I'll start off by requesting no one gives me a solution to this problem. I would just like some direction. Please let me know if I've made a mistake where I've gone wrong and if possible let me know 'why' what I have done is wrong.
Show that there is no rational number $r$ satisfying $2^r = 3$.
If $r$ is rational then we can write $r = a/b : a,b \in$ integers, and $b\neq0$
so, $$ 2^{a/b} = 3,$$ $$log_2(2)^{a/b} = log_2(3)$$ $$a/b=log_2(3)$$ $$a=b*log_2(3)$$
This means $a$ is a multiple of $log_2(3)$ so we can write, $a = m*log_2(3)$ where $m$ is an integer and $m \neq b $
$$m*log_2(3) = b*log_2(3)$$
dividing both sides by $log_2(3)$ gives $m = b$ which is a contradiction. (I think). The logic behind $m \neq b$ is that m can be zero.
This entire thing I've done looks almost absurd though.
Anyways, thanks in advance!
No, there is no contradiction in $m=b.$ The correct proof involves no log, and instead finds a contradiction in $2^a = 3^b.$