# Thread: Prove no rational satisfies 2^x = 3

1. ## Prove no rational satisfies 2^x = 3

Here is the question and i have no idea how to go about it any help would be hugely appreciated.

Prove no rational satisfies

$2^x = 3$

Many thanks

2. Originally Posted by craig in a post he soft deleted and I hard deleted
You would firstly take logs of both sides, $\log_{2} 2^x=\log_{2} 3$.

As $\log_{y} y^x = x$, we can say that:

$x=\log_{2} 3$

Put this into your calculator and you get $x=1.584962501$, not a rational number
On the contrary, 1.584962501 certainly IS a rational number because it is a terminating decimal! $log_2(3)$,which is NOTequal to 1.585962501 is not rational but I suspect it would be necessary to prove that to get credit for this problem!

3. Originally Posted by sebjory
Here is the question and i have no idea how to go about it any help would be hugely appreciated.

Prove no rational satisfies

$2^x = 3$

Many thanks
This is in the prealgebra/algebra section, so i am not sure if a proof like this (a proof by contradiction) will be too complicated. if it is, tell me, we can come up with another.

Assume, to the contrary, that there is a rational number $x$ such that $2^x = 3$.

Then $x = \frac ab$, where $a$ and $b$ are integers, with $b \ne 0$. So we have

$2^{\frac ab} = 3$

$\Rightarrow 2^a = 3^b$

However, this is absurd, since $a$ and $b$ are integers, this equation contradicts the uniqueness of prime factorization (since the left side is always a product of 2's and the right is always a product of 3's).

note that a = b = 0 makes the equation make sense... but b cannot be zero!

4. thankyou proof by contradiction is taught over here in the UK.

5. Originally Posted by sebjory
thankyou proof by contradiction is taught over here in the UK.
oh, you're from the UK, hehe, I was worried. you have a US flag up

i suppose you are also familiar with the uniqueness of prime factorization theorem (you might know it under a different name), but it talks about how each integer can be uniquely expressed as a product of primes with integer powers

6. Yes, we're studying this whole semester on these sorts of theorems about rational numbers (the Completeness axiom etx) and with these come a whole host of proofs...

Do you know any good proof resource where i can learn as many types of proof as possible? (not including Induction since i am fairly happy with my ability to manipulate these sorts of problems)

7. Originally Posted by sebjory
Yes, we're studying this whole semester on these sorts of theorems about rational numbers (the Completeness axiom etx) and with these come a whole host of proofs...

Do you know any good proof resource where i can learn as many types of proof as possible? (not including Induction since i am fairly happy with my ability to manipulate these sorts of problems)
completeness axiom? ...this is prealgebra?!!

i am sure there are resources online that have what you are looking for. i guess wikipedia is always a good place to start, at least for quick reference. but i have never really searched for any such site, as i have several textbooks that cover such material and am fairly happy with them, so i don't know. sorry

8. Originally Posted by Jhevon
This is in the prealgebra/algebra section, so i am not sure if a proof like this (a proof by contradiction) will be too complicated. if it is, tell me, we can come up with another.

Assume, to the contrary, that there is a rational number $x$ such that $2^x = 3$.

Then $x = \frac ab$, where $a$ and $b$ are integers, with $b \ne 0$. So we have

$2^{\frac ab} = 3$

$\Rightarrow 2^a = 3^b$

However, this is absurd, since $a$ and $b$ are integers, this equation contradicts the uniqueness of prime factorization (since the left side is always a product of 2's and the right is always a product of 3's).

note that a = b = 0 makes the equation make sense... but b cannot be zero!
Another argument could be

$2^a$ is always even and $3^b$ is always odd.

Clearly, therefore, $2^a = 3^b$ where a and b are integers is nonsense, giving the contradiction and completing the proof.

9. Originally Posted by Prove It
Another argument could be

$2^a$ is always even and $3^b$ is always odd.

Clearly, therefore, $2^a = 3^b$ where a and b are integers is nonsense, giving the contradiction and completing the proof.
even easier!