## (Ir)rationality

### 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.