Show that if $n > 1$ is not a $k$-th power, then its $k$-th root is irrational, i.e. not of the form $u/v$ for $u, v ∈ N$.

Question

any solutions or suggestions on this one?

An integer $n$ is called a $k$-th power if $n = m^k$ for some $m ∈ Z$. Show that if $n > 1$ is not a $k$-th power, then its $k$-th root is irrational, i.e. not of the form $u/v$ for $u, v ∈ N$.

Answer 1

Assume $\frac{u}{v}$ is in its lowest terms, i.e. $u$ and $v$ have no common factors, then $\frac{u^k}{v^k}$ is in its lowest terms, and will not be an integer unless $v=1$.

Answer 2

If $\sqrt[k]{n} = u/v$, then $v^k n = u^k$. Consider now the power of each prime $p$ that appears in the factorization on both sides. You get a number of the form $ak+c$ on the LHS and a number of the form $bk$ on the $RHS$. Therefore, $ak+c=bk$ and so $c$ is a multiple of $k$. This means that $n$ is a $k$-th power.

Answer 3

Claim 1: Let $c$ and $d$ be integers. If there is a prime $p \ge 2$ that divides $d$ but not $c$, then $\frac{c}{d}$ is not an integer.

Suppose the $k$-th power of $n$ is rational but nonintegral. Then let $a$ and $b$ be relatively prime integers satisfying $(\frac{a}{b})^k = m$. Then as $a$ and $b$ are relatively prime, there is a prime $p \ge 2$ that divides $b$ but not $a$. Thus letting $c=a^k$ and $d=b^k$, that prime $p$ divides $d$ but not $c$ [Note that the prime factors of $m^k$ are the same as that of $m$ for every integer $m$ and every positive integer $k$.] Thus $\frac{c}{d} = \frac{a^k}{b^k} = (\frac{a}{b})^k$ cannot be the integer $n$ after all by Claim 1.

Answer 4

How about proving by the contrapositive. We only care about $m\in \Bbb Q / \{\Bbb Z\}$, so let: $$m=\frac ab, a,b \text{ coprime }, b\neq 1$$

Then $$m^k =\frac{a^k}{b^k}$$ Which is also irreducible. In other words.

$$m\in \Bbb Q / \{\Bbb Z\} \implies m^k\in \Bbb Q / \{\Bbb Z\}$$

We thusly rule out the possibility $m$ is a rational non-integer given integer $m^k$, so it must either be an integer or irrational.

Answer 5

Your claim is equivalent to this:

If the $k$-th root of $n$ is rational, then $n$ is a $k$-th power.

So let $(u/v)^k = n$, i.e. $u^k = n v^k$. We may assume that $\text{gcd}(u,v) = 1$. Then $u^k$ must divide $n$ which means $n = m u^k$ for some $m \in \mathbb{N}$. This yields $u^k = m u^k v^k$, i.e. $1 = m v^k$ which is possible only when $m = 1$ and $v = 1$. Hence $u/v = u \in \mathbb{N}$.

Answer 6

By contrapositive, without using prime factorisation:

Set $x = \sqrt[\uproot{1}\leftroot{-1}n]{a}$, and denote $y$ its decimal part. Suppose $x$ is rational.

Claim: $y = 0$ (so that $\sqrt[\uproot{1}\leftroot{-1}n]{a}$ is an integer).

Let $q$ be the smallest positive integer such that each $qx^i$, for $i = 1, \dots, n-1$, is an integer. Note that $q' = qy$ is an integer, since $qx$ is.

On the other hand, $$ q'x^i = q( x-\lfloor x\rfloor)x^i = qx^{i + 1} - \lfloor x\rfloor qx^i.$$ By hypothesis, in this formula, $qx^i$ and $qx^{i + 1}$ are integers if $i < n-1$ – and if $i = n-1$, $qx^{i + 1} = qa$ is an integer too. There results that each $q'x^i$ is an integer.

However $q$ is the smallest positive integer which has this property, and $0 ≤ q' < q$,so necessarily $q' = 0$, whence $y = 0$.