# Error in proving $e + \pi$ is irrational.

From a proofs class:

Note that $2^{1/n}$ is irrational for all $n>1$, since 2 is prime. Therefore, if $m$ is a positive, rational number, then $m*2^{1/n}$ is irrational for all $n>1$. Let $m = 5$ (although there are other values that will work). Observe that there exists an interval $[a,b], 1<a<b$ for a over $n$ such that $5*2^{1/a} > e +\pi$ and $5*2^{1/b} < e + \pi$. Since $5*2^{1/n}$ is continuous, then there exists an $l$ in $[a,b]$ such that $5*2^{1/l} = e+\pi$. Since $5*2^{1/n}$ is irrational for all $n>1$, it follows that $e + \pi$ is also irrational.

I suspect the error is in the "$2^{1/n}$ is irrational" statement, although the proof we had for that seemed to cover all $n>1$ (not to mention much of this could have been reduced, but like the others we got there's a lot of "fluff" to make the problems more complicated).

I appreciate all and any help. Thank you kindly!

• The only thing I've found so far is a slight working error. $5*2^{\frac{1}{n}}$ is only continuous besides at n=0. – pie314271 Feb 8 '17 at 0:34
• $2^{1/n}$ is always irrational when $n$ in an integer greater than $1$. It is not true for all real numbers. For example $n = \frac{1}{\log_2(3)}$ is irrational, but $2^{\frac{1}{n}} = 3$ is rational. – Winther Feb 8 '17 at 0:34
• Oh, that's what was wrong.. nice! – pie314271 Feb 8 '17 at 0:34

The error is "sort of" in that part. The use of the variable $n$ in the statement "$2^{1/n}$ is irrational for $n > 1$" suggests that integer $n$ are intended, in which case the statement is true. But if $n$ is permitted to be itself irrational - as $l$ is at the end of the "proof" - then $n = \frac{1}{\log_{2}1.5}$ is a clear counterexample (since then $2^{1/n} = 1.5$).
The problem is that $m∗2^{\frac{1}{n}}$ is irrational for all rational $n>1$ . There are irrational $n$s for which $2^\frac{1}{n}$ is rational. Some multiple of log 2 would work. Therefore your $l$ may be one of those irrational exponents that rationalize the result.