Claim: $\log_2(2015)$ is irrational

Proof: Let's assume it is rational. Then:

$$\log_2(2015)=a/b, \quad a,b\in \mathbb Z \setminus \{0\}, b\neq 0$$

Question 1: is it okay how I chose a,b?

then we get: $2^{a/b}=2015 \Leftrightarrow 2^a=2015^b$

Now we know: $2015=31\cdot13\cdot5$ so we get:

$$2^a=(31\cdot13\cdot 5)^b=13^b \cdot 13^b \cdot 5^b \tag{$\ast$}$$

Since we know that any integer can be represented uniquely by a multiplication of primes, we would have two ways of representing the same number in such a way. So it wouldn't be unique. Thus, it isn't rational. Thus, it is indeed irrational.

Question: I don't like the step ($\ast$). I see that the argument at the end is indeed what we want but I'm not sure if it really holds, since we have one exponent for several different primes. E.g. one could also get: $2^p=2^q\cdot 31 \cdot 13 \cdot 5$ Which makes it absolutely clear.

Question: Does ($\ast$) hold?

  • $\begingroup$ This is the standard proof that irrational logarithms are irrational. $\endgroup$ – Arthur Jan 14 '18 at 17:13
  • $\begingroup$ At ($\ast$), you are simply invoking the fundamental theorem of arithmetic. That is the usual proof. $\endgroup$ – Xander Henderson Jan 14 '18 at 17:13
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    $\begingroup$ Seems like everything you do is perfect! But I would instead say that $2^a=31^b\cdot 13^b\cdot 5^5$ forces $a=b=0$, which is a contradiction since $b\neq 0$ $\endgroup$ – cansomeonehelpmeout Jan 14 '18 at 17:14

What you did is correct (other than at a passage where you wrote $13$ instead of $31$). But it is simpler to say that $2^a$ is even and $2015^b$ is odd.

  • $\begingroup$ Thanks. Yeah that would be another way $\endgroup$ – xotix Jan 14 '18 at 17:14

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