2
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

marked as duplicate by JMoravitz, Parcly Taxel, Did real-analysis Nov 5 '16 at 6:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It is 2 sqrt 3. This is twice an irrational number (if you know that) . You could also use prime decomposition: rational numbers have prime decompositions too. $\endgroup$ – Jacob Wakem Nov 5 '16 at 5:05
  • $\begingroup$ @Alephnull 2 sqrt 3 < sqrt 18 $\endgroup$ – mathguy Nov 5 '16 at 5:08
  • $\begingroup$ The best way to prove these constructively is to form the continued fraction. Showing that the continued fraction does not terminate is sufficient. Do it for $\sqrt 2$, it's a lot quicker! $\endgroup$ – Scott Burns Nov 5 '16 at 5:10
  • $\begingroup$ @ahamed 3 sqrt 2. $\endgroup$ – Jacob Wakem Nov 5 '16 at 5:13
  • 2
    $\begingroup$ One finds on wikipedia (en.wikipedia.org/wiki/Square_root_of_2#Constructive_proof) a constructive proof of the irrationality of $\sqrt{2}$. $\endgroup$ – user259242 Nov 5 '16 at 5:13
4
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ Yes but the standard proof of $\sqrt 2 \notin \mathbb{Q}$ uses contradiction. $\endgroup$ – Scott Burns Nov 5 '16 at 5:07
  • $\begingroup$ I am asking for methods that does not use contradiction. $\endgroup$ – mathguy Nov 5 '16 at 5:09
  • $\begingroup$ This does not use contradiction. In fact, one can show that for any two integers $a$ and $b$, $\left| \sqrt{2} - \frac{a}{b}\right| \geq \frac{1}{3b^2}$, without using the law of the excluded middle. $\endgroup$ – user259242 Nov 5 '16 at 5:15
  • $\begingroup$ @user259242 Right ! $\endgroup$ – Duchamp Gérard H. E. Nov 5 '16 at 5:17
  • $\begingroup$ @mathguy I do not use contradiction but only rest on a result which standardly (but not necessarily, see wikipedia) uses it. I bet that this was awaited in your exam $\endgroup$ – Duchamp Gérard H. E. Nov 5 '16 at 5:48
1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ You are saying: If $\sqrt{18}$ is rational then it is an integer. So, assume it is rational. Then it is an integer, which is a contradiction since $18$ is not a perfect square. Thus, $\sqrt{18}$ is irrational. This is a proof by contradiction, no? $\endgroup$ – user259242 Nov 5 '16 at 5:22
  • 1
    $\begingroup$ @user259242 So, assume it is rational. Then it is an integer No, this line of argument said no such. Rather, it said "is either an integer or an irrational number", while noting that the former is trivially false, which left $\sqrt{18}$ having to be irrational. There is no proof by contradiction that I can see there. $\endgroup$ – dxiv Nov 5 '16 at 6:33
  • $\begingroup$ @dxiv Its subtle, but it is in fact a proof by contradiction. Without appealing to the law of the excluded middle, all one can conclude from this proof is that $\sqrt{18}$ is not rational. But then we need LEM to get $\sqrt{18}$ is irrational. $\endgroup$ – user259242 Nov 5 '16 at 7:46
  • $\begingroup$ @user259242: Are you prepared to claim that $\sqrt{18}$ may not be a number? Because irrational is defined as any number that is not rational. $\endgroup$ – LutzL Nov 5 '16 at 11:47
  • $\begingroup$ @LutzL The point of contention relates to the logical difference between proving that a proposition is false, and proving that the negation of that same proposition is true. See Example 2.1 of arxiv.org/pdf/1110.5456v1.pdf $\endgroup$ – user259242 Nov 5 '16 at 11:51
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The fundamental theorem of arithmetic is a very strong result, and I'm afraid you have to prove it by contradiction somewhere... $\endgroup$ – Cave Johnson Nov 5 '16 at 5:24
0
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.