A rational root of $x^{k} = m$ is an integer: if $k \ge 1$, $m \in \mathbb{Z}$, $x \in \mathbb{Q}$ and $x^{k}$ is the image of $m$, then $x$ is the image of an integer

$0 < 1$ in $\mathbb{Z}$, and every integer $y > 0$ satisfies $y \ge 1$: $y = \iota(t)$ with $t \ne 0$, so $1 = \sigma(0) \le t$ and $\iota$ preserves the order.

For integers $u, w$ with $w \ne 0$ and every $t \in \mathbb{N}$: $w^{t} \ne 0$ and $[(u,w)]^{t} = [(u^{t}, w^{t})]$, powers on the left in $\mathbb{Q}$ and on the right in $\mathbb{Z}$. The set of $t$ for which this holds contains $0$, since $w^{0} = 1 \ne 0$ and $[(u,w)]^{0} = 1 = [(1,1)] = [(u^{0}, w^{0})]$; and if it holds at $t$ then $w^{\sigma(t)} = w^{t} w \ne 0$ and $[(u,w)]^{\sigma(t)} = [(u^{t},w^{t})] \cdot [(u,w)] = [(u^{t}u,\, w^{t}w)] = [(u^{\sigma(t)}, w^{\sigma(t)})]$. Induction finishes it.

Suppose first $x = 0$. Since $k \ge 1$, write $k = \sigma(t)$; then $x^{k} = x^{t} \cdot 0 = 0 = j(0)$, so $j(m) = j(0)$ and $m = 0$ by injectivity of $j$; and $x = 0 = j(0)$ is the image of an integer.

Suppose instead $x \ne 0$, and write $x = [(a,b)]$ with $b \ne 0$; then $a \ne 0$. If $b < 0$, replace $(a,b)$ by $(-a,-b)$, which represents the same rational because $a(-b) = (-a)b$; so we may assume $b > 0$.

For a prime $q$, an integer $u$ and $t \ge 1$: if $q \mid u^{t}$ then $q \mid u$. Let $S$ be the set of $t$ for which this implication holds; $0 \in S$ vacuously, since $t \ge 1$ fails there. Suppose $t \in S$ and $q \mid u^{\sigma(t)} = u^{t} u$. By [L9] either $q \mid u^{t}$ or $q \mid u$; in the second case we are done, and in the first, if $t \ge 1$ then $t \in S$ gives $q \mid u$, while if $t = 0$ then $u^{t} = 1$ and $q \mid 1$ is impossible for a prime, since $q > 1 > 0 > -1$ would then be contradicted. So $\sigma(t) \in S$ and $S = \mathbb{N}$.

Put $d := \gcd(a,b)$; since $a \ne 0$ we have $d \ge 1 > 0$, so $d \ne 0$. Put $a_1 := a/d$ and $b_1 := b/d$, so that $a = d a_1$, $b = d b_1$ and $\gcd(a_1,b_1) = 1$.

$a_1 \ne 0$ and $b_1 > 0$: if $a_1 = 0$ then $a = 0$, and if $b_1 \le 0$ then $b = d b_1 \le 0$, both contrary to step 1.4. Moreover $[(a,b)] = [(a_1,b_1)]$, because $a b_1 = (d a_1) b_1 = a_1 (d b_1) = a_1 b$.

By step 1.2, $x^{k} = [(a_1,b_1)]^{k} = [(a_1^{k}, b_1^{k})]$, and this equals $j(m) = [(m,1)]$, so $a_1^{k} \cdot 1 = m\, b_1^{k}$, that is $a_1^{k} = m\, b_1^{k}$.

Suppose $b_1 > 1$ and fix a prime $q$ with $q \mid b_1$. Since $k \ge 1$, write $k = \sigma(t)$; then $b_1^{k} = b_1^{t} b_1$, so $b_1 \mid b_1^{k}$ and hence $q \mid b_1^{k}$ by transitivity. Then $q \mid m\, b_1^{k} = a_1^{k}$, so $q \mid a_1$ by step 2.1.

So $q$ is a common divisor of $a_1$ and $b_1$, which are coprime, hence $q = 1$ or $q = -1$ by [L8] and [L14]; but $q > 1 > 0 > -1$, a contradiction. Therefore $b_1 \le 1$, and $b_1 > 0$ gives $b_1 \ge 1$, so $b_1 = 1$.

Hence $x = [(a_1,1)] = j(a_1)$ is the image of an integer; together with step 1.3 this covers both cases.