Rational and Irrational Values of Powers

Here’s a cute little existence argument that I was exposed to as an undergraduate and have never forgotten. It shows that there must be irrational positive reals \alpha and \beta for which \alpha^\beta is rational. Furthermore, it’s done by showing that one of two very specific pairs of numbers (either \alpha=\sqrt{2}, \beta=\sqrt{2} or \alpha={\sqrt{2}}^{\sqrt{2}}, \beta=\sqrt{2}) satisfies the condition, but without establishing which of the two pairs satisfies the condition.

The argument is elementary. First look at the case \alpha=\sqrt{2}, \beta=\sqrt{2} If \alpha^\beta = {\sqrt{2}}^{\sqrt{2}} is rational, we have an example, and we’re done. If not then {\sqrt{2}}^{\sqrt{2}} must be irrational, and so \alpha^\beta where \alpha = {\sqrt{2}}^{\sqrt{2}} (assumed to be irrational) and \beta=\sqrt{2} is an example of an irrational raised to an irrational. And it’s rational. In fact, it’s 2, because \alpha^\beta =({{\sqrt{2}}^{\sqrt{2}}})^{\sqrt{2}} = {{\sqrt{2}}^{{\sqrt{2}}{\sqrt{2}}}} = {\sqrt{2}}^2 = 2

It’s a great little elementary existence argument, and I think it’s worth being exposed to it. But, it’s also worth knowing that with (much) more advanced techniques we can actually say which of the two choices satisfies the condition (it’s the second choice \alpha={\sqrt{2}}^{\sqrt{2}}, \beta=\sqrt{2}). The technique I’m thinking of is the Gelfond-Schneider theorem, which implies that if \alpha, \beta are positive algebraic irrationals, then \alpha^\beta is irrational: the actual theorem uses weaker conditions on \alpha and \beta and reaches a stronger conclusion about \alpha^\beta And, of course, \sqrt{2} is an algebraic irrational, so the result applies.

And, it’s also worth knowing that there are different examples that can be established using elementary techniques: for example \alpha = \sqrt{2}, \beta = 2 log_2 (3) for which \alpha^\beta =3

Just for fun, here are elementary examples of all possible choices of rational and irrational choices in taking powers:

\alpha^\beta rational irrational
{rational\ }^{rational} 2^3 = 8 2^{1/2} = \sqrt{2}
{irrational\ }^{rational} {\sqrt{2}}^2 = 2 {\sqrt{2}}^3 = 2 \sqrt{2}
{rational\ }^{irrational} 2^{log_2 3} = 3 2^{({log_2 3})/2} = \sqrt{3}
{irrational\ }^{irrational} {\sqrt{2}}^{2 log_2 3} = 3 {\sqrt{2}}^{log_2 3} = \sqrt{3}

All that you need to prove the above results is that the three numbers \sqrt{2}, \sqrt{3}, log_2 3 are irrational, and there are elementary arguments for establishing each of these. For completeness, I’ll sketch proofs for the three results.

\sqrt{2} is irrational. Assume it’s not. Then there are integers m,n for which \sqrt{2} = \frac{m}{n} i.e. for which m^2 = 2 n^2 (and, of course, n \ne 0). We show that both m,n have to be even numbers, and that’s impossible for all m,n with n \ne 0 satisfying \sqrt{2} = \frac{m}{n} because by removing common factors from any given m,n we arrive at a simplified m,n for which one of the two numbers is odd. To show that both m,n are even, we first have that m is even because m^2 is even (it’s 2 n^2) and writing m = 2 k we have that 4 k^2 = 2 n^2 and so 2 k^2 = n^2 which tells us that n^2 is even and so n is even. QED

The above argument has been known for over two thousand years, and has been repeated in numerous places, including many popular mathematics books. I would recommend G.H. Hardy’s “A Mathematician’s Apology” if you want to find another source: Hardy calls it “Pythagoras’ proof”. With extremely minor modifications, the same argument can be used to prove that \sqrt{3} is irrational, as shown below.

\sqrt{3} is irrational. Assume it’s not. Then there are integers m,n for which \sqrt{3} = \frac{m}{n} i.e. for which m^2 = 3 n^2 (and, of course, n \ne 0). We show that both m,n have to be divisible by 3, and that’s impossible for all m,n with n \ne 0 satisfying \sqrt{3} = \frac{m}{n} because by removing common factors from any given m,n we arrive at a simplified m,n for which one of the two numbers is not divisible by 3. To show that both m,n are divisible by 3, we first have that m is divisible by 3 because m^2 is divisible by 3 (it’s 3 n^2) and writing m = 3 k we have that 9 k^2 = 3 n^2 and so 3 k^2 = n^2 which tells us that n^2 is divisible by 3 and so n is divisible by 3. QED

log_2 3 is irrational. Assume it’s not. Then there are integers m,n for which log_2 3 = \frac{m}{n} i.e. for which 3  = 2^{m/n} i.e. for which 3^n  = 2^m (and, of course, n \ne 0). We show that both m,n have to be zero, and that contradicts n \ne 0. If m,n are both non-negative, then 3^n is an odd number and the only way that 2^m can be equal to an odd number is if 2^m = 1 and so m=0, from which it immediately follows that 3^n = 1 and so n=0 and we’re done. The other three (partially overlapping) cases (m non-negative and n non-positive, m non-positive and n non-negative, m non-positive and n non-positive) all can be ruled out using similar reasoning (on 2^a 3^b = 1 for a,b non-negative integers in some cases). QED

And, while I’m on the subject, here are some elementary examples for \alpha + \beta and \alpha \cdot  \beta

\alpha + \beta rational irrational
{rational\ }+{\ rational} 2+3=5 cannot happen
{irrational\ }+{\ rational} cannot happen ({\sqrt{2}}-1) + 1 = \sqrt{2}
{rational\ }+{\ irrational} cannot happen 1+({\sqrt{2}}-1) = \sqrt{2}
{irrational\ }+{\ irrational} (3-\sqrt{2})+ \sqrt{2} = 3 \sqrt{2} + \sqrt{2} = 2\sqrt{2}
\alpha \cdot \beta rational irrational
{rational\ }\cdot{\ rational} 2\cdot3=6 cannot happen
{irrational\ }\cdot{\ rational} \sqrt{2} \cdot 0 = 0 \sqrt{2} \cdot 2 = 2\sqrt{2}
{rational\ }\cdot{\ irrational} 0 \cdot \sqrt{2} = 0 2 \cdot \sqrt{2} = 2\sqrt{2}
{irrational\ }\cdot{\ irrational} (\sqrt{3} -1 )( \sqrt{3} +1)= 2 \sqrt{2} \cdot \sqrt{3} = \sqrt{6}

Author: Walter Vannini

Hi, I'm Walter Vannini. I'm a computer programmer and I'm based in the San Francisco Bay Area. Before I wrote software, I was a mathematics professor. I think about math, computer science, and related fields all the time, and this blog is one of my outlets. I can be reached via walterv at gbbservices dot com.

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