# Prove $\log_{4}6$ is irrational

Thanks for taking the time to verify my approach and as well as my answer.

Background:

• B.S. in Business from a 4-year university taking CS courses online
• I would like some help with a basic proof from MIT's 6.042J Mathematics for Computer Science course.

The question is: Prove $$\log_{4}6$$ is irrational.

We prove the contradiction.

• Suppose $$\log_{4}6$$ is rational (i.e. a quotient of integers) $$\log_{4}6 = m/n$$
• So we must have m, n integers without common prime factors such that $$4^{m/n} = 6$$
• We will show that m and n are both even $$(4^{m/n})^{n} = 6^{n}$$
• So $$4^{m} = 6^{n}$$
• We then divide the two base numbers by their common factor, $$2$$, which gives us:

$$2^{m} = 3^{n}$$

• Since the product of two even numbers must be even AND the product of two odd numbers must be odd, $$2^{m}$$ and $$3^{n}$$ are not equivalent and therefore $$m/n$$ must not be rational.

Q.E.D. We conclude that $$\log_{4}6$$ is irrational.

• Until $4^m=6^n$, you are ok, but after that you go astray. To continue on the path you started on, from $4^m=6^n$, deduce $2^{2m}=2^n3^n$ hence $2^{2m-n}=3^n$. And now, how would you conclude? – Did Dec 13 '18 at 16:13

It's fine until you reach that equality $$4^m=6^n$$. But you can't just divide by their common factor $$2$$. Are you dividing by $$2^m$$ or by $$2^n$$?
You can say that, since $$n>0$$, then $$3\mid6^n$$. But $$3\nmid4^m$$.
Your proof works until you reach $$4^m=6^n.$$ However, you don't really need to divide by any factor of $$2.$$
Since $$\log_46$$ is positive, as 6 is greater than 4, $$m$$ and $$n$$ are positive integers. This means that the right hand side ($$6^n$$) is divisible by $$3.$$ However, the left hand side, $$4^m$$ can be prime factorized as $$2^{2m},$$ and therefore is not divisible by $$3,$$ meaning that $$4^m\neq 6^n,$$ yielding contradiction.