Question

Here is a fun exercise:
\[\log_{2}7.\]Prove that the above expression is irrational.

Answer 1

Why is it always that a proof is considered a "fun" exercise?

Answer 2

to try to get someone to try it :)

Answer 3

7 <> 2^p/2^q where p and q are integers ?

Answer 4

It's not immediately obvious 2^p/2^q ≠ 7, but you're on the right track.

Perhaps I should post the proof?

Answer 5

Yeah, that would be helpful

Answer 6

Proof by contradiction:

Assume that the expression is rational. This implies that
\[\log_{2}7=\frac{p}{q},\]\[p, q \in \mathbb{Z},\]\[q \neq 0.\]By simple arithmetic,
\[\log_{2}7=\frac{p}{q},\]\[q\log_{2}7=p,\]\[\log_{2}7^{q}=p,\]\[7^{q}=2^{p}.\]Note that the LHS of this expression is always odd whilst its RHS is always even. This is a contradiction since an odd number cannot be equal to an even number, and therefore,
\[\log_{2}7\]is irrational. ∎

Answer 7

nice, thanks :)

Answer 8

really good

Answer 9

classic proof by contradiction

Answer 10

I hate proof by contradiction