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Laws of rational exponents

Statement

Let $a, b \in R$ with $a, b > 0$ and let $r, s \in Q$ , with rational powers as in Rational powers $a^{r}$ of a positive base. Then:

$a^{r} > 0$ .
$a^{r + s} = a^{r} a^{s}$ .
$(a b)^{r} = a^{r} b^{r}$ ; in particular $(a b)^{1 / N} = a^{1 / N} b^{1 / N}$ for every natural $N \geq 1$ .
$a^{- r} = (a^{r})^{- 1} = 1 / a^{r}$ .
$(a^{r})^{s} = a^{r s}$ .

Claims 2 and 3 persist in the supplementary case of Rational powers $a^{r}$ of a positive base: for $a, b \geq 0$ and rationals $r, s > 0$ (Order on the rationals) one still has $(a b)^{r} = a^{r} b^{r}$ and $a^{r + s} = a^{r} a^{s}$ . The two identities degenerate differently, and it is worth saying how. In the product identity, a zero base on either side makes both sides $0$ . In the addition identity only the base $a$ occurs, so it degenerates only when $a = 0$ , and then both sides are $0$ ; when $a > 0$ that identity holds with no hypothesis on $b$ at all.

Facts & Assumptions

Given: Reals $a, b > 0$ and rationals $r, s$ .

[L1]

Definition and well-definedness (Rational powers $a^{r}$ of a positive base, Rational powers do not depend on the representative): for ANY representative $r = m / N$ with $m \in Z$ and $N \geq 1$ natural, $a^{r} = (a^{1 / N})^{m}$ ; and $a^{1 / N}$ is the unique $x \geq 0$ with $x^{N} = a$ (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ), which is $> 0$ when $a > 0$ .

[L2]

Laws of integer exponents (Laws of integer exponents, Integer powers $a^{m}$ ): for $x, y \neq 0$ and integers $j, k$ , $x^{j + k} = x^{j} x^{k}$ , $(x^{j})^{k} = x^{j k}$ , $(x y)^{j} = x^{j} y^{j}$ and $x^{- j} = (x^{j})^{- 1}$ .

[L3]

Positivity and injectivity: $x > 0$ implies $x^{j} > 0$ for every NATURAL $j$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , claim 1), and hence for every integer $j$ , since $x^{- j} = (x^{j})^{- 1}$ (Laws of integer exponents, claim 2) and the inverse of a positive element is positive (Inverses of positives are positive, and reciprocation reverses order); and $x \mapsto x^{N}$ is injective on ${x \geq 0}$ for $N \geq 1$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , claim 2).

[L4]

Rational arithmetic (Arithmetic on the rationals, The rationals as equivalence classes of pairs of integers): any two rationals can be written with a common positive denominator, $m / N + k / N = (m + k) / N$ , $- (m / N) = (- m) / N$ , and $(m / n) (p / q) = (m p) / (n q)$ .

[L5]

The order on $Q$ (The rationals form a totally ordered field, Order on the rationals) is compatible with addition, so $r > 0$ and $s > 0$ imply $r + s > 0$ .

[L6]

The supplementary clause of Rational powers $a^{r}$ of a positive base: $0^{t} = 0$ for every rational $t > 0$ , while $0^{t}$ is left undefined for rational $t < 0$ and the convention $0^{0} = 1$ of Integer powers $a^{m}$ is untouched. In a field, a product with a factor $0$ is $0$ (Multiplication by zero: $0 \cdot a = 0$ ).

Proof

technique · direct

1.1L4

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

1.2L1L2L3

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

2.1step 1.1L1L3

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

2.2step 1.1step 1.2L1L2

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

3.1step 1.1step 2.1L1L2

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

3.2step 1.1step 2.1L1L2

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

3.3step 2.1L1L2L3L4

Claim 5: write $r = m / n$ and $s = p / q$ with $n, q \geq 1$ , and put $x : = a^{1 / (n q)}$ , so $x > 0$ and $x^{n q} = a$ ; then $(x^{q})^{n} = x^{q n} = a$ 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 $z^{q} = (x^{m})^{q} = (x^{q})^{m} = (a^{1 / n})^{m} = a^{r}$ , so $z$ is the nonnegative $q$ -th root of $a^{r}$ , that is $z = (a^{r})^{1 / q}$ ; therefore $(a^{r})^{s} = ((a^{r})^{1 / q})^{p} = z^{p} = (x^{m})^{p} = x^{m p} = (a^{1 / (n q)})^{m p} = a^{(m p) / (n q)} = a^{r s}$ .

3.4step 2.2L6

The supplementary nonnegative case, product identity: let $a, b \geq 0$ 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 $a b = 0$ 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.

4.1step 3.1L5L6

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 $0 \cdot 0 = 0$ .

5.1step 2.1step 3.1step 3.2step 2.2step 3.3step 3.4step 4.1∎

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

Laws of rational exponents

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources