# prove $\log_{10} (2)$ is irrational [duplicate]

Possible Duplicate:
About irrational logarithms

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

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## marked as duplicate by J. M., Aryabhata, Robin Chapman, Pete L. Clark, kennytmNov 15 '10 at 7:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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. – Carl Mummert Nov 12 '10 at 3:39

## 3 Answers

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

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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.

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Qwirk I wish you would not give such full responses to incomplete questions. Try giving a hint. – futurebird Nov 12 '10 at 3:55
@a little don - noted, however think the hard work is still to come. – Juan S Nov 12 '10 at 3:59

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 $a, b \in \mathbb{N}$.

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You didn't actually use gcd(p,q)=1, as far as I can see. – Oscar Cunningham Nov 14 '10 at 20:01
thanks. I edited the post. – user02138 Nov 14 '10 at 20:15