CTK Insights

18 Apr

A Representation of Rational Numbers

One of the most amusing instances of indirect proof - proof by contradiction - is the establishment of the existence of two irrational numbers a and b such that a^b is rational. Indeed, \sqrt{2} is irrational. Then if \sqrt{2}^\sqrt{2} is rational the problem is solved with a=b=\sqrt{2}. Otherwise, it is solved with a=\sqrt{2}^\sqrt{2} and b=\sqrt{2} because then a^b=2.

Stan Dolan has recently proved that practically all rational numbers in the interval (1/e,\infty) are in the form a^a, where a is irrational.

The claim is really elementary if we consider a smaller interval (1,\infty). On this interval function f(x)=x^x is clearly monotone increasing. To prove that this is also true for x\gt 1/e requires a little calculus: f'(x) = (1+\mbox{ln}x)x^x and is positive for x\gt 1/e.

Function f maps continuously interval (1/e,\infty) onto interval I=({(1/e)}^{1/e},\infty). Let r\in I be rational. There is a such that r=a^a. Assume a is also rational. More accurately, assume r =\frac{b}{c} and a=\frac{n}{m}, both fractions in lowest terms. Then {(n/m)}^{n/m}=b/c. This implies

n^{n}c^{m}=m^{n}b^m.

If m=1, r=n^n. We shall show that this is an exceptional case. In other words, an assumption that m\gt 1 leads to a contradiction. Since m and n are mutually prime, every factor of m^n divides c^m. The converse is also true because c and b are also mutually prime. It follows that

c^m=m^n.

Assume p is a common prime factor of c and m. We'll write

c=p^{i}k and m=p^{j}l

where both k and l are prime to p. Substitution gives

p^{im}k^m=p^{jn}l^n.

In particular, im=jn. Further, p^{j}|m so that p^{j}|jn. But m and n are assumed to be mutually prime, implying p^{j}|j. We thus led to

2^{j}\le p^{j}\le j.

But this is a contradiction, for, for all positive integer j, 2^{j}\gt j.

To sum up, the identity a^a=r, with a rational r is possible in two cases only: when a is an integer or irrational. Put differently, every rational number in the interval I, except for those in the form n^n, with integer n, are in the form a^a with a irrational.

References

  1. S. Dolan, A rational number of the form a^a with a irrational, The Mathematical Gazette, v 96 n 535 March 2012, 106-108

3 Responses to “A Representation of Rational Numbers”

  1. 1
    Matrix67: My Blog » Blog Archive » 经典证明:几乎所有有理数都是无理数的无理数次方 Says:

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    经典证明:几乎所有有理数都是无理数的无理数次方 | Mining Time (Digests) Says:

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    经典证明:几乎所有有理数都是无理数的无理数次方 | 好豆饼 Says:

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