rational Briggsian logarithms of integersRationalBriggsianLogarithmsOfIntegers2007-12-05 13:56:282008-06-13 13:03:34Theoremirrational% this is the default PlanetMath preamble. as your knowledge
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\DeclareMathOperator{\lb}{lb}\textbf{Theorem.}\, The only positive integers, whose Briggsian logarithms are rational, are the \PMlinkname{powers}{GeneralAssociativity}\, $1,\,10,\,100,\,\ldots$\, of ten.\, The logarithms of other positive integers are thus irrational (in fact, transcendental numbers).\, The same concerns also the Briggsian logarithms of the positive fractional numbers.\\
{\em Proof.}\, Let $a$ be a positive integer such that
$$\lg{a} = \frac{m}{n} \in \mathbb{Q},$$
where $m$ and $n$ are positive integers.\, By the definition of logarithm, we have\,
$\displaystyle10^{\frac{m}{n}} = a$,\, which is \PMlinkname{equivalent}{Equivalent3} to
$$10^m = 2^m\cdot 5^m = a^n.$$
According to the fundamental theorem of arithmetics, the integer $a^n$ must have exactly $m$ prime divisors $2$ and equally many prime divisors $5$.\, This is not possible otherwise than that also $a$ itself consists of a same amount of prime divisors 2 and 5, i.e. the number $a$ is an integer power of 10.
As for any rational number $\displaystyle\frac{a}{b}$ (with\, $a,\,b \in \mathbb{Z}_+$), if one had
$$\lg{\frac{a}{b}} = \frac{m}{n} \in \mathbb{Q},$$
then
$$\left(\frac{a}{b}\right)^n = 10^m,$$
and it is apparent that the rational number $\displaystyle\frac{a}{b}$ has to be an integer, more accurately a power of ten.\, Therefore the logarithms of all fractional numbers are irrational.\\
\textbf{Note.}\, An analogous theorem concerns e.g. the binary logarithms ($\lb{a}$).\, As for the natural logarithms of positive rationals ($\ln{a}$), they all are transcendental numbers except\, $\ln1 = 0$.