Proof that $\log_{10} 2$ is irrational. I am student of high school and have a question from my math book.
Q. Prove that $\log 2$ is irrational.
I had a proof using contradiction as follows
Let
$$\log 2=\frac{p}{q}$$
$$10^\frac{p}{q} = 2$$
$$10^p = 2^q$$
$$(2 \times 5)^p = 2^q$$
$$5^p = 2^{q-p}$$
Now since a power of 2 can never be equal to power of 5 it leads to a contradiction and hence the assumption is false.
This proof is okay, but I was thinking of any other proof of it which could make me more comfortable than that proof.
So I only request a rigorous proof of irrationality of $\log 2$, using simple high school math or else using calculus and other stuff(I don't mind the way you use to solve the problem but interested in alternative proof of the irrationality of $\log 2$.
 A: If $\log_{10}(2)$ is rational, there must be two naturals $n,m$ such that
$$n\log_{10}(2)=m.$$
Taking the antilogarithm,
$$2^n=10^m.$$
But no power of $2$ ends in $0$ !
A: Okay so $\log 2$ is the $x$ so that $10^x = 2$
Therefore $\frac {10^x}2 =1$
So $2^{x-1}5^{x} = 1$
Let's assume $x$ is rational and can be written as a fraction $\frac pq$.
If $x=\frac pq$ then $x -1 = \frac pq - 1= \frac pq -\frac qq = \frac {p-q}q$.
So $2^{\frac {p-q}q}5^{\frac pq} = 1$
So $2^{p-q} 5^p = 1^q = 1$
So $5^p = \frac 1{2^{p-q}}= 2^{q-p}$
If $q>p$ then $q-p > 0$ and $2^{q-p}$ is an integer and a positive integer power of $2$ and greater than $1$ ans $5^p > 1$ and $p$ is a positive power integer of $5$. But a power of $2$ only has $2$ as a prime factor and a power of $5$ only has $5$ as prime factors.  SO this is impossible.
If $q < p$ then $q-p < 0$ and $2^{q-p}= \frac 1{2^{p-q}}$ is a reciprocal of an integer.  So $5^p$ is not an integer and $p$ must be negative. So $5^p = \frac 1{5^{|p|}}$ and $2^{q-p} = \frac 1{2^{|q-p|}}$ so $5^{|p|} = 2^{|q-p|}$. And this is the same case as the paragraph above and impossible as powers of $2$ have only $2$ as prime factors and powers or $5$ only have $5$.
And if $q = p$ then $\frac pq = x = 1$.  That would mean $\log 2 = 1$.  That would mean $10^1 = 2$ or $10 =2$. That isn't true.
None of the options are possible so $\log 2$ is either irrational or simply does not exist.....
Which, I guess, means we have to prove there does exist a real number so that $10^x =2$....  That may have been a basic assumption.  The usually idea is that $10^0 =1 < 2$ and $10^1 > 2$ and $10^{\frac 12} \approx 3.16 > 2$ and if we take it as a general assumption about real numbers and $10^x$ that it doesn't "jump about" there has to be some number in there where $10^x = 2$.  I think that was a basic assumption in the question that $\log 2$ actually exists.
.....
Now this proof is exactly the same as the one you gave.  But maybe I put int in terms from concepts that do follow one after one and a way that follows one's natural thought process.
IMO I prefer the proof as given.  But they are the same.
