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I'm trying to prove that

If $n$ is any natural number then either $\sqrt{n}$ is an integer or it is irrational.

I'm not attaching the lecture notes from which I've taken this statement as I think it's under copyright. My attempt to prove it is as follows:

Suppose $n$ is an arbitrary natural number. Then, either $n$ is a perfect square or not a perfect square. If $n$ is a perfect square, then $n = m^2$ for some $m \in \mathbb{N}$, i.e. $m = \sqrt{n}$ is natural and hence an integer. If $n$ is not a perfect square, suppose for the sake of contradiction that $\sqrt{n}$ is not irrational, i.e. $\sqrt{n} = \frac{p}{q}$ for some $p, q \in \mathbb{N}$. Then $n = \frac{p^2}{q^2}$, but this contradicts the assumption that $n$ is natural. Hence, $\sqrt{n}$ is either an integer or an irrational number. $\square$

I wanted to know if my proof is valid and if there is a way to prove the second part directly instead of resorting to contradiction.

Zim
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Viktor Glushkov
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    The second case doesn't look correct. How do you know that $q$ doesn't equal $1$ in this case? You should need to use the fact that $n$ is not a perfect square. – Trevor Norton Jun 05 '24 at 14:56
  • Would it work to argue that if $q = 1$ then $n$ is a perfect square and so the case is already covered by the first part of the proof, while if $q \neq 1$, this leads to contradiction? And $q$ is either $1$ or not, so all possibilities are exhausted. – Viktor Glushkov Jun 05 '24 at 15:00
  • That seems a bit messy to me. You may want to expand the assumption that "$n$ is not a perfect square" in a more rigorous way using the fundamental theorem of arithmetic (i.e. if $n = p_1^{\alpha_1}\cdots p_n^{\alpha_n}$ what can we say about one of the $\alpha_i$s?). It'll also make things cleaner if you assume that $p$ and $q$ don't have common factors (since we could reduce the fraction if they did). – Trevor Norton Jun 05 '24 at 15:05
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    "Then $n=\frac{p^2}{q^2}$, but this contradicts the assumption that n is natural". No, it doesn't. $4 = \frac {10^2}{5^2}$. How is that a contradiction? – fleablood Jun 05 '24 at 15:10
  • You really only need the second case. Suppose $\sqrt{n}=\frac{p}{q}$ in lowest terms. Then $n=\frac{p^2}{q^2}$. Since $n$ is an integer $q=1.$ Consequently, $\sqrt{n}=p$ is an integer. – Steen82 Jun 05 '24 at 15:11
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    The things you need are that 1) A rational number can be written in "lowest terms" and 2) If $p$ and $q$ have no factors in common, then $p^k$ and $q^k$ won't either. Those are the core concepts. With those, that there are no integers with non-integer rational roots comes as a near freebie. – fleablood Jun 05 '24 at 15:17
  • @Steen82 How does $n$ being and integer imply $q=1$? $7$ is an integer and $7 =\frac {35}5$ but $5\ne 1$. – fleablood Jun 05 '24 at 15:23
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    "there is a way to prove the second part directly instead of resorting to contradiction." As the definition of irrational is itself a contradiction, a contradiction is desirable and not to be avoided. However a "direct" prove would be to let $\sqrt{n}=\frac pq$ where $p$ is an integer and $q$ is not an integer that shares any factors with $p$. Then $n=\frac {p^2}{q^2}$ then as $p^2$ shares no factors with $q^2$ and $q\ne 1$ then $q^2$ can not be an integer and $q$ can not have been integer as so $\frac pq$ is irrational. But that's weird and convoluted. – fleablood Jun 05 '24 at 15:38
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    Also in all of this you are assuming that somehow $\sqrt n$ exists at all, either as a rational or irrational number? How do you know that? And this level you really have no way of describing or proving the existence of specific irrational number (and for that matter I doubt you have even demonstrated any irrational numbers "exist"), you probably can't prove $\sqrt{n}$ is irrational. The best you can prove is no rational number is the $\sqrt{n}$. – fleablood Jun 05 '24 at 15:43
  • Does this answer your question? Proof verification: Prove $\sqrt{n}$ is irrational. – Afntu Jun 05 '24 at 16:52
  • @fleablood Thank you for the comments. So the assumption that $p$ and $q$ are in lowest terms suffices to fix the problems you raised, if I understand correctly? Then the argument would be as follows: suppose $\sqrt{n} = p/q$ for some $p, q \in \mathbb{N}$, $q \neq 0$, and also assume the fraction is in lowest terms. Then $p^2/q^2$ is also in lowest terms, and so the only way for it to be a positive integer is if $q = 1$, in which case $n$ is a perfect square. Hence $\sqrt{n}$ is either a positive integer or an irrational number. – Viktor Glushkov Jun 05 '24 at 17:35
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    Thats it mor or les. Problem I have is that a lot dep.s on what basic ground work has been laid. Can we assume roots exist at all as irrational reals if they do not exist as rationals? Do we know that fractions can always be expressed in lowest terms? Uniquely? Do we know prime factorisation is unique? If $q^2$ and $p^2$ don't have factors in common, do we know if we can reexpress it as in terms of two numbers that do? And if $p$ and $q$ have no factors in common how do we know $p^2$ and $p^2$ don't either. All those can be answered but whether we have to is very course specific. – fleablood Jun 05 '24 at 18:41
  • @fleablood you are missing the in lowest terms condition in my previous comment. – Steen82 Jun 05 '24 at 19:56
  • @Steen82 I didn't miss that. You have $\frac pq$ in lowest case. But how do you know $\frac {p^2}{q^2}$ is also in lowest case?... At least I think that's what I remember being my point. Although why I didn't just say that, I don't remember. I think in some way I noticed you noting $\frac pq$ in lowest temrs and somehow I didn't really relate $\frac {p^2}{q^2}$ being related to $\frac pq$ as fractions in anyway. I really don't remember too much. But I still maintain we need to be dilligent in noting all cases of "lowest terms" – fleablood Jun 05 '24 at 21:09

2 Answers2

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In fact/general, the $n$th root of a positive integer is either an integer or an irrational number.

Suppose $\sqrt[n]N=\frac ab$ in lowest terms, i.e., $a$ and $b\neq1$ are coprime (relatively prime). It means, $\frac{a^n}{b^n}=N$. Since $a$ and $b$ are relatively prime, $a^n$ and $b^n$ are also relatively prime ($\forall n\in\mathbb{N}, n\geq2$) by the fundamental theorem of arithmetic. If $a$ has certain prime factors, multiplying $a$ times itself doesn’t yield new primes. So, the prime factorizations of $a^n$ and $b^n$ have no common primes because $a$ and $b$ don’t. Hence, $\frac{a^n}{b^n}$ can not be an integer unless $b^n=1$. That is to say, no fraction (non-integer in lowest terms), when raised to a power, can produce an integer. Thus, the $n$th root of a positive integer is either an integer or an irrational number.

Alternatively, by the rational root theorem, every rational root of the polynomial $x^n-N$ would have a numerator (say $a$) that divides $N$ (constant term) and a denominator (say $b$) that divides $1$ (leading coefficient), i.e., $b=\pm1$. Hence, the only possible rational roots are $\pm a$, an integer factor of $N$ (integral root theorem as a special case). It can not be a (non-integer) fraction. It means $\sqrt[n]N$ is either an integer or an irrational number.

I hope this helps you!

I would like to share an interesting proof (see below).

Theorem: If $n$ is a positive integer and is not a perfect square, then $\sqrt n$ is irrational.
Proof: Suppose $\sqrt{n} = a/b$ for positive integers $a,b$ with no common factor greater than $1$. Then $b/a = \sqrt{n}/n$, and so $a/b = (bn)/a$. Since the first fraction is in the lowest terms, the numerator and denominator of the second fraction must be a common integer multiple, say $c$, of the numerator and denominator of the first. Hence, $a = cb$, and therefore, $\sqrt{n} = c$, that is, $n$ is a perfect square.

I learned this proof from a one-paragraph insert in the American Mathematical Monthly (vol. 115, June-July 2008, p. 524) written by Geoffrey C. Berresford. I love it. I hope you find it interesting.

Pustam Raut
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Correct proof

Suppose $n$ is an arbitrary natural number. Then either $n$ is a perfect square or it is not.

If $n$ is a perfect square, then $n = m^2$ for some $m \in \mathbb{N}$, so $\sqrt{n} = m$ and hence is a positive integer.

If $n$ is not a perfect square, suppose for the sake of contradiction that $\sqrt{n}$ is rational, i.e. $\sqrt{n} = p/q$ for some $p, q \in \mathbb{N}$, and let us also assume that the fraction is in lowest terms, i.e. $p$ and $q$ are coprime. Then $n = p^2/q^2$ and $p^2$ and $q^2$ are also coprime. So $p^2/q^2$ is never a positive integer unless $q = 1$ (i.e. unless $n = p^2$), which falls under the perfect-square case covered earlier. Hence either $\sqrt{n}$ is an integer or it is irrational. $\square$

Viktor Glushkov
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