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I would like to know whether I have proved this statement correctly using contradiction and if not get some tips or pointers on how to improve it/make it correct.

We are asked to prove that there are infinitely $n \in \mathbb{Z^+}$ such that $\sqrt{n}$ is irrational.

Proof: $$\text{We assume, to the contrary, that there are infinitely many } n \text{ } (n \in Z^+) \text{ such that } \sqrt{n} \text{ is rational.} \\\text{Then, } \sqrt{n}=\frac{a}{b} \text{, } n=\frac{a^2}{b^2} \text{ where } a \text{ and } b \text{ are integers and } \frac{a}{b} \text{ has been reduced to lowest terms.} \\\text{Hence, } a^2=nb^2 \text{ and } b^2=\frac{1}{n}a^2\\\text{We know if } n\mid a^2 \text{ then } n\mid a \text{. It follows that } a=nk, k\in\mathbb{Z} \\ b^2=\frac{1}{n}(nk)^2=nk^2 \text{ and thus } n\mid b^2 \text{ and thus } n \mid b \\ \text{Hence, } n = \frac{nb^2}{nk^2} \\ \text{This is a contradiction given that we have assumed }a \text{ and } b \text{ to be reduced to lowest terms. }\blacksquare$$ ps. This isn't homework. Answers are in the back of my book but I want to improve.

Cro-Magnon
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    No, your first statement is wrong. Your first statement should be "we assume to the contrary that there are finitely many natural numbers $n$ such that $\sqrt{n}$ is irrational". – stressed out Jul 09 '18 at 06:36
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    Also, $n|a^2$ does not in general imply $n|a$. Counterexample: $n=4,,a=2$. – J.G. Jul 09 '18 at 06:38
  • @stressedout Can you elaborate on how to correctly identify the hypothesis and conclusion in this statement? I have a hard time doing this when the statement isn't in the form of if ..., then ... – Cro-Magnon Jul 09 '18 at 06:40
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    For completeness regarding what @J.G. said, in the other direction, it is (trivially) true: If $n|a$ then $n|a^2$ – Joaquin Liniado Jul 09 '18 at 06:40
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    The best way to negate a statement is to ask yourself when it's false? If I claim that something holds for an infinite number of things and I want to disprove it, I should show that only a finite number of things with that property can exist. Right? Now look at the flaw of your negation: We know that there exists an infinite number of odd integers. But we also know that there exists an infinite number of even integers! Do you see the problem with your negation now? – stressed out Jul 09 '18 at 06:45
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    Why don't you use a direct proof ? You can for example use the irrationality of $\sqrt{2}$ to easily show the irrationality of every number $\sqrt{2k^2}=k\sqrt{2}$ where $k$ is a positive integer immediately proving the given statement. – Peter Jul 09 '18 at 06:45
  • @stressedout Ah, yes thank you! That's obvious.. Back to the drawing board. – Cro-Magnon Jul 09 '18 at 06:50
  • @Peter My book asked me to prove it via contradiction rather than by direct proof. – Cro-Magnon Jul 09 '18 at 06:50
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    Well, Peter's argument can be modified to a proof by contradiction. A valid proof can assume that only a finite number of something with property blah blah can exist and then shows that you can always find a new thing satisfying your conditions that isn't in your list! – stressed out Jul 09 '18 at 06:52
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    @Cro-Magnon If you show that $\sqrt{n}$ is irrational if and only if $n$ is not a perfect square the given statement becomes equivalent to "There are infinite many positive integers that are not a perfect square". So you only have to refute that there are only finite many positive integers that are not a perfect square. – Peter Jul 09 '18 at 06:55
  • @Peter I see. This really helped me. I'm going to ponder this for a while and redo the proof. Thank you! – Cro-Magnon Jul 09 '18 at 06:59

2 Answers2

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What is the statement you are trying to prove, saying? It is saying that the number of elements of the set $\{n \in \mathbb N : \sqrt n \ \mbox{is irrational}\}$, is infinite.

What is the opposite of this statement? Naturally, that the number of elements of that set is not infinite, in which case it is finite. So, the contradictory statement is this :

The set $\{n \in \mathbb N : \sqrt n \ \mbox{ is irrational}\}$ is finite. Or, there are only finitely many natural numbers $n$ such that $\sqrt n$ is irrational.

Now, you have to assume this is true, and derive a contradiction.

To derive a contradiction to this statement, you have to realize, that every non-empty finite set of natural numbers has a maximum element. We know, from a classical proof by contradiction, that $\sqrt 2$ is irrational, so $2$ is an element of our subject set.

So, if the set of natural numbers $n$ such that $\sqrt n$ is irrational has a maximum, then let $L$ be that maximum number. So $\sqrt L$ is irrational.

But, as Peter mentioned in his comment, $2 \sqrt L$ is also irrational, and this is equal to $\sqrt{4L}$, so $4L$ also belongs to the set of natural numbers whose square roots are irrational... can you take it from here?

Note that Peter's comment may also be used to produce infinitely many members of the set, so the two arguments , contradiction and direct proof, are not very different from one another.

Sarvesh Ravichandran Iyer
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  • Of course, phrasing this as a proof by contradiction is an enormous obfuscation of what you're doing. You're just directly constructing an infinite sequence of integers with no rational square root. – Jack M Jul 09 '18 at 07:58
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An idea for a different kind of proof:

It is known that $\sqrt{n}$ rational if and only if $n$ is a perfect square.

Indeed, assume that $\sqrt{n} = \frac{a}{b}$ with $\gcd(a,b) = 1$. Hence there exist $\alpha, \beta \in \mathbb{Z}$ such that $\alpha a - \beta b = 1$.

Now $$0 = (a-b\sqrt{n})(\beta + \alpha \sqrt{n}) = \beta a - \alpha bn + (\alpha a - \beta b) \sqrt{n} = \beta a - \alpha bn + \sqrt{n}$$

And hence $\sqrt{n} \in \mathbb{Z}$ so $n$ is a perfect square. The converse is obvious.

Now it just remains to show that the set $\{n \in \mathbb{N} : n \text{ is not a perfect square}\}$ is infinite.

For this notice that numbers of the form $3k^2+2$ for $k \in \mathbb{N}$ are never perfect squares.

mechanodroid
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