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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.

mathguy
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  • 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. – Jacob Wakem Nov 05 '16 at 05:05
  • @Alephnull 2 sqrt 3 < sqrt 18 – mathguy Nov 05 '16 at 05:08
  • 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! – user383778 Nov 05 '16 at 05:10
  • @ahamed 3 sqrt 2. – Jacob Wakem Nov 05 '16 at 05:13
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    One finds on wikipedia (https://en.wikipedia.org/wiki/Square_root_of_2#Constructive_proof) a constructive proof of the irrationality of $\sqrt{2}$. – user259242 Nov 05 '16 at 05:13
  • @user259242 That's a good idea, but how can we, without contradiction, define the concept of valuation, such that we may claim the valuation of $2a^2$ is odd while that of $b^2$ is even? We define it using "any finite set of natural numbers has a maximum", which is usually proved by contradiction. – Cave Johnson Nov 05 '16 at 05:30
  • @ScottBurns but how can one prove that a continued fraction does not terminate? It might terminate after 2 million terms and what then? :q For $sqrt{2}$ it might be easy to observe that we end up with what we've started, but it might not be so easy for other numbers. – BarbaraKwarc Jul 25 '17 at 03:32
  • see the accepted answer for https://math.stackexchange.com/questions/1310014/what-is-the-most-rigorous-proof-of-the-irrationality-of-the-square-root-of-3 It covers the proof for all non-perfect squares and uses contrapositive instead of contradiction. – Mark Joshi Sep 08 '17 at 02:50

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

Duchamp Gérard H. E.
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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.

Panglossian Oporopolist
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    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? – user259242 Nov 05 '16 at 05:22
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    @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. – dxiv Nov 05 '16 at 06:33
  • @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. – user259242 Nov 05 '16 at 07:46
  • @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. – Lutz Lehmann Nov 05 '16 at 11:47
  • @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 https://arxiv.org/pdf/1110.5456v1.pdf – user259242 Nov 05 '16 at 11:51
  • @user259242 Irrationals are defined as numbers which are not rational. But if you truly mean to push this into a discussion about Brouwer's intuitionism then I'll just leave it at that. – dxiv Nov 05 '16 at 16:37
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I suppose you could grind out the continued fraction for $\sqrt{18}$ and show or observe that it's with non-zero period .

Duchamp Gérard H. E.
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B. Goddard
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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.

D_S
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  • The fundamental theorem of arithmetic is a very strong result, and I'm afraid you have to prove it by contradiction somewhere... – Cave Johnson Nov 05 '16 at 05:24