-4
$\begingroup$

Possible Duplicate:
About irrational logarithms

Please help proving that $\log_{10}(2)$ is irrational.

  • 3
    I'm going to downvote this question based on the complete lack of information. You haven't said why you're interested, what context you're working in, or what you have tried.2010-11-12

3 Answers 3

12

Hint: if $10^{a/b} = 2$ then $10^a = 2^b$.

4

Prove by contradiction. So start by assuming

$$\log_{10}{2}=\frac{m}{n}$$, where $m$ and $n$ are integers.

$$10^\frac{m}{n}=2$$

$$10^m=2^n$$

and derive a contradiction.

  • 0
    Qwirk I wish you would not give such full responses to incomplete questions. Try giving a hint.2010-11-12
  • 1
    @a little don - noted, however think the hard work is still to come.2010-11-12
2

We argue by contradiction. Suppose $\log_{10} 2 = \frac{p}{q}$ is rational with $q > p > 0$. Then $ 2^{q} = 10^{p} = 2^{p} \ 5^{p}$, so $2^{q - p} = 5^{p}$. Since $q > p$ and $p > 0$, it follows that $2^{q - p} \equiv 0 \mod 5$, which is impossible since no power of $2$ ends in $0$ or $5$. Hence $\log_{10} 2$ is irrational.

This method generalizes easily to prove the irrationality of many reals of the form $\log_{b} \ a$ for $a, b \in \mathbb{N}$.

  • 0
    You didn't actually use gcd(p,q)=1, as far as I can see.2010-11-14
  • 1
    thanks. I edited the post.2010-11-14