Prove there is no rational number r such that $2^r = 3$ Prove there is no rational number r such that $2^r = 3$.
I am wondering if my proof is correct.
$\mathbf{Proof:}$ We will provide a proof by contradiction. Assume there is a rational number r such that $2^r = 3$. This means by definition $r=\frac{p}{q}$$\;$ $p,q\in\mathbb{Z}$ with p and q having no common factors. We write $2^{\frac{p}{q}}=3$. Raise both sides to the $q^{th}$ power to get $2^p=3^q$. We have two cases to take care of, $\;r=0$ and $r\not = 0$. The first case being $\;r=0\;$, if $\;r = 0$ then p has to be zero because if q was zero then we wouldn't be able to complete the operation. If $\;r = 0$ then we have that $1=3\;$ which is a contradiction. For $r \not= 0$ we have two different cases, $r>0$ and $r<0$. First we will take care of the $r>0$ case. If $r>0$ then $p,q>0$ and we have $2^p=3^q$ which says an even number is equal to an odd number which is a contradiction. Finally, we take care of the $r<0$ case. If $r<0$ then $r=-\frac{p}{q}\; p,q\in\mathbb{Z}^+$ which implies $2^{-\frac{p}{q}}=3$$\;\Rightarrow $$\; \frac{1}{2^p}=3^q$$\;\Rightarrow\;$$1=2^p3^q $ which is a contradiction because $\:6\leq2^p3^q\:$ and $6\not\leq 1$.
Thanks y'all for the help.
 A: You need to consider if $r < 0$.  And you need to redo if $r = 0$ correctly.
If $r > 0$ then there are $p, q \in \mathbb Z^+$ where $r =\frac pq$ and, although we can claim $p, q$ have no common factors that is not relevant or necessary.  Your argument was PERFECT.  $2^{\frac pq} =3 \implies 2^p = 3^q$ but LHS is even and RHS is odd.  Beautiful!
If $r = 0$ you kind of botch it.  You say  $2^0 = 3^q$  so $3^q = 1$ is odd which.... is not a contradiction.  More to the point:  If $r = 0$ then $2^r = 2^0=1$ which.... is not equal to $3$  That's all there is to it.
And to consider $r < 0$, if $r < 0$ then there are $p,q \in \mathbb Z^+$ so that $r =-\frac pq$ so $2^{-\frac pq} = \frac 1{2^{\frac pq}} = 3$ so raise both sides to the $q$ power and get $\frac 1{2^p} = 3^q$ and LHS is less than $1$ while RHS is more than $1$.
A: There are several logical mistakes in your argument:
($1$) $2^p=3^q$ does not imply $3^q$ is a multiple of $2$. You need to consider the case $p=0$ and $p \neq0$ separately.
($2$) "In the case that $2^p$ is odd we have an even number being equal to an odd number which is also a contradiction"
This statement is not correct. When $p=0$, we have an odd number equals an odd number. The fallacy comes from the assumption in ($1$)
Edit: as comment by Graham Kemp points out, you also need to prove the case where $p<0$ and $q>0$, which is trivial but need to be stated.
A: A direct sort of answer $2^r=3 \implies r=\log_2 3 \notin Q$
