Show $e^x$ is irrational for rational $x \neq 0$

Question

I want to show that if $x$ is rational and nonzero then $e^x$ is irrational.

Clearly $e^{\frac{r}{s}} = \frac{p}{q} \Rightarrow q^s e^r = p^s$, but this doesn't seem helpful. The usual proof that $e$ is irrational doesn't look like it can be extended either.

You get $e^r=p^s/q^s$ which is a rational number by assumption. Note that this time, $r$ is an integer, and may even be assumed positive. So you're left with proving that $e^r$ is irrational for all natural numbers $r$, instead of rational numbers $r$. That seems a bit less daunting to me. — Arthur, Jan 12 '16 at 16:11
See here page 27: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.167.5285&rep=rep1&type=pdf — Gregory Grant, Jan 12 '16 at 16:12
Hint: It suffices to show $e^n$ is irrational for all integer exponents $n\neq 0$. — hardmath, Jan 12 '16 at 16:20

score 0 · Answer 1 · answered Jan 12 '16 at 16:14

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You are on the right track.

Knowning that $e$ is transcendental, the algebraic equation $q^sz^r-p^s=0$ cannot have $e$ as a root.

answered Jan 12 '16 at 16:14

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Yeah but how do you know $e$ is transcendental? See Theorem 1 page 30 of this book. They spend pages to prove this, so it can't be that easy: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.167.5285&rep=rep1&type=pdf – Gregory Grant Jan 12 '16 at 16:16
Transcendence of $e$ is taken for granted and we show that it implies the irrationality of $e^x$ for $x$ rational. – Jan 12 '16 at 16:24
I don't see the transcendental assumption in the statement of the question. – Gregory Grant Jan 12 '16 at 16:27
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So: maybe (?) the point of the question is that proving $e^x$ irrational for all rational $x$ is easier than proving $e$ transcendental. – GEdgar Jan 12 '16 at 16:31

Show $e^x$ is irrational for rational $x \neq 0$

1 Answers1