Laws of rational exponents

Choose a common denominator: there are a natural $N \ge 1$ and integers $m, k$ with $r = m/N$ and $s = k/N$; then $r + s = (m+k)/N$ and $-r = (-m)/N$.

Roots of a product: for $a, b > 0$ and $N \ge 1$ the element $a^{1/N} b^{1/N}$ is positive and satisfies $\big(a^{1/N} b^{1/N}\big)^{N} = \big(a^{1/N}\big)^{N}\big(b^{1/N}\big)^{N} = ab$, so by uniqueness of the nonnegative $N$-th root $\big(ab\big)^{1/N} = a^{1/N} b^{1/N}$.

Claim 1: $a^{r} = \big(a^{1/N}\big)^{m}$ with $a^{1/N} > 0$, and a positive element has positive integer powers, so $a^{r} > 0$.

Claim 3: $(ab)^{r} = \big((ab)^{1/N}\big)^{m} = \big(a^{1/N} b^{1/N}\big)^{m} = \big(a^{1/N}\big)^{m}\big(b^{1/N}\big)^{m} = a^{r} b^{r}$, using the root-of-a-product identity and then the integer product law.

Claim 2: $a^{r+s} = \big(a^{1/N}\big)^{m+k} = \big(a^{1/N}\big)^{m}\big(a^{1/N}\big)^{k} = a^{r} a^{s}$, the middle equality being the integer addition law applied to the nonzero base $a^{1/N}$.

Claim 4: $a^{-r} = \big(a^{1/N}\big)^{-m} = \Big(\big(a^{1/N}\big)^{m}\Big)^{-1} = \big(a^{r}\big)^{-1}$.

Claim 5: write $r = m/n$ and $s = p/q$ with $n, q \ge 1$, and put $x := a^{1/(nq)}$, so $x > 0$ and $x^{nq} = a$; then $\big(x^{q}\big)^{n} = x^{qn} = a$ with $x^{q} > 0$, so $x^{q} = a^{1/n}$ by uniqueness of the nonnegative $n$-th root; putting $z := x^{m}$ we get $z > 0$ and $z^{q} = \big(x^{m}\big)^{q} = \big(x^{q}\big)^{m} = \big(a^{1/n}\big)^{m} = a^{r}$, so $z$ is the nonnegative $q$-th root of $a^{r}$, that is $z = \big(a^{r}\big)^{1/q}$; therefore $\big(a^{r}\big)^{s} = \Big(\big(a^{r}\big)^{1/q}\Big)^{p} = z^{p} = \big(x^{m}\big)^{p} = x^{mp} = \big(a^{1/(nq)}\big)^{mp} = a^{(mp)/(nq)} = a^{rs}$.

The supplementary nonnegative case, product identity: let $a, b \ge 0$ and let $r > 0$ be rational; if $a > 0$ and $b > 0$ this is step 2.2, and otherwise $a = 0$ or $b = 0$, so $ab = 0$ and the left side is $0^{r} = 0$, while the right side $a^{r}b^{r}$ has a factor $0^{r} = 0$ and is therefore $0$ as well.

The supplementary nonnegative case, addition identity: the identity $a^{r+s} = a^{r}a^{s}$ involves the base $a$ only, so nothing need be assumed about $b$; for $a > 0$ it is step 3.1 verbatim, which uses only $a > 0$, and both sides are then positive rather than $0$; for $a = 0$ the exponents satisfy $r + s > 0$, so the left side is $0^{r+s} = 0$ and the right side is $0 \cdot 0 = 0$.

All five claims hold for positive bases and arbitrary rational exponents, together with the two supplementary identities for nonnegative bases and positive rational exponents.