in this case log means $log_{e}$.
What I've learned before is that to prove a number irrational use a contradiction proof by assuming it's rational. Do I do that in this case? If so, how do I start the proof?
Asked
Active
Viewed 153 times
1
-
2Are you allowed to use the fact that $e$ is transcendental? – carmichael561 Sep 24 '16 at 20:59
2 Answers
2
As you said. Assume that $$\log q=\frac ab,$$ with $a,b\in\mathbb Z\setminus\{0\}$. Then $q=e^{a/b}$. Then $e=q^{b/a}$. So $e$ would be a root of $$ p(x)=x^a-q^b. $$ But $e$ is transcendental (i.e., not algebraic).
Martin Argerami
- 205,756
0
This is equivalent to showing that $e^{x}$ is irrational if $x$ is a non-zero rational number. There are many proofs available for this property of $e$ the first of which was given by Lambert based on continued fraction expansion of $\tanh x$ (the same technique yields irrationality of $\pi$). Another common proof was given by Ivan M. Niven in his wonderful book Irrational Numbers. And a totally different proof was given by C. L. Siegel. As mentioned in Martin Argerami's answer the result is also a simple corollary of the fact that $e$ is transcendental.
Paramanand Singh
- 87,309