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I tried to prove $\ln(2)$ is irrational using the method of contradiction knowing that $e$ is irrational:

Let $$\ln2=\frac{p}{q}$$ be in simplest form where $p$ and $q$ are positive integers.

Now we have $$e^{\frac{p}{q}}=2$$ $$2^q=e^p$$ Now $2^q$ is always an even positive integer.

How can we reason that $e^p$ can never be an integer?

Community
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Umesh shankar
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1 Answers1

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Well, if one accepts the fact that $e$ is transcendental over $\Bbb Q$, then $e^p$ cannot be rational, since if

$e^p = r \in \Bbb Q, \tag 1$

then $e$ would be a root of

$x^p - r \in \Bbb Q[x]; \tag 2$

i.e., $e$ would be an algebraic, not a transcendental, number.

Robert Lewis
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