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(This is from a proof by contradiction, so that's why the equality does not actually hold. Edited for brevity; I don't think I've omitted anything pertinent to my questions.)

[...] The monotonicity of $f$ implies that $2^u<3^v$ if and only if $3^u<6^v$, $u$, $v$ being positive integers. Taking logarithms this means that $\frac{v}{u}>\log_2 3$ if and only if $\frac{v}{u}>\log_3 6$. Since rationals are dense, it follows that $\log_2 3 = \log_3 6$. This can be written as $\log_2 3 = \frac{1}{\log_2 3}+1$, and so $\log_2 3$ is the positive solution of the quadratic equation $x^2−x −1=0$ [...]

  • Question 1

When he concludes that:

Since rationals are dense, it follows that $\log_2 3 = \log_3 6$.

What is the reason for this? Wouldn't this only be true if $\frac{v}{u}$ goes to $0$?

  • Question 2

[$\log_2 3 = \log_3 6$] can be written as $\log_2 3 = \frac{1}{\log_2 3}+1$.

This one is the most confusing. How did he jump from $x=y$ to $x = \frac{1}{x}+1$?

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@Austin: That's why the question starts out by explaining that this is from a proof by contradiction. – joriki Jun 24 '11 at 0:28
On Question 1: No, $v/u$ doesn't have to go to $0$ -- it follows that $\log_2 3$ and $\log_3 6$ are equal because if they were not, we could find a rational between them, which would contradict one of the two inequalities. – joriki Jun 24 '11 at 0:31
@joriki why would it contradict the inequalities? — wouldn't, say, $\frac{v}{u}>\log_2 3>\log_3 6$, or vice versa, hold without contradicting either inequalities? – fakaff Jun 24 '11 at 3:16
That's not between them. – joriki Jun 24 '11 at 4:13
But isn't that still consistent with the two inequalities? That's what I don't understand: How does either inequality require that there not be a rational between them? — For example, if I say $x>2$, $x>1$, then $x>2>1$ is perfectly OK; one would not then assume that $1=2$. Then why is it not OK for $\frac{v}{u}>\log_2 3>\log_3 6$? — How did you arrive at the conclusion that there cannot be a rational between them? – fakaff Jun 24 '11 at 4:34
up vote 2 down vote accepted

Here is help for Question 2:
First note that by the change of base formula, we have that $$\log_{3}6=\frac{\log_{2}6}{\log_{2}3}.$$

Also, $$\log_{2}6=\log_{2}3+\log_{2}2.$$

Combining the above results should resolve all the issues.

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Got it, thanks. – fakaff Jun 24 '11 at 0:49
you are welcome...:) – Nana Jun 24 '11 at 1:07

Nana's answer is fine. Alternatively, $${1\over\log_23}+1=\log_32+1=\log_32+\log_33=\log_36$$

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