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About irrational logarithms
Please help proving that $\log_{10}(2)$ is irrational.
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3$\begingroup$ 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. $\endgroup$– Carl MummertNov 12, 2010 at 3:39
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3 Answers
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Hint: if $10^{a/b} = 2$ then $10^a = 2^b$.
answered Nov 12, 2010 at 3:40
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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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$\begingroup$ Qwirk I wish you would not give such full responses to incomplete questions. Try giving a hint. $\endgroup$ Nov 12, 2010 at 3:55
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1$\begingroup$ @a little don - noted, however think the hard work is still to come. $\endgroup$– Juan SNov 12, 2010 at 3:59
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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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$\begingroup$ You didn't actually use gcd(p,q)=1, as far as I can see. $\endgroup$ Nov 14, 2010 at 20:01