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Prove that the logarithm of 3 base 10 is irrational

The Fundamental Theorem of Arithmetic is that every integer is a product of primes.

So far I have,

Suppose $\log_{10}(5)$ is rational. Then suppose $\log_{10}(5) = \frac {p}{q}$ for some positive integers $p$ and $q$ with $\frac {p}{q}$ in lowest terms and $p< q$. Exponentiating both sides using 10 as the base we get, $5=10^{p/q}$. Take both sides to the qth power. We get $5^q=10^p=2^p*5^p$. Then we get $5^{q-p}=2^p$.

But I'm not sure if this has anything to do with the Fundamental Theorem of Arithmetic.

If you have another way of doing this that would be great too.

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    Is it because since each integer has a unique prime factorization, but the above equality shows otherwise (its a number which is a power of$5$and 2). So there are no integers satisfying this. Hence, contradicting log10(5) is rational2012-12-09

2 Answers 2

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$5^{q-p}$ is odd and therefore $p=0$. Hence $5^{q-p}=1$, thus $q-p=0$ therefore $q=0$ contradicting the fact $q$ is nonzero

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Hint: What FTA actually says is that every integer can be written uniquely as a product of primes. Given this, think about what the statement $5^q = 2^p\times5^p$ tells us.

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    Well the point is that the product *is* unique, we know that from FTA. This means that the number of $2$s and $5$s must be the same on each side. Hence $p = 0$ (by matching the $2$s) and $p = q$ (by matching the $5$s) so $p=q=0$. But this is a contradiction of our assumptions about $p$ and $q$ since $q$ is non-zero. Does that help?2012-12-10