Powers, nth roots and rational exponents – Operations on the Set of Real Numbers (Senior 4 Mathematics) | Ganzaa

Operations on the Set of Real Numbers

Lessons

Senior 4 · Mathematics

Operations on the Set of Real Numbers

5 lessons

Lesson 2 of 5

Powers, nth roots and rational exponents

25 min

A sheet that keeps its shape

Fold a sheet of A4 paper in half across its long side and you get A5, a smaller sheet of exactly the same shape. Two A4 sheets side by side make one A3. That only works for one ratio of length to width. If a sheet has width 1 and length $r$ , its half has width $\frac{r}{2}$ and length 1, and "same shape" means

\frac{r}{1} = \frac{1}{\frac{r}{2}} \quad\Longrightarrow\quad r^2 = 2 \quad\Longrightarrow\quad r = \sqrt{2} .

The international standard ISO 216 fixes this ratio for the A series and adds one more rule: the largest sheet, A0, has an area of $1\,\mathrm{m^2}$ . Finding the size of A0 and A4 from these two rules needs a fourth root and powers such as $2^{-\frac{9}{4}}$ . This lesson extends powers to fractional exponents and roots to any index, with the conditions that make them safe.

Definition

Power with an integer exponent: for a real number $a$ and a positive integer $n$ , $a^n = a \times a \times \cdots \times a$ ( $n$ factors). For $a \ne 0$ we also define

a^0 = 1, \qquad a^{-n} = \frac{1}{a^n} .

In $a^{n}$ , $a$ is the base and $n$ the exponent (or index). Example: $2^{-3} = \frac{1}{8}$ and $\left(\frac{2}{5}\right)^{-2} = \frac{25}{4}$ .

Important

Laws of exponents. For real $a, b$ (non-zero wherever a zero or negative exponent appears) and integers $m, n$ :

Law	Example
$a^m \times a^n = a^{m+n}$	$3^5 \times 3^{-2} = 3^3 = 27$
$\frac{a^m}{a^n} = a^{m-n}$	$\frac{x^7}{x^{10}} = x^{-3}$
$(a^m)^n = a^{mn}$	$(5^{-1})^{2} = 5^{-2} = \frac{1}{25}$
$(a b)^{n} = a^{n} b^{n}$	$(2 x)^{3} = 8 x^{3}$
$\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$	$\left(\frac{3}{y}\right)^{2} = \frac{9}{y^2}$

The same laws hold for rational exponents, provided the bases are positive.

Definition

nth root: let $n \ge 2$ be an integer.

If $n$ is odd, every real number $a$ has exactly one real $n$ th root, written $\sqrt[n]{a}$ : the real number whose $n$ th power is $a$ .
If $n$ is even and $a \ge 0$ , $\sqrt[n]{a}$ is the non-negative number whose $n$ th power is $a$ . If $n$ is even and $a \lt 0$ , there is no real $n$ th root.

In $\sqrt[n]{a}$ , $n$ is the index and $a$ the radicand; the whole expression is a radical. For $n = 2$ the index is not written.

Examples: $\sqrt[3]{-125} = -5$ because $(- 5)^{3} = - 125$ ; $\sqrt[4]{81} = 3$ , not $\pm 3$ ; $\sqrt[4]{-16}$ is not a real number.

Warning

An even root of an even power gives an absolute value

\sqrt[n]{a^n} = \begin{cases} |a| & \text{if } n \text{ is even,} \\ a & \text{if } n \text{ is odd.} \end{cases}

So $\sqrt[4]{(-3)^4} = \sqrt[4]{81} = 3$ , while $\sqrt[5]{(-3)^5} = -3$ . Writing $\sqrt{x^2} = x$ without knowing that $x \ge 0$ is one of the commonest A-level errors.

Rational exponents

We want the law $(a^m)^n = a^{mn}$ to keep working when an exponent is a fraction. If $a^{\frac{1}{n}}$ is to make sense, then $\left(a^{\frac{1}{n}}\right)^n = a^{1} = a$ , so $a^{\frac{1}{n}}$ must be a number whose $n$ th power is $a$ . That forces the definition, for $a \gt 0$ and integers $m$ and $n \ge 2$ :

a^{\frac{1}{n}} = \sqrt[n]{a}, \qquad a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m} .

In practice, take the root first (the numbers stay small), then the power.

Why $a \gt 0$ ? With a negative base the "same" exponent can give different answers: $\frac{1}{3} = \frac{2}{6}$ , yet $(-8)^{\frac{1}{3}} = -2$ while $\sqrt[6]{(-8)^2} = \sqrt[6]{64} = 2$ . To avoid such contradictions, fractional exponents are used only with positive bases; for a negative radicand with an odd index, write the radical $\sqrt[3]{-8}$ instead.

Worked example 1: evaluating rational powers

Evaluate without a calculator: (a) $32^{\frac{3}{5}}$ (b) $27^{-\frac{2}{3}}$ (c) $\left(\frac{16}{81}\right)^{-\frac{3}{4}}$ (d) $0.001^{\frac{2}{3}}$ .

Solution

$32^{\frac{3}{5}} = \left(\sqrt[5]{32}\right)^3 = 2^3 = 8$ .
$27^{-\frac{2}{3}} = \frac{1}{27^{\frac{2}{3}}} = \frac{1}{\left(\sqrt[3]{27}\right)^2} = \frac{1}{3^2} = \frac{1}{9}$ .
A negative exponent inverts the fraction: $\left(\frac{16}{81}\right)^{-\frac{3}{4}} = \left(\frac{81}{16}\right)^{\frac{3}{4}} = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$ .
$0.001 = 10^{-3}$ , so $0.001^{\frac{2}{3}} = 10^{-3 \times \frac{2}{3}} = 10^{-2} = 0.01$ .

Properties of radicals

Each property below is a law of exponents written in radical form. Take $a, b \ge 0$ (when an index is odd, any real $a, b$ will do), $b \ne 0$ in a quotient, and positive integers $m, n, p$ with $n, p \ge 2$ .

Property	Radical form	Exponent form
Product	$\sqrt[n]{ab} = \sqrt[n]{a}\,\sqrt[n]{b}$	$(ab)^{\frac{1}{n}} = a^{\frac{1}{n}} b^{\frac{1}{n}}$
Quotient	$\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}$	$\left(\frac{a}{b}\right)^{\frac{1}{n}} = \frac{a^{\frac{1}{n}}}{b^{\frac{1}{n}}}$
Power	$\left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}$	$\left(a^{\frac{1}{n}}\right)^m = a^{\frac{m}{n}}$
Root of a root	$\sqrt[p]{\sqrt[n]{a}} = \sqrt[np]{a}$	$\left(a^{\frac{1}{n}}\right)^{\frac{1}{p}} = a^{\frac{1}{np}}$
Change of index	$\sqrt[np]{a^{mp}} = \sqrt[n]{a^m}$	$a^{\frac{mp}{np}} = a^{\frac{m}{n}}$

A radical is in simplest form when the radicand has no factor that is a perfect $n$ th power, there is no fraction under the root, and the index is as small as possible.

Worked example 2: simplest form

Simplify (a) $\sqrt{72}$ (b) $\sqrt[3]{54x^4}$ (c) $\sqrt[4]{48}$ (d) $\sqrt[6]{a^4}$ for $a \ge 0$ .

Solution Look for the largest perfect power of the index inside the radicand.

$72 = 36 \times 2$ , so $\sqrt{72} = \sqrt{36}\,\sqrt{2} = 6\sqrt{2}$ .
$54x^4 = 27x^3 \times 2x$ , so $\sqrt[3]{54x^4} = \sqrt[3]{27x^3}\,\sqrt[3]{2x} = 3x\sqrt[3]{2x}$ . (Odd index: no condition on $x$ .)
$48 = 16 \times 3 = 2^4 \times 3$ , so $\sqrt[4]{48} = 2\sqrt[4]{3}$ .
Change of index with $p = 2$ : $\sqrt[6]{a^4} = a^{\frac{4}{6}} = a^{\frac{2}{3}} = \sqrt[3]{a^2}$ .

Worked example 3: comparing radicals with different indices

Arrange $\sqrt{2}$ , $\sqrt[3]{3}$ and $\sqrt[5]{5}$ in increasing order without a calculator.

Solution Write them as powers and raise all three to the power 30, the lowest common multiple of 2, 3 and 5. Since $t \mapsto t^{30}$ is increasing for $t \gt 0$ , the order does not change.

\left(2^{\frac{1}{2}}\right)^{30} = 2^{15} = 32\,768, \qquad \left(3^{\frac{1}{3}}\right)^{30} = 3^{10} = 59\,049, \qquad \left(5^{\frac{1}{5}}\right)^{30} = 5^{6} = 15\,625 .

So $\sqrt[5]{5} \lt \sqrt{2} \lt \sqrt[3]{3}$ . (Check: about 1.380, 1.414 and 1.442.)

Worked example 4: the size of A0 and A4

Use the two ISO 216 rules (ratio $\sqrt{2}$ , A0 area $1\,\mathrm{m^2}$ ) to find the dimensions of A0 and of A4 in millimetres.

Solution Let A0 have width $w$ metres, so its length is $\sqrt{2}\,w = 2^{\frac{1}{2}} w$ . Its area is

2^{\frac{1}{2}} w^2 = 1 \;\Longrightarrow\; w^2 = 2^{-\frac{1}{2}} \;\Longrightarrow\; w = 2^{-\frac{1}{4}} \approx 0.8409\,\mathrm{m}, \qquad \text{length} = 2^{\frac{1}{2}} \cdot 2^{-\frac{1}{4}} = 2^{\frac{1}{4}} \approx 1.1892\,\mathrm{m}.

Halving a sheet keeps its shape and halves its area, so every length is multiplied by $\sqrt{\frac{1}{2}} = 2^{-\frac{1}{2}}$ . A4 is four halvings from A0:

\text{width} = 2^{-\frac{1}{4}} \left(2^{-\frac{1}{2}}\right)^4 = 2^{-\frac{9}{4}} \approx 0.2102\,\mathrm{m}, \qquad \text{length} = 2^{\frac{1}{4}} \left(2^{-\frac{1}{2}}\right)^4 = 2^{-\frac{7}{4}} \approx 0.2973\,\mathrm{m}.

To the nearest millimetre: A0 is $841\,\mathrm{mm} \times 1189\,\mathrm{mm}$ and A4 is $210\,\mathrm{mm} \times 297\,\mathrm{mm}$ , exactly the sizes printed in the standard.

Tip

When radicals with different indices are multiplied or divided, switch to exponents, use the laws, and switch back at the end: $\sqrt{x}\,\sqrt[3]{x} = x^{\frac{1}{2} + \frac{1}{3}} = x^{\frac{5}{6}} = \sqrt[6]{x^5}$ for $x \ge 0$ .

Numberphile: Root 2 (about 10 min). Watch for why √2 cannot be written as a fraction, and for its link with A-series paper.

✦

Practice

Work without a calculator unless told otherwise.

Evaluate: (a) $125^{\frac{2}{3}}$ (b) $16^{-\frac{3}{4}}$ (c) $\left(\sqrt[5]{-32}\right)^3$ .
Simplify: (a) $\sqrt[3]{-54}$ (b) $\sqrt[4]{162}$ (c) $\sqrt{x^6 y^3}$ for $y \ge 0$ .
Write as a single power of $x$ , for $x \gt 0$ : (a) $\sqrt{x}\,\sqrt[3]{x^2}$ (b) $\frac{\sqrt[4]{x^3}}{\sqrt{x}}$ (c) $\sqrt[3]{\sqrt{x}}$ .
Reduce the index: (a) $\sqrt[8]{a^6}$ , $a \ge 0$ (b) $\sqrt[12]{64}$ .
Which is larger, $\sqrt[4]{5}$ or $\sqrt[3]{3}$ ? Justify without a calculator.
Find $\sqrt[6]{(-2)^6}$ and $\sqrt[5]{(-2)^5}$ , and explain why the answers differ.
Show that the dimensions of A3 are $297\,\mathrm{mm} \times 420\,\mathrm{mm}$ to the nearest millimetre. (A calculator may be used for the last step.)
Solve $x^{\frac{3}{2}} = 27$ for $x \gt 0$ .

Show answersHide answers

(a) $5^{2} = 25$ (b) $\frac{1}{2^3} = \frac{1}{8}$ (c) $(- 2)^{3} = - 8$ .
(a) $-3\sqrt[3]{2}$ (b) $162 = 81 \times 2$ , so $3\sqrt[4]{2}$ (c) $\sqrt{x^6 y^2 \cdot y} = |x|^3\, y\sqrt{y}$ (the bars are needed: $x$ may be negative).
(a) $x^{\frac{1}{2} + \frac{2}{3}} = x^{\frac{7}{6}}$ (b) $x^{\frac{3}{4} - \frac{1}{2}} = x^{\frac{1}{4}}$ (c) $x^{\frac{1}{6}}$ .
(a) $\sqrt[4]{a^3}$ (b) $64 = 2^{6}$ , so $\sqrt[12]{2^6} = \sqrt{2}$ .
Raise both to the power 12: $\left(5^{\frac{1}{4}}\right)^{12} = 5^3 = 125$ and $\left(3^{\frac{1}{3}}\right)^{12} = 3^4 = 81$ . So $\sqrt[4]{5} \gt \sqrt[3]{3}$ .
$\sqrt[6]{64} = 2$ but $\sqrt[5]{-32} = -2$ : an even root is never negative, so $\sqrt[6]{a^6} = |a|$ , while an odd root keeps the sign, $\sqrt[5]{a^5} = a$ .
A3 is three halvings from A0: width $2^{-\frac{1}{4}} \cdot 2^{-\frac{3}{2}} = 2^{-\frac{7}{4}} \approx 0.2973\,\mathrm{m}$ , length $2^{\frac{1}{4}} \cdot 2^{-\frac{3}{2}} = 2^{-\frac{5}{4}} \approx 0.4204\,\mathrm{m}$ : $297\,\mathrm{mm} \times 420\,\mathrm{mm}$ .
Raise both sides to the power $\frac{2}{3}$ : $x = 27^{\frac{2}{3}} = 3^2 = 9$ . Check: $9^{\frac{3}{2}} = 3^3 = 27$ . ✓

Quick check

Key points

An odd root exists for every real number; an even root only for $a \ge 0$ , and it is never negative.
$\sqrt[n]{a^n} = |a|$ for even $n$ and $a$ for odd $n$ .
For $a \gt 0$ : $a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}$ , and all the laws of exponents apply.
To compare or combine radicals with different indices, bring them to a common index.

Sources and credits

Written by Ganzaa: the explanations, the diagram, the worked examples, the practice items and the questions.
The A-series rules and sheet sizes are from ISO 216, Wikipedia contributors, Wikipedia, CC BY-SA 4.0, adapted by Ganzaa.
Video: Root 2, Numberphile, embedded from YouTube.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1A sheet that keeps its shape

2Rational exponents

3Properties of radicals

A sheet that keeps its shape

Rational exponents

Properties of radicals