Prove $\log_{4}6$ is irrational

Question

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.

Answer 1

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 $3\nmid4^m$.

Answer 2

Your proof works until you reach $4^m=6^n.$ However, you don't really need to divide by any factor of $2.$

Instead, you can add:

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.