# Irrationality of roots.
Let $b$ and $n$ both be positive integers. We claim that $\sqrt[n]{b}$ is either an integer or irrational, and it cannot be a non-integer rational.
Let us give Niven's proof of this fact, using the **rational root theorem**, to which we remind us here.
**Rational root theorem.** $\quad$ Suppose an integer polynomial $a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}$ has a rational root $p / q$, with $p,q$ integers and $\gcd(p,q) = 1$. Then $p \mid a_{0}$ and $q \mid a_{n}$.
**Proof.** $\quad$ Substituting the rational root $x = p / q$ into the polynomial, we get $$
a_{n} \left( \frac{p}{q} \right)^{n} + a_{n-1} \left( \frac{p}{q} \right)^{n-1} + \cdots + a_{0} = 0.
$$By clearing the denominator, we get $$
a_{n} p^{n} + a_{n-1}p^{n-1} q + \cdots + a_{0}q ^{n} = 0
$$
By factoring, we see that $$
a_{n}p^{n} = -q(a_{n-1}p^{n-1}+\cdots+a_{0}q^{n-1}),
$$so $q$ divides $a_{n}p^{n}$. But $\gcd(p,q) = 1$, hence $q \mid a_{n}$. And we can factor the other way, where $$
a_{0}q^{n} = -p(a_{n}p^{n-1}+\cdots + a_{1}q^{n-1}),
$$and similarly we see $p$ divides $a_{0}q^{n}$, which means $p \mid a_{0}$, as $\gcd(p,q)=1$. We have as claimed. $\blacksquare$
Now we prove our claim.
**Proof of claim.** $\quad$ Take $b$ to be a positive integer, and consider the polynomial $x^{n} - b$. Note that $\sqrt[n]{b}$ is a root of the polynomial. Suppose $\sqrt[n]{b} = p / q$ is rational with $\gcd(p,q)=1$, then $p \mid b$ and $q \mid 1$. Hence $q = \pm1$ and $\sqrt[n]{b} = p/q$ is an integer. $\blacksquare$