1. ## Proofs.

Prove that if n ∈ Z and log2 n is rational, then log2 n is an integer.

2. ## Maybe

Originally Posted by pearlyc
Prove that if n ∈ Z and log2 n is rational, then log2 n is an integer.
I am misunderstanding or you are trying to prove the opposite...but $ln(2)$ is transcendental therefore irrational...

3. Honestly, I really have no idea what the question is asking for! I do have a clue that it's related to proof by contradiction though. Anyone can help?

4. How can you prove something that's not true? Are you sure what you typed the question correctly?

Prove that if $n \in \mathbb{Z}$ and $\log_{2} n$ is rational, then $\log_{2} n$ is an integer.
n = 5. $\log_{2} 5 = 2.3219 ...$. Not an integer.

5. Hi! Yes, I have double checked the question and it is exactly what I typed. It is an assignment question. Gotta know how to do this!

6. Then I don't know what we can do. Can't prove something that isn't true as I've shown for $n = 5$ .

7. Apparently, you are supposed to approach the question with 'Proof by Contradictions' but I'm not very sure of myself on how to apply proof by contradiction as I don't quite understand what is it about either.

8. ## hmm

Originally Posted by pearlyc
Prove that if n ∈ Z and log2 n is rational, then log2 n is an integer.
I would start with $n=2^{\frac{p}{q}}$ then work from there

9. Yes, that's exactly how I started it as well.

And I am stucked at,

n^q = 2^p.

I don't know where to go from here.

10. Hello,

Yep, n^q=2^p

n=a/b, as n is in Z.

-> a^q=(2b)^p

Now, we have to show that there exists k in Z such as p=kq

Perhaps it can help..

11. Thanks, but I still don't know how to proceed from there. Sighhh. This is so difficult