Prove there is no rational number r such that $2^r = 3$

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.

3 Answers

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

Add your own answers!

Related Questions

How can you derive an algorithm for dividing natural numbers?

0  Asked on December 3, 2020 by honza-prochazka


Expansion of the Frobenius norm

1  Asked on December 3, 2020 by wuannnn


On the definition of an algebra

1  Asked on December 3, 2020 by hakuna-matata


What is a polynomial approximation?

3  Asked on December 2, 2020 by brilliant


Permutation involving different outcomes

1  Asked on December 2, 2020 by adam-mansfield


Ask a Question

Get help from others!

© 2022 All rights reserved. Sites we Love: PCI Database, MenuIva, UKBizDB, Menu Kuliner, Sharing RPP, SolveDir