Prove that $\log_2 3$ is irrational!

Question

So, I have been told that for every $x\in\mathbb{N}$, $\ln{x}$ is irrational by using the fact that $e$ is transcendental number.

But, how to prove that $\log_{2}{3}$ is irrational number? My idea is to rewrite the form

$$\log_2 3 = \frac{\ln 3}{\ln 2}$$

But, if both $x, y$ irrational, it isn't necessary that $\frac x y$ irrational. Please, help me!

Suppose $\log_2 3 = \frac{a}{b}$, with $a, b$ integers. Then $2^{a/b} = 3$, so $2^a=3^b$. The LHS of the last equation is even, and the RHS is odd, a contradiction. — Dustan Levenstein, Nov 24 '16 at 15:38
It is NOT true that if $x$ and $y$ are irrational, then $x/y$ is irrational. For example, if $x=2\pi$ and $y=\pi$, then $x$ and $y$ are irrational and $x/y$ is rational. $\qquad$ — Michael Hardy, Nov 24 '16 at 15:52

score 10 · Accepted Answer · answered Nov 24 '16 at 15:51

10

Suppose $\log_2 3 = \dfrac m n$ for some $m,n\in\{1,2,3,\ldots\}.$

Then $2^{m/n} = 3.$

Therefore $2^m = 3^n.$

Therefore an even number equals an odd number.

answered Nov 24 '16 at 15:51

Michael Hardy

1

Thanks ^^ I never thought that way. – fahadh4ilyas Nov 24 '16 at 15:54

Bill Dubuque · Answer 2 · 2016-11-24T17:09:37.353

More generally the logs of primes $\rm\,p_i\,$ are linearly independent over $\,\Bbb Q\,$ since if

$$\rm\ \: c_1 \log\,2 \, +\, c_2 \log\, 3\, + \cdots +\, c_n\log\,p_n =\ 0,\ \: c_i\in\mathbb Q\:,\: $$

then multiplying by a common denominator we can assume all $\rm\ c_i \in \mathbb Z,\:$ so, exponentiating $$\rm\: 2^{\,\large c_1} 3^{\,\large c_2}\cdots p_n^{\,\large c_n} = 1\ \Rightarrow\ c_i = 0\ \text{ for all }\rm\:i$$

OP is the special case $\rm\, n = 2\ $ since $\rm\,c = \log_2 3 = \log 3/\log 2\,\Rightarrow\, c\log 2 - \log 3 = 0$

Prove that $\log_2 3$ is irrational!

2 Answers2

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