)^n

Then F^{(k)}_n(1) is divisible by n+1! and F^{(k)}_n(0) is divisible by n!  F^{(k)}_n(0) is divisible by n+1! when k ≠ n, but not when k = n if n+1 is not a power of two.

ProofF^{(k)}_n(t)=(-1)^k\frac{d^kF_n(t)}{d(1-t)^k}.  Since F_n(t) is a polynomial in 1-t whose term of least degree is (1-t)^{2n}, F^{(k)}_n(1)=0 when k<2n.  When 2n≤ k≤ 4n, F^{(k)}_n(t)=b_kk!+f(1-t).  n+1! divides k!, so F^{(k)}_n(1)/(n+1!) is an integer.

F_n(t) is a polynomial in t whose term of least degree is tn, so F^{(k)}_n(0)=0 when k < n.  F^n_n(0) = 2^nn! so if n+1 is not a power of two, it is not divisible by n+1!  When k > n, F^{(k)}_n(t) = akk! + f(t), where f(t) is either zero or has a factor t.  Hence F^{(k)}_n(t) is divisible by k! and hence by n+1!

 

[1] I Niven, ‘Irrational Numbers’, Wiley, 1956 pp 16-19

Hermite’s proof that exp(x) is irrational for rational non-zero x

July 14, 2009

This post follows on from the previous one.

The following method was introduced by Hermite, in a more general form, in the lead-up to his proof that e is transcendental.  It can be found on pp 153 and 154 of his Oeuvres III.  Although Hermite’s proof postdates that of Lambert by more than a century, I give it first because it is somewhat simpler.

The idea is to produce an expression that enables us to use the following:

Let x be a real number.  If  for any given real number 1>ε>0, there exist integers h and k such that 0 <|kx-h|<ε, then x is irrational.

Consider the integral ∫ehtFn(t)dt with h a positive integer and Fn(t) a polynomial in t with integer coefficients, and whose degree is 2n.  If we integrate repeatedly by parts we obtain

h^{2n+1}\int e^{ht}F_n(t)dt=e^{ht}\sum_{k=0}^{2n}(-1)^kh^{2n-k}F^{(k)}_n(t)

If we make this a definite integral from t = 0 to t = 1 we obtain the following expression, the right hand side of which has the desired form

h^{2n+1}\int^1_0 e^{ht}F_n(t)dt=e^h\sum_{k=0}^{2n}(-1)^kh^{2n-k}F^{(k)}_n(1)-\\ \indent\sum_{k=0}^{2n}(-1)^kh^{2n-k}F^{(k)}_n(0)=e^hk_n-h_n

with hn, kn integers.

Now suppose we take Fn(t) = tn(1 – t)n.  This polynomial has the properties that 1 ≥ Fn(t) ≥ 0 if 1 ≥ t ≥ 0, and (see Lemma below) F^{(k)}_n(1) and F^{(k)}_n(0) are divisible by n!.  Thus we have

            h\frac{(h^2)^n}{n!}\int^1_0e^{ht}F_n(t)dt=e^hk^\prime_n+h^\prime_n

with k’n(h) = kn(h)/n! and h’n(h) = hn(h)/n! integers.  The expression on the left is greater than zero.  Since the integrand is positive but less than eh, the integral is also positive but less than eh.  Suppose n1 is the smallest integer greater than h2.  Therefore for n > n1 the expression on the left is equal to ABn with A, B positive constants and B < 1.  This expression can be made arbitrarily small by taking n large enough, so by the theorem above, eh is irrational if h is a positive integer.

We can easily extend this result to show that ex is irrational if x is non-zero and rational.  Let x be positive and rational and equal to a/b, with a, b positive integers.  Then if ex = ea/b is rational, then (ea/b)b = ea is rational so by the counterpositive, if ea is irrational then ex is irrational.  If e-x = 1/ex is rational then ex is rational so by the counterpositive, if ex is irrational then e-x is irrational.

By the following:

Let f(x) be a function whose argument is a set of real numbers and let it have an inverse function f -1(y) where y is the set of numbers f(x).  Let f be such that if x is rational then f(x) is irrational.  Then if y is rational f -1(y) is irrational

taking f(x) = ex and the set x to be the set of real numbers excluding zero, f-1(y) = ln y and y the set of real numbers greater than zero but excluding one, then ln y is irrational if y is rational.

Lemma:  Let F_n(t)=t^n(1 - t)^n.  Then F^{(k)}_n(0) and F^{k)}_n(1) are divisible by n!

Proof:  In the first case, since the term of least degree in F_n(t) is tn, F^{(k)}_n(0)= 0 when k < n.  When 2n ≥ k ≥ n, F^{(k)}_n(t) = a_kk! + f(t), where f(t) is either zero or has a factor t.  Thus F^{(k)}_n(0) is divisible by k! and hence by n! 

In the second case, since F_n(t)=(1-t)^n(1-(1-t))^n and F^{(k)}_n (t)=(-1)^k\frac{d^k}{d(1-t)^k}F_n(t), we have F^{(k)}_n(1)=0 when k < n.  When 2n ≥ k ≥ n, F^{(k)}(t) = b_kk!+f(1-t) with f(1 – t) either zero or having a factor 1 – t.  Hence F^{(k)}(1) is divisible by n!

When is a real number irrational?

July 13, 2009

This post follows on from the previous one.

Hermite’s and Lambert’s methods make use of the fact that if x is a real number and if, for any given real number 0<ε<1, there exist integers h and k such that 0<|kx-h|<ε, then x is irrational.

Theorem 1:  if x is rational then there exists a positive integer n such that for every pair of integers h, k where k≠0 and h/k≠x, |kx-h|≥1/n, with n being independent of h, k

Proof:  The theorem is true if x = 0 because then |kx-h|=|h|≥1.  Otherwise suppose x = a/b with a, b ≠ 0 and g.c.d.(a,b) = 1.  Then |kx-h|=|ka-hb|/|b|.  The numerator is an integer and cannot be zero because if it were, we would have ka-hb = 0 and h/k=a/b=x.  Therefore the numerator is ≥1.  Setting the denominator equal to n we have the result.

Corollary:  Let x be a real number.  If  for any given real number 1>ε>0, there exist integers h and k such that 0 <|kx-h|<ε, then x is irrational.

Proof:  since |kx-h|≠0, h/k≠x.  Since ε<1, k≠0.  The rest follows by the counterpositive to the Theorem.

If the converse of the Corollary were true the Corollary would be a necessary condition for a number to be irrational and thus, at least in theory, always available as an avenue of proof that a number is irrational.  We now show:

Corollary (Converse):  if x is irrational then there exist integers h and k such that 0<|kx-h|<ε for any given real number 1>ε>0

Proof:  The proof uses the ‘pigeonhole principle’.  In what follows,  [x] means the integer part of x.  Assume x is irrational.  We can also assume without loss of generality that x is positive.  Let Q be a positive integer and consider the Q+1 numbers  x-[x], 2x-[2x], … , (Q+1)x-[(Q+1)x].  These are all irrational and less than 1.  They are distinct since if nx-[nx]=mx-[mx] then unless n=m, x=([nx]-[mx])/(n-m) which is rational, a contradiction.  We divide the interval from 0 to 1 into Q equal intervals of length 1/Q.  Since each of the Q+1 numbers must fit inside one of these intervals, and there are only Q intervals, one interval must contain two of them, nx-[nx] and mx-[mx] say, and these are a distance of 1/Q or less apart.

Theorem 2:  Let f(x) be a function whose argument is a set of real numbers and let it have an inverse function f -1(y) where y is the set of numbers f(x).  Let f be such that if x is rational then f(x) is irrational.  Then if y is rational f -1(y) is irrational.

Proof:  If  f -1(f(x)) = x is rational then f(x) is irrational.  By the counterpositive, if y = f(x) is rational, f -1(y) is irrational.

The proposition at the head of the post is essentially as used by Hermite in an 1873  paper ‘Sur l’irrationalité de la base des logarithmes hyperboliques’ [1].  However the proofs of the theorem and Corollaries are largely derived from sections 11.1 and 11.3 of Hardy and Wright [2].

[1] Hermite, Oeuvres III,  pp 127-129.  These can be accessed online through the website of the University of Michigan Historical Math Collection. 

[2]  G. H. Hardy and E. M. Wright, ‘An Introduction to the Theory of Numbers’, Fifth edition, Oxford Science Publications (a Sixth Edition has recently appeared, but I continue to work from my copy, the Fifth).  

The irrational nature of exponential and trigonometric functions of rational numbers, and their inverses

July 12, 2009

It is well-known that if x is non-zero and rational, then exp(x) and tan x are irrational (see, for example, the Mathworld article ‘Irrational Number’). From the former result, exp(1) = e is irrational, and from the latter, since tan pi/4 = 1 is not irrational, pi/4 and hence pi is not rational.

Perhaps one day someone will find a general method for determining, from the coefficients in a function’s Taylor series, whether the function possesses the property of mapping rational numbers almost exclusively to irrational numbers, but that day seems a long way off.

The purpose of this series of posts is to revisit Hermite’s, Niven’s and Lambert’s examinations of the exponential function, sin, cos and tan, and their inverses.

Above all, I will try to make the motivation of the various proofs clear.  The underlying methods are actually quite simple.  Hermite’s method, extended by Niven, uses repeated integration by parts of a product of the relevant function and a carefully selected polynomial.  Lambert’s method uses an iterative method to generate a sequence of rational fraction approximations to tanh x and tan x, from which the results on irrationality can be deduced fairly simply.

I will start (in the next post) with a couple of simple, general theorems, then I will explore Hermite’s and Niven’s methods, and finally Lambert’s method.