1.2.5. Fractional Indices

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Name	Status	Author	Last Modified
1.2.5. Fractional Indices	Ready to use	Merryn Horrocks	28/06/2023 05:34
1.2.5 Task: Simplify Fractional Indices	Ready to use	Merryn Horrocks	08/07/2023 07:17

Question statement

Fractional Indices

So far we have used indices that are whole numbers. We now consider fractional powers. Consider the expression $\displaystyle{\left(16^{\frac{1}{2}}\right)^2}$ . Using the third law of indices, $(a^m)^n=a^{mn}$ , we can write

$\left( 16^{\frac{1}{2}} \right)^2 = 16^{ \frac{1}{2} \times 2 } = 16^1 = 16.$

So $\displaystyle{16^{\frac{1}{2}}}$ is a number which when squared equals $16$ , that is $4$ or $−4$ . In other words $\displaystyle{16^{\frac{1}{2}}}$ is a square root of $16$ . There are always two square roots of a non-zero positive number, and we write $\displaystyle{16^{\frac{1}{2}}=\pm 4}$

Key Point 10

In general $\displaystyle{a^{\frac{1}{2}}}$ is a square root of $a$ , $a\geq 0$

Similarly

$\left(8^{\frac{1}{3}}\right)^3=8^{\frac{1}{3}\times 3}=8^1=8$

so that $\displaystyle{8^{\frac{1}{3}}}$ is a number which when cubed equals $8$ . Thus $\displaystyle{8^{\frac{1}{3}}}$ is the cube root of $8$ , that is $\sqrt[3]{8}$ , namely $2$ . Each number has only one cube root, and so $8^{\frac{1}{3}} = 2$

In general

Key Point 11

In general $\displaystyle{a^{\frac{1}{3}}}$ is the cube root of $a$

More generally we have

Key Point 12

The $n$ th root of $a$ is denoted by $\displaystyle{a^{\frac{1}{n}}}$

When $a<0$ the $n$ th root only exists if $n$ is odd.

When $a>0$ the positive $n$ th root is denoted by $\sqrt[n]{a}$ .

If $a<0$ the negative $n$ th root is $-\sqrt[n]{|a|}$ .

Your calculator will be able to evaluate fractional powers, and roots of numbers. Check that you can obtain the results of the following Examples on your calculator, but be aware that calculators normally give only one root when there may be others.

Example 22

Evaluate (a) $\displaystyle{144^{\frac{1}{2}}}$ , $\quad$ (b) $\displaystyle{125^{\frac{1}{3}}}$

Solution

(a) $\displaystyle{144^{\frac{1}{2}}}$ is a square root of $144$ , that is $\pm 12$ .

(b) Noting that $5^3=125$ , we see that $\displaystyle{125^{\frac{1}{3}}}=\sqrt[3]{125}=5$

Example 23

Evaluate (a) $\displaystyle{32^{\frac{1}{5}}}$ , $\quad$ (b) $\displaystyle{32^{\frac{2}{5}}}$ , $\quad$ (c) $\displaystyle{8^{\frac{2}{3}}}$

Solution

(a) $\displaystyle{32^{\frac{1}{5}}}$ is the $5$ th root of $32$ , that is $\sqrt[5]{32}$ . Now $2^5=32$ and so $\sqrt[5]{32}=2$ .

(b) Using the third law of indices we can write $\displaystyle{32^{\frac{2}{5}} = 32^{\left(2\times\frac{1}{5}\right)}=(32^{\frac{1}{5}})^2}$ .Thus

$32^{\frac{2}{5}} = ((32)^{\frac{1}{5}})^2=2^2=4$

$8^{\frac{2}{3}} = 8^{\left( 2\times\frac{1}{3}\right)}= (8^{\frac{1}{3}})^2 = 2^2=4$

Note the following alternatives:

$8^{\frac{2}{3}} = (8^{\frac{1}{3}})^2 = (8^2)^{\frac{1}{3}}$

Example 24

Write the following as a simple power with a single index:

(a) $\displaystyle{\sqrt{x^5}}$ , $\quad$ (b) $\displaystyle{\sqrt[4]{x^3}}$

Solution

(a) $\displaystyle{\sqrt{x^5}=(x^5)^{\frac{1}{2}}}$ . Then using the third law of indices we can write this as $\displaystyle{x^{\left(5\times\frac{1}{2}\right)}=x^{\frac{5}{2}}}$ .

(b) $\displaystyle{\sqrt[4]{x^3} = (x^3)^{\frac{1}{4}}}$ . Using the third law we can write this as $\displaystyle{ x^{\left( 3\times\frac{1}{4} \right)} = x^{ \frac{3}{4} } }$ .

Example 25

Show that $\displaystyle{ z^{-\frac{1}{2}}=\frac{1}{\sqrt{z}} }$ .

Solution

$z^{-\frac{1}{2}} = \frac{1}{z^\frac{1}{2} } = \frac{1}{\sqrt{z}}$

Give any introductory information the student needs.

			Name	Type	Generated Value

q1_expr	string	\frac\{\sqrt[3]\{h\}\}\{(h^\{1
q1_letter	string	h
q1v1	integer	3
q1r1	string	[3]
q1_num	string	\sqrt[3]\{h\}
q1v2	rational	1/4
q1v3	rational	-5
q1_den	string	(h^\{1/4\}h^\{-5\})
q1_ans	expression	h^(61/12)

			Name	Type	Generated Value

q2_expr	expression	4^(5/6)
q2_base	integer	4
q2_idx	rational	5/6
q2_idxden	integer	6
q2_idxnum	integer	5

			Name	Type	Generated Value

q3Letter	string	c
q3idxSurds	list	[ 1/2, 1/3, 1/4 ]
q3idxFracs	rational	-1/4
q3T1idx	integer	0
q3T1idxStr	string	0
q3T1Latex	string
q3T2idx	rational	-1/4
q3T2idxStr	string	-1/4
q3T2Latex	string	c^\{-1/4\}
q3T3idx	integer	0
q3T3idxStr	string	0
q3T3Latex	string
q3T4Idx	rational	1/4
q3T4idxStr	string	1/4
q3T4Latex	string	\sqrt[4]{c\}
q3_num	string	c^\{-1/4\}
q3_den	string	\sqrt[4]{c\}
q3_expr	string	\frac\{c^\{-1/4\}\}\{\sqrt[4]{
q3_ans	expression	c^(-1/2)

			Name	Type	Generated Value

q4_letter	string	d
q4i_arr	list	List of 8 items
q4indices	list	[ -1/5, 5, 4, 1/5, -1/4, 3 ]
q4a_expr	expression	(d^(-1/5))^5
q4b_expr	expression	d^4*d^(1/5)
q4c_expr	expression	d^(-1/4)/d^3
q4a_ans	expression	d^(-1)
q4b_ans	expression	d^(21/5)
q4c_ans	expression	d^(-13/4)

			Name	Type	Generated Value

alphabet

list

List of 21 items

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
latex("\\frac\{" + q1_num+"\}\{"+q1_den + "\}")

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Generated value: `string`

\frac\{\sqrt[3]\{h\}\}\{(h^\{1/4\}h^\{-5\})\}

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This question is used in the following exam:

HELM Book 1.2 Indices by Merryn Horrocks in HELM books.

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Merryn Horrocks 1 year, 9 months ago

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Fractional Indices

Key Point 10

Key Point 11

Key Point 12

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1.2.5. Fractional Indices

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

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Comment

Checkpoint description

Merryn Horrocks 1 year, 9 months ago

Merryn Horrocks 2 years, 10 months ago

Merryn Horrocks 2 years, 10 months ago

Fractional Indices

Key Point 10

Key Point 11

Key Point 12

Error in variable testing condition

Error

Warning

Generated value: string

Parts

Mathematical expression

Advanced settings

Checking accuracy

Variable value generators

Pattern restriction

Variables

String restrictions

Generated value: `string`