Our proof works by showing a series of “If x is rational so is y”. Suppose \(S\) is rational, then the following sequence of numbers must be rational too.

But we know that \(\sqrt{21}\) is not rational and hence a contradiction.

We know that \(\frac{\ln (12)}{\ln (18)}=\log_{18}(12) .\) Suppose this is rational, say \(=\frac{a}{b}\) where \(a, b\) are integers with \(b \neq 0\). Then \(18^{\frac{a}{b}}=12,\) so \(18^{a}=12^{b} .\) By factoring into primes this gives \(3^{2 a} 2^{a}=3^{b} 2^{2 b},\) which by unique factorization can happen only if \(2 a=b\) and \(a=2 b\). But this gives \(a=b=0\), a contradiction.

False-True-True

Here is one of several possible ways. \(x^{2014}-x^{2009}=x^{2009}\left(x^{5}-1\right)\) and \(x^{2004}\left(x^{5}-1\right)\) are integers, which we may assume to be nonzero (else \(x=0\) or 1 and we are done). Dividing, we get that \(x^{5}\) is rational. Now dividing the integer \(x^{2004}\left(x^{5}-1\right)\) by the rational number \(x^{5}-1,\) we see that \(x^{2004}\) is rational. since 2004 and 5 are coprime, \(x\) is rational as well. (E.g., \(x^{5}\) is rational, so \(\left(x^{5}\right)^{401}=x^{2005}\) is rational. Now divide by the rational number \(x^{2004}\).) Let \(x=\frac{a}{b}\) with \(a, b\) coprime integers.

Consider the integer \(\displaystyle \frac{a^{2000}}{b^{2009}}-\frac{a^{2004}}{b^{2004}}=\frac{a^{2009}-b^{5} a^{2004}}{b^{2000}}\). If a prime \(p\) divides the denominator, it must divide the numerator as well.

Now \(p \mid b,\) so \(p \mid b^{5} a^{2004},\) so \(p \mid a^{2009}\) and finally \(p \mid a,\) a contradiction. Thus \(b=1,\) i.e., \(x\) is an integer.