How to prove $\sqrt{18}$ is irrational without using proof by contradiction?

Question

I want to know how to prove $\sqrt{18}$ is irrational using method other than proof by contradiction. I have always been taught to prove irrationality using proof by contradiction. So when I was asked this in my exam I was really surprised. Can anyone think of other methods to prove this? Please help. Thank you.

Answer 1

$\sqrt{18}=3\sqrt{2}$ and $\sqrt{2}\notin \mathbb{Q}$ (for proofs of this last point not using contradiction, see wikipedia for example).

Answer 2

Interesting question.

There's a short proof using the rational root theorem. (Credit - Wikipedia)

The theorem essentially say that if $q(x)$ is a monic polynomial (https://en.wikipedia.org/wiki/Monic_polynomial), then any rational root of the aforementioned polynomial must be an integer or an irrational number.

Conveniently, take the polynomial: $q(x)=x^2-18$

According to the theorem, it follows that $\sqrt{18}$ is either an integer or an irrational number. Because it is not an integer (for 18 is not a perfect square, i.e. 18 is not the square of an integer), it is irrational.

Answer 3

In general, if $x$ is a positive integer, and $\sqrt[q]{x}$ is not an integer, then it will be irrational. You can argue this without using proof by contradiction as follows.

Order the prime numbers as $p_1 = 2, p_2 = 3, p_3 = 5,...$ By the fundamental theorem of arithmetic, every positive rational number $x$ can be identified uniquely with a sequence $(n_1,n_2,n_3,...)$ of integers such that $n_i = 0$ for all but finitely many $i$. Specifically $(n_1,n_2,n_3,...)$ is such that

$$x = 2^{n_1} 3^{n_2} 5^{n_3} \cdots$$

which is a finite product. For example, $(-1,-1,-1,0,0,...)$ corresponds to the rational number $\frac{1}{30}$.

If $x,y$ are positive rational numbers, $(n_1, n_2, ...)$ is the sequence corresponding to $x$, and $(m_1, m_2, ...)$ is the sequence corresponding to $y$, then $(n_1+m_1,n_2+m_2,...)$ is the sequence corresponding to $xy$.

In particular, $y^2$ corresponds to the sequence $(2m_1,2m_2,...)$.

A positive rational number is an integer if and only if all the terms in its sequence are nonnegative.

Every positive integer is uniquely expressible as a product of primes, so its sequence, which uniquely identifies it, will consist of nonnegative terms.

From these two facts, we see that the only integers which are squares of rational numbers are those for which the correspondence sequence consists of nonnegative even terms, in which case their square roots are also integers.

Answer 4

I suppose you could grind out the continued fraction for $\sqrt{18}$ and show or observe that it's with non-zero period .