$\sqrt a$ is either an integer or an irrational number.

Question

I got this interesting question in my mind:

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

Can we extend this result? That is, can it be shown that if $a,b \in \mathbb N$, then $a^{1/b}$ is an integer or an irrational number?

Answer 1

These (standard) results are discussed in detail in

http://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.

Answer 2

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 $x^t$. 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)]

Answer 3

Below I give a simple conceptual proof of irrationality of certain square roots - from first principles. Call a natural $\rm\,d > 0\,$ a denominator of a rational $\rm\:r\:$ 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. $\ \displaystyle\rm\sqrt{a}\; = \frac{a}{\sqrt{a}} \ \Rightarrow\ \frac{c}d = \frac{a\:d}c \ \Rightarrow\ d\:$ divides $\rm\: 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 $\;\rm n\,r,\: m\,r\in \mathbb Z \ \Rightarrow\: (n\!-\!m)\:r\in \mathbb Z.\:$ Now apply

Lemma' $\ \ $ If a set of positive integers $\rm\: M\:$ satisfies $\rm\ n > m \,\in\, M \ \Rightarrow\ \: n-m\, \in\, M$
then every element of $\rm\:M\:$ is a multiple of the least element $\rm\:\bar m \,\in\, M\:.$

Proof $\ \: $ If not there is a least nonmultiple $\rm\:n\in M,\:$ contra $\rm\:n-\bar m \,\in\, M\:$ is a nonmultiple of $\rm\:\bar m$

Remark $\:$ The above Lemma is quite fundamental to factorization. I frequently refer to it by the suggestive moniker unique fractionization, since it is equivalent to uniqueness of factorizations into irreducibles (indeed 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/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).

Answer 4

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, $n > 1$.

Answer 5

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.

Answer 6

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.

Answer 7

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. the gcd $\rm\,(a,b)\,$ of integers $\rm\,a,b\,$ may be expressed as an integral linear combination of the given integers: $\rm\:\ (a,b)\ =\ a\ d - b\ c\:.\ $

THEOREM $\quad \rm r = \sqrt{n}\;\;$ 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{#0C0}{\bf n}\:\Rightarrow\ 0\ =\ (a\!-\!br)\, (c\!+\!dr) \ =\ ac\!-\!bd\color{#0C0}{\bf n} \:+\: \color{#c00}{\bf 1}\cdot r \ \Rightarrow\ r \in \mathbb{Z}\ \ $ QED

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}\ \ $ QED

Nowadays my favorite proof is the $1$-line gem using Dedekind's conductor ideal. As I explained at length elsewhere, it beautifully encapsulates the denominator descent in ad-hoc "elementary" irrationality proofs.

See also my other post here. That post concisely proves the theorem after first proving from scratch the fundamental lemma that the least positive denominator $\rm\:b\:$ of a fraction divides every other denominator, i.e. $\rm\ a/b = c/d\ \Rightarrow\ b\ |\ d\ $ if $\rm\ a/b\ $ is in lowest terms. The irrationality proof follows immediately from this principal denominator ideal theorem ("unique fractionization"), namely

Proof $\ \ $ Suppose $\displaystyle\;\rm \sqrt{n} \:= \frac{a}b,\;\;$ least $\rm b>0.\ \ $ $\displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \ \Rightarrow\ \frac{a}b = \frac{nb}a \ \Rightarrow\ b\:|\:a\ \Rightarrow\ \sqrt{n}\ \in\ \mathbb Z$

Answer 8

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.

Answer 9

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.

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