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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, KennyTM Nov 15 '10 at 7:48

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What have you tried? What's the context? (ie. what theorems can you use?) –  a little don Nov 12 '10 at 3:31
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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 3

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. –  a little don Nov 12 '10 at 3:55
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@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
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thanks. I edited the post. –  user02138 Nov 14 '10 at 20:15

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