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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.

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    $\begingroup$ 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? $\endgroup$ – Did Dec 13 '18 at 16:13
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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 $3\nmid4^m$.

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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.

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