# How to prove: if $a,b \in \mathbb N$, then $a^{1/b}$ is an integer or an irrational number?

It is well known that $$\sqrt{2}$$ is irrational, and by modifying the proof (replacing 'even' with 'divisible by $$3$$'), one can prove that $$\sqrt{3}$$ is irrational, as well. On the other hand, clearly $$\sqrt{n^2} = n$$ for any positive integer $$n$$. It seems that any positive integer has a square root that is either an integer or irrational number.

1. How do we prove that if $$a \in \mathbb N$$, then $$\sqrt a$$ is an integer or an irrational number?

I also notice that I can modify the proof that $$\sqrt{2}$$ is irrational to prove that $$\sqrt[3]{2}, \sqrt[4]{2}, \cdots$$ are all irrational. This suggests we can extend the previous result to other radicals.

1. Can we extend 1? That is, can we show that for any $$a, b \in \mathbb{N}$$, $$a^{1/b}$$ is either an integer or irrational?
• It doesn't really make sense to say something is easy to see but you can't prove it. "It is easy to see this" is math jargon for "I have an easy proof of this". Sep 12, 2010 at 11:50
• @muad, anyone saying "it is easy to see" about the Jordan curve theorem only displays the fact that s/he has not thought about the thing enough and, more importantly, not even tried to actually prove it. Sep 12, 2010 at 12:05
• It's not obvious, although the geometric model strongly suggests it when the curve is simple (e.g. a circle). In the present case, however, there is no geometric model that might suggest that no natural number has a non-integer rational square root, so the analogy is not very strong. Sep 12, 2010 at 12:21
• mathoverflow.net/questions/32011/direct-proof-of-irrationality Sep 12, 2010 at 13:11
• One could say that the Jordan curve theorem seems obvious, but actually isn’t. One of the main aspects of mathematical training, I’d say, is learning to chase down your intuitions and either turn them into proofs (justifying that yes, something really was obvious!) or finding the weakness in the intuition (realising that actually something isn’t so obvious after all). Nov 25, 2010 at 22:06

Theorem: If $$a$$ and $$b$$ are positive integers, then $$a^{1/b}$$ is either irrational or an integer.

If $$a^{1/b}=x/y$$ where $$y$$ does not divide $$x$$, then $$a=(a^{1/b})^b=x^b/y^b$$ is not an integer (since $$y^b$$ does not divide $$x^b$$), giving a contradiction.

I subsequently found a variant of this proof on Wikipedia, under Proof by unique factorization.

The bracketed claim is proved below.

Lemma: If $$y$$ does not divide $$x$$, then $$y^b$$ does not divide $$x^b$$.

Unique prime factorisation implies that there exists a prime $$p$$ and positive integer $$t$$ such that $$p^t$$ divides $$y$$ while $$p^t$$ does not divide $$x$$. Therefore $$p^{bt}$$ divides $$y^b$$ while $$p^{bt}$$ does not divide $$x^b$$ (since otherwise $$p^t$$ would divide $$x$$). Hence $$y^b$$ does not divide $$x^b$$.

[OOC: This answer has been through several revisions (some of the comments below might not relate to this version)]

• @Douglas S. Stones: Am I correct to say that the Wiki proof you link to proves that square root of a number is either an integer or an irrational number?
Dec 5, 2013 at 18:47
• There is something that confuses me in this proof. The proof of the lemma (from "Unique prime..." to "...does not divide x^b.") says at a point: "...while p^bt does not divide x^b (since otherwise p^t would divide x).". But isn't this what we are trying to prove? [To make sure, set p^t = a in the phrase: "...while a^b does not divide x^b (since otherwise a would divide x)."] Doesn't this make the proof invalid (i.e. using what we want to prove as a fact in the proof itself)? Oct 12, 2015 at 14:31
• @ThanasisPapoutsidakis He is using a special case of the lemma for prime powers to prove the general lemma. But I agree it is not sufficiently explained; my own justification of the statement would go via the ${\rm ord}_p$ function, which defeats the purpose and makes the proof subordinate to other proofs on this page. Jun 15, 2016 at 21:00
• For the record, there was once discussion in the comments here about the (lack of) rigor in this answer (esp. prior versions) - which was continued in this question. Though the answer has been improved since the original version, it still leaves much to be desired in terms of rigor (e.g. it does not always explicitly refer to the crucial properties invoked, such as Euclid's Lemma or existence & uniqueness of prime factorizations). Aug 6, 2019 at 0:30
• One of the major goals of a course in number theory is to teach how to make rigorous such arithmetical intuition so that one is not led astray in more general number rings. Aug 6, 2019 at 0:30

These (standard) results are discussed in detail in

http://alpha.math.uga.edu/~pete/4400irrationals.pdf

This is the second handout for a first course in number theory at the advanced undergraduate level. Three different proofs are discussed:

1. A generalization of the proof of irrationality of $$\sqrt{2}$$, using the decomposition of any positive integer into a perfect $$k$$th power times a $$k$$th power-free integer, followed by Euclid's Lemma. (For some reason, I don't give all the details of this proof. Maybe I should...)

2. A proof using the functions $$\operatorname{ord}_p$$, very much along the lines of the one Carl Mummert mentions in his answer.

3. A proof by establishing that the ring of integers is integrally closed. This is done directly from unique factorization, but afterwards I mention that it is a special case of the Rational Roots Theorem.

Let me also remark that every proof I have ever seen of this fact uses the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations) in some form. [Edit: I have now seen Robin Chapman's answer to the question, so this is no longer quite true.] However, if you want to prove any particular case of the result, you can use a brute force case-by-case analysis that avoids FTA.

• The remark in your last paragraph is misleading since the results under discussion are much weaker than the property of being a UFD. Indeed, even the stronger property of being integrally closed is much weaker than UFD. One can deduce that Z is integrally closed from more general results that don't (immediately) imply that Z is a UFD. Sep 12, 2010 at 16:26
• @BD: I certainly agree that for a general domain, being integrally closed is much weaker than being a UFD. But I'm having trouble thinking of an argument that one could give in an elementary number theory course which would show the integers are integrally closed but not that they are a UFD. What do you have in mind? Sep 12, 2010 at 23:09
• Why do you think that Robin's answer has any relevance to your remark? As I've argued at length elsewhere (via conductors) such descent-based proofs are just unwindings of ideal-theoretic proofs, so they really do invoke the Euclidean property or some essentially equivalent property of Z. Sep 14, 2010 at 2:06
• From your handout, when $n$ is not a perfect $k$th power, $\sqrt[k]{n}\not\in\Bbb Q$. What happen with $\sqrt[k]{r}$ for rational and positive $r$?
– leo
Jun 11, 2014 at 5:08
• This link is now broken.
– Shai
Mar 25 at 7:22

As muad points out, you can also obtain this as an easy consequence of the Rational Root Theorem: if $a_nx^n+\cdots+a_0$ is a polynomial with integer coefficients, and $\frac{p}{q}$ is a rational root with $\gcd(p,q)=1$, then $p|a_0$ and $q|a_n$ (plug in, clear denominators, factor out).

So if you look at the polynomial $x^b-a$, with $b$ and $a$ positive integers, then a rational root must be of the form $\frac{p}{q}$, with $\gcd(p,q)=1$, and $q|1$. Thus, it must be an integer. So if it has a rational root, then the root is integral.

• Yeah! Niven's Proof! Oct 5, 2013 at 3:48

Here is a simple conceptual proof of irrationality of certain square roots - from first principles. Call a natural $$\rm\,d > 0\,$$ a denominator of $$\rm\:r\in\Bbb Q\:$$ if $$\rm\:r = c/d\:$$ for some $$\rm\:c\in\mathbb Z,\:$$ i.e.  if $$\rm\ dr\in \mathbb Z.$$

Theorem $$\ \ \rm r = \sqrt{a}\$$ is integral if rational,$$\:$$ for all $$\:\rm a\in\mathbb{N}$$

Proof $$\ \$$ Put $$\ \ \displaystyle\rm r = \frac{c}d ,\;$$ with $$\rm\; 0 < d\:$$ least. $$\ \displaystyle\rm\sqrt{a}\; = \frac{a}{\sqrt{a}} \ \Rightarrow\ \frac{c}{\color{#c00}d} = \frac{a\:d}{\color{#c00}c} \, \Rightarrow\ \color{#c00}d\:$$ divides $$\rm\: \color{#c00}c, \$$ by

Lemma $$\;$$ The least denominator of a rational $$\:\rm r\:$$ divides every denominator of $$\rm\:r\:.$$

Proof $$\rm\ \, n > m\$$ denominators $$\, \Rightarrow\,$$ so is $$\rm\ n\!-\!m\$$ by $$\;\rm nr, mr\in \mathbb Z \, \Rightarrow\, (n\!-\!m)r\in \mathbb Z.\,$$ Now apply

Lemma' $$\ \$$ Let $$\:\rm S\ne\{\,\} \,$$ be a set of integers $$>0\,$$ closed under subtraction $$> 0,\,$$ i.e. for all $$\rm\,n,m\in S, \,$$ $$\rm\ n > m\ \Rightarrow\ n-m\, \in\, S.\,$$ All elements of $$\rm\,S\,$$ are multiples of the least $$\rm\:\ell = \min\, S.$$

Proof $${\bf\ 1}\,\$$ If not there is a least nonmultiple $$\rm\,n\in S,\,$$ contra $$\rm\,n-\ell \in S\,$$ is a nonmultiple of $$\rm\,\ell.$$

Proof $${\bf\ 2}\,\rm\,\ \ S\,$$ closed under subtraction $$\rm\,\Rightarrow\,S\,$$ closed under remainder (mod), when it is $$\ne 0,$$ since mod is just repeated subtraction, i.e. $$\rm\, a\ mod\ b\, =\, a - k b\, =\, a-b-b-\cdots -b.\,$$ Hence $$\rm\,n\in S\,$$ $$\Rightarrow$$ $$\rm\, (n\ mod\ \ell) = 0,\,$$ else it's in $$S$$ and smaller than $$\rm\,\ell,\,$$ contra mimimality of $$\rm\,\ell.$$

Remark $$\$$ In a nutshell, two applications of induction yield the following inferences

$$\rm\begin{eqnarray} S\ closed\ under\ {\bf subtraction}\!\! &\Rightarrow&\rm S\ closed\ under\ {\bf mod} = remainder = repeated\ subtraction \\ &\:\Rightarrow\:&\rm S\ closed\ under\ {\bf gcd} = repeated\ mod\ (Euclid's\ algorithm) \end{eqnarray}$$

Interpreted constructively, this yields the extended Euclidean algorithm for the gcd.

The above Lemma is quite fundamental to factorization. I frequently refer to it by the suggestive moniker unique fractionization in order to highlight its equivalence to uniqueness of factorizations into irreducibles (one easily verifies that it is equivalent to Euclid's Lemma, which implies that irreducibles are prime). The structure implicit in the Lemma is a denominator or order ideal. Exploiting this structure, the proof easily generalizes to show that rational roots of monic integer coefficient polynomials must be integers, i.e. $$\:\mathbb Z\:$$ is integrally-closed (cf. the monic case of the Rational Root Test). In fact, this generalizes much further, employing Dedekind's key notion of a conductor ideal, to a one-line proof that PIDs are integrally closed. For much more on this see my post here and especially the posts linked there, and their links $$\ldots$$ (it is a beautiful web of ideas - mostly all due to Dedekind - as Noether often rightly remarked).

• Why $12x^3-3$ isnt a square number? Jan 23 at 12:04

Below is a simple proof of irrationality of square-roots that I discovered as a teenager (inspired by a proof of Dedekind). It employs the Bezout identity for the gcd, i.e. $$\rm\,\gcd(a,b)\,$$ is an integral linear combination of the integers $$\rm\,a,b,\,$$ i.e. $$\rm\,\gcd(a,b)\, =\, a d - b c\,$$ for some integers $$\,\rm c,d$$.

Theorem $$\ \ \ \rm r = \sqrt{n}\;\;$$ is integral if rational, $$\:$$ for $$\:\rm n\in\mathbb{N}$$

Proof $$\ \$$ Note that $$\rm\ r = a/b,\ \ \gcd(a,b) = 1\ \Rightarrow\ ad\!-\!bc \,=\, \color{#C00}{\bf 1}\;$$ for some $$\:\rm c,d \in \mathbb{Z}\$$ by Bezout.

$$\rm\color{#C00}{That\,}$$ and $$\rm\: r^2\! = \color{#0a0}{ n}\:\Rightarrow\ 0 \,=\, (a\!-\!br)\, (c\!+\!dr) \, =\, ac\!-\!bd\color{#0a0}{ n} \:+\: \color{#c00}{\bf 1}\cdot r,\$$ thus $$\ r \in \mathbb{Z}$$

This generalizes to roots of monic quadratic polynomials (and to higher degree, see here).

Theorem $$\,\$$ If $$\rm\,\ r^2 =\, \color{#0A0}{m\ r + n}\ \,$$ for $$\rm\ m,n\in\mathbb Z\$$ then $$\rm\ r\in \mathbb Q\ \Rightarrow\ r\in\mathbb Z$$

Proof $$\quad \rm r = a/b\in \mathbb Q,\ \ \gcd(a,b) = 1\ \Rightarrow\ ad\!-\!bc \,=\, \color{#C00}{\bf 1}\;$$ for some $$\:\rm c,d \in \mathbb{Z}\$$ by Bezout.

So $$\rm\,\ 0\, =\, (a\!-\!br)\: (c\!+\!dr)\, =\, ac\! -\! bd(\color{#0A0}{m\:r\!+\!n})+\color{#C00}{\bf 1}\cdot r \, =\, ac\!-\!adm\!-\!bdn + r \ \Rightarrow\ r \in \mathbb{Z}$$

Alternatively denominator descent can be done by the Division Algorithm instead of Bezout.

Theorem $$\, \$$ If $$\rm \ n\in\Bbb Z_{\phantom{\frac{i}.}}$$ and $$\rm \ r = \sqrt{n}\in \Bbb Q\$$ then $$\rm \ r\in \Bbb Z$$

\rm \begin{align}\\[-2em] & {\bf Proof}\,\ \ {\rm Deny,\ \,so}\,\ \ \smash[t]{\overbrace{ \rm r = {\small \frac{a}b}}^{\!\!\!\rm\large\color{#c00}{\ b\:\!r\ =\ a}}},\ \ {\rm and}\,\ \ \rm \color{#b0f}{b\nmid a}\,\ {\rm by}\,\ r\not\in\Bbb Z. {\rm\, wlog}\,\ b = \color{#0a0}{\rm least}\ {\rm denom}. \\[.1em] &\rm \color{#c00}ar\, =\, (\color{#c00}{br})r\, =\, bn.\ \ a = qb+\bar a,\,\ \color{#b0f}{0 < \bar a} < b\ \ \text{via \rm \,a\div b\, (Division Algorithm)}\\[.1em] &\rm\!\!\vphantom{\dfrac{}{|_{|_{|_|}}}}\smash[b]{\underbrace{\bar a r\! =\! a\color{#c00}r\!-\!q\color{#c00}{br}}_{\rm\Large (\bar a \ =\ a\, -\, q b)\:\!r}}\!\! = bn\!-\!q\color{#c00}a := j\in\Bbb Z\,\Rightarrow\, r = \frac{j}{\color{#b0f}{\bar a} },\,\ \text{contra }\, to\, \ b = \text{ \rm \color{#0a0}{\rm least}\, denom}.\end{align}

Remark $$\$$ It is instructive to compare the various denominator descents employed.

The denom descent in the first proof is $$\rm\,a,b\,$$ denom $$\rm\,\Rightarrow\, \gcd(a,b)\,$$ [$$= 1$$]  denom.

In the last proof the denom descent is: $$\rm\,a,b\,$$ denom $$\rm\,\Rightarrow\, a\ {\rm mod}\ b\,=:\, \bar a\:$$ denom.

In this answer, the denom descent is: $$\rm\, a\!>\!b\,$$ denom $$\rm\,\Rightarrow\ a\:\!-\:\!b\$$ denom.

These are all essentially special cases of proofs that the ideal $$\,a\Bbb Z + b\Bbb Z = \Bbb Z\,$$ using fast or slow descent based on the (Euclidean) Division algorithm (with remainder), i.e. specializations of a proof that ideals are principal in a Euclidean domain (compare posts on denominator ideals).

That the numerator $$\rm\,a = br\,$$ is also a denom of $$\rm\,r\,$$ generalizes to any algebraic integer $$\rm\,r,\,$$ most efficiently by using Dedekind's conductor ideal. This generalizes the above proofs into a slick one-line proof that PIDs are integrally-closed, as I explained at length elsewhere. It beautifully abstracts the denominator descent that governs ad-hoc "elementary" irrationality proofs.

• so that $(a-br)(c+dr)=ac-bdn+r$ and since $ac-bdn$ is an integer and $a-br=0$, we get that $r=-(ac-bdn)\in\mathbb{Z}$.
– robjohn
Aug 18, 2012 at 15:11
• @Peter The $\rm\color{#C00}{leading}$ coef of the product, viewed as a polynomial in $\rm\:r,\:$ is $\rm\:ad\!-\!bc.\:$ We want this $= \color{#C00}{\bf 1},$ since the goal of the proof is: $\,$ given that $\rm\:r\:$ is a root of a monic polynomial $\rm\:r^2-a = 0\:$ to deduce a lower-degree monic polynomial also having $\rm\:r\:$ as a root. The inductive step is clearer if you look at said higher-degree generalization. There, using the same degree-reduction step, by induction, we eventually reach a monic linear polynomial $\rm\:r - n = 0\:$ so $\rm\:r=n\in \Bbb Z.\:$ Aug 18, 2012 at 15:34
• I see that originally the word elsewhere was a link to MathOverflow posts. But it was edited to another link - but this link now simply redirects to Classroom Resources on nctm.org. Some version is available in the Wayback Machine. Apr 15 at 9:47
• @Martin Thanks for notifying me. I have updated the link to an answer here. Apr 15 at 10:44

The general theorem is that a natural number $a$ has a rational square root if and only if the multiplicity of every prime factor of $a$ is even. For example $2^43^611^2$ has a rational square root but $5^411^3$ does not.

Moreover, if a natural number has a rational square root, that square root is always obtained by halving the multiplicity of each prime factor, and so the square root is also a natural number.

The same principle works for $n$th roots, $n > 1$.

• Right -- this argument can be stated very cleanly using the ord_p functions, and is one of the proofs I give in the notes I linked to in my answer. (This is probably my favorite proof, since it showcases the usefulness of the ord_p's.) Sep 12, 2010 at 14:55

To answer Pete's comment as to how to prove integral closure of $\mathbb{Z}$ without using the UFD property.

Let $a/b$ be a rational ($a$, $b\in\mathbb{Z}$) which is integral over $\mathbb{Z}$. Let $R=\mathbb{Z}[a/b]$. Then $R$ is a finitely-generated $\mathbb{Z}$-module. It follows that $b^n R\subseteq\mathbb{Z}$ for some $n$. We reduce to proving the lemma that if $R$ is a ring with $\mathbb{Z}\subseteq R\subseteq N^{-1}\mathbb{Z}$ for some nonzero integer $N$ then $R=\mathbb{Z}$.

There are various ways of proving this. For instance if $R\ne\mathbb{Z}$ there is an element of $R$ strictly between two consecutive integers (this is the division algorithm) and so an element $x$ of $R$ strictly between $0$ and $1$. If $y$ is the least such number then considering $y^2$ gives a contradiction. Alternatively, $M=xN$ is an integer and $R\subseteq M^{-1}\mathbb{Z}$ so we can always replace $N$ by a smaller integer etc.

• Thanks, Robin. I will note that you used the division algorithm, which is also what you use to prove that $\mathbb{Z}$ is a UFD. (Admittedly, if you hadn't said that you used it then I wouldn't have seen it.) I guess the question is: is this in any sense easier than just proving FTA? I'm not sure... Sep 13, 2010 at 7:27
• You'll certainly need something like the division algorithm. The argument will work for Euclidean domains too. Note how easy it is to use the division algorithm without being aware of it :-) Sep 13, 2010 at 9:31
• @Robin: Agreed! Sep 13, 2010 at 9:49
• The slickest way to do this is a one-line proof using Dedekind's notion of conductor ideal. See one of my posts here and chase the links to sci.math for full details. Sep 14, 2010 at 1:46

So... we can prove something pretty general here: the only rational roots of polynomials $x^n + \cdots + a_0$ with integer coefficients are integers.

Indeed, suppose $p/q$ is a rational root, in lowest form, then $p^n = -q(a_0q^{n-1} + a_1pq^{n-2} + \cdots + a_{n-1}p^{n-1})$. Now, if $q>1$, then any prime divisor of $q$ also divides $p^n$, and hence $p$. But this contradicts our assumption that $p/q$ is in lowest form, so we conclude that $q=1$, so the root is integral.

• This is simply the monic case of the well-known rational root test which has already been mentioned above a few times. In more technical terms one simply says that Z is integrally closed (in its fraction field), i.e. fraction that is integral over Z (root of a monic polynomial) already lies in Z. Sep 14, 2010 at 1:53
• I saw people mention the rational root test... but then they just applied it to polynomials like x^b-a ... the point is that the same works for monic polynomials in general. That's all :) Sep 14, 2010 at 1:55
• They applied it to $x^b - c$ simply because that's all that is required here. I don't recall anyone ever implying that it works only for such binomials. Sep 14, 2010 at 1:57
• No worries, wasn't trying to be confrontational here. I'll remove my last sentence if it makes you feel better Sep 14, 2010 at 2:07
• No problem, I was just trying to help ensure that you understood what was above. Sep 14, 2010 at 2:08

Proposition: For natural number $a$, $\sqrt{a}$ is irrational or an integer.

Proof: Suppose $\sqrt{a}$ is not an integer, but that it is rational. So $\sqrt{a} = \frac{p}{q}$ for some integer $p$ and natural number $q$. Thus $q \sqrt{a} = p$. That is, there exists at least one natural number $q$ such that $q \sqrt{a}$ is an integer. Let $m$ be the smallest natural number such that $m\sqrt{a}$ is an integer.

Since $\sqrt{a}$ is not an integer, there is an integer $b$ such that $b < \sqrt{a} < b+1$. Thus $0 < \sqrt{a} - b < 1$ and $0 < m(\sqrt{a} -b) < m$.

Consider the number $k = m(\sqrt{a} -b)$. First note that $k = m\sqrt{a} - mb$, which is an integer since both $m \sqrt{a}$ and $mb$ are integers.

Also, $k\sqrt{a} = (m \sqrt{a} - mb)\sqrt{a} = ma - bm\sqrt{a}$. Again this is an integer since $ma$, $b$ and $m\sqrt{a}$ are integers.

However, $0 < k < m$, so k is in fact a natural number less than $m$ such that $k\sqrt{a}$ is an integer.

This is a contradiction. So $\sqrt{a}$ cannot be rational if it is not an integer. $\blacksquare$

Note: This proof was inspired by "What Pythagoras could have done", Yoram Sagher, American Mathematical Monthly, 95 (1988) (doi - paywalled)

Proposition: For natural numbers $a$ and $t$, $a^{1/t}$ is either irrational or an integer.

Proof: Suppose $a^{1/t}$ is not an integer, but it is rational. Therefore all the integer powers of $a^{1/t}$ are also rational. Consider the list of rational numbers $a^{i/t}$ for $i = 1$, ..., $t-1$. These can be written with a single common denominator, and multiplying them all by this denominator will produce a list of integers. Let $m$ be the smallest natural number such that $ma^{i/t}$ is an integer for all of $i = 1$, ..., $t-1$. (We have already argued that such a number exists.)

Now $a^{1/t}$ is not an integer, so there exists an integer $b$ such that $b < a^{1/t} < b+1$. Then $0 < a^{1/t} - b < 1$ and so $0 < m(a^{1/t} -b) < m$. Consider the number $k = m(a^{1/t} - b)$.

Now $k = ma^{1/t} - mb$, which is an integer since $ma^{1/t}$ and $mb$ are integers.

Also, for any $i = 1$, ..., $t-1$, $k a^{i/t} = ma^{(i+1)/t} - bma^{i/t}$. This is an integer since $ma^{i/t}$ and $ma^{(i+1)/t}$ are both integers. (Note that for $i=t-1$, $a^{(t-1+1)/t} = a^{t/t} = a$, so $ma^{(i+1)/t}$ is still an integer for $i = t-1$.)

However, $0 < k < m$, so in fact $k$ is a natural number less than $m$ such that $ka^{i/t}$ is an integer for $i =1$, ..., $t-1$. $\blacksquare$

Note: this proof was inspired by "Irrationality without number theory", Richard Beigel, American Mathematical Monthly, 98 (1991) (doi - paywalled)

• +1. But why not prove only the extended result/your 2nd proposition? May 17, 2018 at 7:02
• I prove both because it is easier to understand the second proof after having first understood the first proof. Also the original poster asked for both, so I provided both. May 27, 2018 at 9:57

Definition: An algebraic integer is a solution of a monic polynomial with integer coefficients. This set is closed by sums and products and such.

Theorem: If an algebraic integer is rational, then it is an integer.

Proof: Apply rational roots theorem.

This theorem proves that $\sqrt[b]{a}$ is irrational unless $a$ is a $b$-th power.

My personal version:

If $\dfrac pq$ is an irreducible fraction, so is $\left(\dfrac pq\right)^b=\dfrac{p^b}{q^b}$, as $p$ and $q$ have no common prime factor.

So a power of a fraction cannot give an integer.

• This can equivalently be stated: if $(p,q)=1$ ($p,q$ integers), then $(p^a,q^a)=1$. (More generally, $(p^a,q^b)=1$.) Jun 15, 2016 at 21:14

Here's a proof that uses the same idea found in user65203's answer. It relies on the following theorem:

For all integers $$n\geq 1$$ and $$k\geq 1$$, $$\sqrt[n]{k}$$ is rational if and only if $$k$$ is the $$n$$th power of an integer, that is, $$k=q^n$$ for some integer $$q$$.

This implies that $$\sqrt[b]{a}$$ is either an integer or irrational because "$$a=q^b$$ for some integer $$q$$" is either true or it isn't. If it is, then $$\sqrt[b]{a}=q$$, so $$\sqrt[b]{a}$$ is an integer. If it isn't, then $$\sqrt[b]{a}$$ can't be rational because the prior theorem implies that $$a=q^b$$ is a necessary condition for $$\sqrt[b]{a}$$ to be rational.

To prove the aforementioned theorem, it is sufficient to prove these two conditionals:

• If $$k=q^n$$ for some integer $$q$$, then $$\sqrt[n]{k}$$ is rational.
• If $$\sqrt[n]{k}$$ is rational, then $$k=q^n$$ for some integer $$q$$

Fix arbitrary integers $$n\geq 1$$ and $$k\geq 1$$. The first of these conditionals is trivial:

$$k=q^n\implies\sqrt[n]{k}=q$$

which shows that $$\sqrt[n]{k}$$ is an integer, and hence rational. To prove the second, we'll need to establish two preliminary lemmas:

Lemma 1: if $$a$$ and $$b>0$$ are coprime integers and $$b$$ divides $$a$$, then $$b=1$$.

Proof: if $$a$$ and $$b>0$$ are coprime, then the only positive integer that simultaneously divides them is $$1$$. $$b$$ clearly divides $$b$$, and (by assumption) $$b$$ divides $$a$$, so $$b$$ is a positive integer that simultaneously divides $$a$$ and $$b$$. Since $$1$$ is the only number with this property, $$b$$ must equal $$1$$.

Lemma 2: if $$a>0$$ and $$b>0$$ are coprime integers, then $$a^m$$ and $$b^n$$ are coprime for any positive integers $$m,n$$.

Proof: it is known that two integers are coprime if and only if there is no prime that simultaneously divides them. By the fundamental theorem of arithmetic, we can write

$$a=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$$ $$b=q_1^{l_1}q_2^{l_2}\cdots q_s^{l_s}$$

where $$p_1$$, $$p_2$$, ...,$$p_r$$, $$q_1$$, $$q_2$$, ...,$$q_s$$ are primes, and $$k_1$$, $$k_2$$, ...,$$k_r$$, $$l_1$$, $$l_2$$, ...,$$l_s$$ are the corresponding exponents appearing in the factorization.

Fix an arbitrary pair of positive integers $$(m,n)$$ and raise $$a$$ and $$b$$ to the $$m$$th and $$n$$th power, respectively. We get

$$a^m=\left(p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}\right)^m=p_1^{mk_1}p_2^{mk_2}\cdots p_r^{mk_r}$$ $$b^n=\left(q_1^{l_1}q_2^{l_2}\cdots q_s^{l_s}\right)^n=q_1^{nl_1}q_2^{nl_2}\cdots q_s^{nl_s}$$

It can be seen that $$a^m$$ has the same prime factors as $$a$$, and $$b^n$$ the same ones as $$b$$. Since $$a$$ and $$b$$ are coprime, they don't share a prime factor, so the prior observation implies that there is no prime that simultaneously divides $$a^m$$ and $$b^n$$. It follows that $$a^m$$ and $$b^n$$ are coprime. Generalizing from the arbitrary $$(m,n)$$, we conclude that $$a^m$$ and $$b^n$$ are coprime for every pair of positive integers $$(m,n)$$.

We are now in a position to prove the converse. Assume that $$\sqrt[n]{k}$$ is rational. We can write

$$\sqrt[n]{k}=\frac{p}{q}$$

for integers $$p$$ and $$q$$, which we may assume to be positive and coprime since $$\sqrt[n]{k}>0$$ and every fraction is equivalent to a reduced fraction.

Now,

\begin{align*} \sqrt[n]{k}=\frac{p}{q} &\iff k=\left(\frac{p}{q}\right)^n\\ &\iff k=\frac{p^n}{q^n}\\ &\iff p^n = kq^n \end{align*}

and since $$p$$ and $$q$$ are coprime, we know from Lemma 2 that $$p^n$$ and $$q^n>0$$ are coprime. But the equation $$p^n = kq^n$$ implies that $$q^n$$ divides $$p^n$$. Thus, $$q^n=1$$ from Lemma 1, so the equation $$p^n = kq^n$$ reduces to $$k=p^n$$, which is what we were aiming to show.

Q.E.D.

Generalize this argument to any integers $$n\geq 1$$ and $$k\geq 1$$, and we're done!

A generalisation of Euclid's lemma states that if $$n$$ divides $$ab$$ and $$n$$ is coprime to $$a$$, then $$n$$ divides $$b$$. Using this lemma, it is easy to prove directly that if $$x$$ is rational and $$x^2$$ is an integer, then $$x$$ is an integer.

Suppose that $$\left(\frac{p}{q}\right)^2=m$$, where $$p$$ and $$q$$ are coprime, and $$m$$ is an integer. Then $$p^2=mq^2$$, so $$q$$ divides $$p^2$$. By the above lemma, $$q$$ divides $$p$$. But $$p$$ and $$q$$ are coprime, so $$q=1$$ and $$m=p^2$$.

The rational root theorem: Given a polynomial $$a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_2 x^2 + a_1 x + a_0$$

with integer coefficients and $$a_n \ne 0$$, then each rational solution written as $$x = \frac pq$$, where $$\gcd(p,q) = 1$$, satisfies
$$\circ \quad p$$ divides $$a_0$$.
$$\circ \quad q$$ divides $$a_n$$.

$$x = \sqrt a$$ if and only if $$x$$ is a root of the polynomial $$x^2 - a.$$

If $$x = \frac pq$$ is going to be a rational root, then we need $$p|a$$ and $$q|1$$. In other words, we must have $$x = \pm p$$ where $$p$$ is an integer divisor of a.

We end up with $$p^2 = a$$. Hence, either $$a$$ is a perfect square or the square root of $$a$$ is irrational.

The extension to $$\sqrt[p]{a}$$ is obvious.