When is a real number irrational?

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).  


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