## (Ir)rationality

#### Problems

### Rationality preserving operations

A11, 2010

Using the fact that \(\sqrt{n}\) is an irrational number whenever \(n\) is not a perfect square, show that \(S=\sqrt{3}+\sqrt{7}+\sqrt{21}\) is irrational.

## Solution

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.

\(7\sqrt{3}+3\sqrt{7}+\sqrt{21}\) | Square the given number and subtract the integer part |

\(6\sqrt{3}+2\sqrt{7}\quad\) | Subtract \(S\) from the above number. Since \(S\) is assumed to be rational, this number must be rational too. |

\(\sqrt{3}\sqrt{7}\quad\quad\) | Square the above number and remove the integer part. |

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

### Irrational fraction

A3, 2012

Prove that \(\frac{\ln (12)}{\ln (18)}\) is irrational.

## Solution

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.

### Summations

A1, 2014

Let \(\alpha, \beta\) and \(c\) be positive numbers less than \(1,\) with \(c\) rational and \(\alpha, \beta\) irrational.

(a) The number \(\alpha+\beta\) must be irrational.

(b) The infinite sum \(\sum_{i=0}^{\infty} \alpha c^{i}=\alpha+\alpha c+\alpha c^{2}+\cdots\) must be irrational.

(c) The value of the integral \(\int_{0}^{\pi}(\beta \cos x+c) d x\) must be irrational.

## Solution

False-True-True

### A polynomial integer

B2, 2014

Let \(x\) be a real number such that \(x^{2014}-x^{2004}\) and \(x^{2009}-x^{2004}\) are both integers. Show that \(x\) is an integer.

Hint: it may be useful to first prove that \(x\) is rational.

## Solution

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.