Prove that $\log_2 3$ is irrational

Question

Prove that $\log_2 3$ is irrational

Seemingly simple homework assignment. Was never the best with logarithms, how would I go about proving?

Answer 1

Let $\log_23 = p/q$ where $p\in\mathbb{Z}$ and $q\in\mathbb{N}$ (since surely $\log_23 > 0$ you may directly assume that $p\in\mathbb{N}$ as well.)

Now it must hold

$$ 2^p = 3^q$$

But note that one side is even and the other one is odd!

Hence $\log_23$ is not rational!

Answer 2

You can't because it isn't. Assume $\log_2(3)=\frac pq$, then$ 3=2^{\frac pq}, 3^q=2^p$ so???

Answer 3

Theorem $\ \color{blue}{\gcd(a,b)=1}\ \Rightarrow\ \log_{\,b}(a)\not\in \Bbb Q,\,\ $ for $\ \,1<a,b\in \Bbb N$

Proof $\ $ Else $\, \log_{\,b} a = c/d\,\Rightarrow\, a = b^{c/d}\Rightarrow a^\color{#c00}d = b^\color{#c00}c\,$ contra $\,\color{blue}{\gcd(a,b)=1},\,$ since $\,\color{#c00}{d,c} \ge 1.$

Answer 4

Assume, $x := log_2(3) = \frac{a}{b}$

As x > 0 , we can assume that a and b are positive integers.

Now $$2^x=3$$

leads to $$2^\frac{a}{b}=3$$

Therefore $$2^a=3^b$$

If a and b are positive, the left side is even, the right side odd. A clear contradiction.