Mathematics Asked by yastown on September 16, 2020

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}$$;$ $p,qinmathbb{Z}$ with p and q having no common factors. We write $2^{frac{p}{q}}=3$. Raise both sides to the $q^{th}$ power to get $2^p=3^q$. We have two cases to take care of, $;r=0$ and $rnot = 0$. The first case being $;r=0;$, if $;r = 0$ then p has to be zero because if q was zero then we wouldn’t be able to complete the operation. If $;r = 0$ then we have that $1=3;$ which is a contradiction. For $r not= 0$ we have two different cases, $r>0$ and $r<0$. First we will take care of the $r>0$ case. If $r>0$ then $p,q>0$ and we have $2^p=3^q$ which says an even number is equal to an odd number which is a contradiction. Finally, we take care of the $r<0$ case. If $r<0$ then $r=-frac{p}{q}; p,qinmathbb{Z}^+$ which implies $2^{-frac{p}{q}}=3$$;Rightarrow $$; frac{1}{2^p}=3^q$$;Rightarrow;$$1=2^p3^q $ which is a contradiction because $:6leq2^p3^q:$ and $6notleq 1$.

Thanks y’all for the help.

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 $1$.

Correct answer by fleablood on September 16, 2020

A direct sort of answer $2^r=3 implies r=log_2 3 notin Q$

Answered by Popular Power on September 16, 2020

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.

Answered by cr001 on September 16, 2020

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