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.

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