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Hello guys it would be a massive help for me if you could take a look at my proof and point out any errors. I want to know whether its coherent and whether I need more or less detail.

Suppose $2^r = 3$ where $r$ is some rational number. Thus we can express $r$ as $p/q$ for positive integers $p,q$ (this is because $r$ cannot be negative otherwise $2^r<1$, $q$ cannot be $0$ for $r$ to be defined, and for $p=0$, we have $r=0, $ and $2^0=1 \neq 3)$. Raising both sides of the equation to the power of $q$, we have $2^p =3^q$. Since $2$ and $3$ are co-prime, it follows that $2$ raised to the power of any positive integer can never be a multiple of $3$, so $2^p\neq 3^q$. However it directly follows from the premise $2^r=3$ that $2^p=3^q$. We have arrived at a contradiction and thus it is not possible for any rational number to fulfill the equation $2^r = 3$.

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    $\begingroup$ A perfectly valid proof. A slight improvement is to say that $p,q$ can assumed to be both positive (they could also be both negative, but we can than multiply them both with $-1$). That neither of them can be $0$ ($q=0$ is forbidden anyway) and the signs cannot differ follows immediately from $r>0$ $\endgroup$
    – Peter
    Commented Jan 16, 2022 at 10:19
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    $\begingroup$ But @Peter, those concerns are already addressed in the post or am I overlooking something? $\endgroup$
    – Manatee Pink
    Commented Jan 16, 2022 at 10:26
  • $\begingroup$ @ManateePink I am not the one wanting to be this accurate ! I would accept the claim that a rational number $r>0$ can be represented as a fraction $\frac{p}{q}$ with positive integers $p,q$ as obvious. But the author apparently wants to be a little more formal. $\endgroup$
    – Peter
    Commented Jan 16, 2022 at 11:40

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I would say your proof is perfectly fine. If you want to be really pedantic, you might want to elaborate on why 2 and 3 are coprime, but I personally feel like that would be overkill.

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  • $\begingroup$ In fact that would be like shooting rockets on ants. The only prime factor of $2^p$ is $2$ and the only prime factor of $3^q$ is $3$, this is obvious and need not be pointed out. Moreover, the argument can be made even simpler : $2^p$ is even and $3^q$ is odd and there is no number being both even and odd. $\endgroup$
    – Peter
    Commented Jan 16, 2022 at 11:42
  • $\begingroup$ One says that something is an overkill when one uses a very powerful/advanced/deep theorem to prove a rather simple statement. I think that's not what you meant. But I agree on that proving that $2$ and $3$ are comprime would be an exaggeration $\endgroup$
    – jjagmath
    Commented Mar 22, 2022 at 23:21

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