# Surds – Indices

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# Surds – Indices

### Introduction

Surds - Indices deals with problems on surds and indices involved with laws of surds and laws of indices respectively.

### Methods

Surds:
Surds are the  irrational numbers that contain the radical sign $\sqrt{}$. The set of irrational numbers contain numbers such as $\sqrt{2}$, $\sqrt[3]{2}$, $\pi$, etc.

• Surds can be added or subtracted only if they are like surds. Example: $\sqrt{2}$ + 5$\sqrt{2}$ = 6$\sqrt{2}$

• Surds can be multiplied using $\sqrt{x}$ x $\sqrt{x}$= $x$ and $\sqrt{x}$ x $\sqrt{y}$ = $\sqrt{xy}$

• To rationalize the denominator of a fraction, we need to convert it into equivalent fraction. Example: $\frac{1}{\sqrt{7}}$ = $\frac{1}{\sqrt{7}}$ x $\frac{\sqrt{7}}{\sqrt{7}}$ = $\frac{\sqrt{7}}{7}$

• A surd can be expressed in index form as a fractional index. Example: $\sqrt[n]{a}$ = $a^{(\frac{1}{n})}$ here $\sqrt[n]{a}$ is in surd form and $a^{(\frac{1}{n})}$ is in index form.

Indices
Indices are a useful way expressing large numbers with the help of powers or also known as indices. Exponent is another commonly used term for a power. In the below example the power/exponent of 4 is 6.
Example: 4096 is obtained by multiplying 4 by itself for 6 times and so 4096 can be expresses as:$4^6$  where 4 x 4 x 4 x 4 x 4 x 4 = 4096

• Laws of indices can be applied only to the expressions having equal bases. Example: $2^6$, $2^4$, $2^8, etc$

Example 1: Show that for any positive real number p, the expression $a^{-p}$ is equivalent to $\frac{1}{a^{p}}$.
Solution:
We proceed with the following manipulation – $a^{-p}$ = $a^{0 - p}$
Using Law 2 i.e. $\frac{a^{m}}{a^{n}}$ = $a^{(m - n)}$, we can rewrite the above expression as -
$\frac{a^{0}}{a^{p}}$
= $\frac{1}{a^{p}}$, which is the required result.

Example 2: Simplify and evaluate $(\frac{16}{81})^{-(\frac{3}{4})}$
Solution:
Using the laws of indices and some manipulation –
$(\frac{16}{81})^{-(\frac{3}{4})}$ = $\frac{1}{(\frac{16}{81})^{\frac{3}{4}}}$
= $(\frac{81}{16})^{\frac{3}{4}}$
= $((\frac{81}{16})^{\frac{1}{4}})^{3}$
= $(\frac{81^{\frac{1}{4}}}{16^{\frac{1}{4}}})^{3}$
= $(\frac{3}{2})^{3}$
= $\frac{3^{3}}{2^{3}}$
= $\frac{27}{8}$

Example 3: Simplify the expression $y = x^{a - b} \times x^{b - c} \times x^{c - a} \times x^{-a - b}$
Solution:
Using the laws of indices:
$y = x^{a - b} \times x^{b - c} \times x^{c - a} \times x^{-a - b}$
$y = x^{(a - b) + (b - c) + (c - a) + (-a - b)}$
$y = x^{-a - b}$
$y = \frac{1}{x^{a + b}}$

Laws of Surds/ Rules of Surds:
Example: Simplify $\sqrt{18}$
Solution:
Since 18 = 9 x 2 = $3^{2} \times 2$, as 9 is the largest perfect square factor of 18.
$\sqrt{18}$ = $\sqrt{3^{2} \times 2}$
= $\sqrt{3^{2}}$ x $\sqrt{2}$ (Using the rule $\sqrt{(a \times b)}$ = $\sqrt{a}$ x $\sqrt{b}$)
= 3$\sqrt{2}$

Example: Simplify $\sqrt{\frac{12}{121}}$
Solution:
$\sqrt{\frac{12}{121}}$ = $\frac{\sqrt{12}}{\sqrt{121}}$ (Using the rule $\sqrt{\frac{a}{b}}$ = $\frac{\sqrt{a}}{\sqrt{b}}$)
= $\frac{\sqrt{2^{2} \times 3}}{11}$ (Since 4 is the largest perfect square factor of 12)
= $\frac{\sqrt{2^{2} \times \sqrt{3}}}{11}$ (Using the rule $\sqrt{(a \times b)}$ = $\sqrt{a}$ x $\sqrt{b}$)
= $\frac{2\sqrt{3}}{11}$

Example: Rationalise $\frac{5}{\sqrt{7}}$
Solution:
$\frac{5}{\sqrt{7}}$ = $\frac{5}{\sqrt{7}}$ x $\frac{\sqrt{7}}{\sqrt{7}}$ (Multiply both numerator and denominator by $\sqrt{7}$)
= $5\frac{\sqrt{7}}{7}$

Example: Simplify $5\sqrt{6} + 4\sqrt{6}$
Solution:
$5\sqrt{6} + 4\sqrt{6}$ = (5 + 4)$\sqrt{6}$ (Using the rule $a\sqrt{c} \pm b\sqrt{c} = (a \pm b)\sqrt{c}$)
= $9\sqrt{6}$

Following this rule enables to rationalise the denominator.
Example: Rationalise $\frac{3}{2 + \sqrt{2}}$
Solution:
$\frac{3}{2 + \sqrt{2}}$ = $\frac{3}{2 + \sqrt{2}}$ x $\frac{2 - \sqrt{2}}{2 - \sqrt{2}}$ (Multiply the numerator and denominator by $2 - \sqrt{2}$)
= $\frac{6 - 3\sqrt{2}}{4 - 2}$
= $\frac{6 - 3\sqrt{2}}{2}$

Following this rule enables to rationalise the denominator.
Example: Rationalise $\frac{3}{2 - \sqrt{2}}$
Solution:
$\frac{3}{2 - \sqrt{2}}$ = $\frac{3}{2 - \sqrt{2}}$ x $\frac{2 + \sqrt{2}}{2 + \sqrt{2}}$ (Multiply the numerator and denominator by $2 + \sqrt{2}$)
= $\frac{6 + 3\sqrt{2}}{4 - 2}$
= $\frac{6 + 3\sqrt{2}}{2}$

### Formulae

1. Laws of Indices/ Laws of Exponents:

• $a^m$ x $a^n$ = $a^{m + n}$

• $\frac{a^m}{a^n}$ = $a^{m - n}$

• $({a^m})^{n}$ = $a^{mn}$

• $(ab)^m$ = $a^m$ x $b^m$

• $(\frac{a}{b})^n$ = $\frac{a^n}{b^n}$

• $a^0$ = 1

2. Laws of Surds:

• $\sqrt[n]{a}$ = $a^{(\frac{1}{n})}$

• $\sqrt[n]{ab}$ = $\sqrt[n]{a}$ x $\sqrt[n]{b}$

• $\sqrt[n]{\frac{a}{b}}$ = $\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

• $(\sqrt[n]{a})^n$ = a

• $\sqrt[m]{\sqrt[n]{a}}$ = $\sqrt[mn]{a}$

• $(\sqrt[n]{a})^m$ = $\sqrt[n]{a}^m$

### Samples

1 Find $\sqrt{{8}\sqrt{{8}\sqrt{{8}\sqrt{8.....∝}}}}$ = ?
Solution:
Assume $\sqrt{{8}\sqrt{{8}{\sqrt{{8.....}}}}}$ = $x$
Now, consider $\sqrt{{8}\sqrt{{8}\sqrt{{8}\sqrt{8.....∝}}}}$
⇒$\sqrt{8 x}$ = $x$
⇒$x^2$ = 8$x$
⇒$x$ = 8
Therefore, the value of $\sqrt{{8}\sqrt{{8}\sqrt{{8}\sqrt{8.....∝}}}}$ = 8

2. Simplify the value of $(125)^{(\frac{2}{3})}$
Solution:
Consider$(125)^{(\frac{2}{3})}$
=$(5^3)^{(\frac{2}{3})}$
=$5^2$
=25
Therefore, the value of $(125)^{(\frac{2}{3})}$ = 25

3. Find the value of [$(8)^{\frac{5}{3}}$ + $(8)^{\frac{-5}{3}}$]?
Solution:
Consider [$(8)^{\frac{5}{3}}$ + $(8)^{\frac{-5}{3}}$]
= [$(2^3)^{\frac{5}{3}}$ + $(2^3)^{\frac{-5}{3}}$]
= [$(2^5)$ + $(2^{-5})$]
= $(2^5)$ + $\frac{1}{(2^5)}$
= $\frac{2^5 + 1}{2^5}$
= $\frac{1024 + 1}{32}$
= $\frac{1025}{32}$
Therefore, [$(8)^{\frac{5}{3}}$ + $(8)^{\frac{-5}{3}}$] = $\frac{1025}{32}$

4. What will be the quotient if $(x^{-1}-1)$ is divided by $(x - 1)$?
Solution:
Given that
Dividend = $(x^{-1}-1)$
Divisor = $(x - 1)$
Now, Consider $\frac{(x^{-1}-1)}{(x - 1)}$
= $\frac{\frac{1}{x} -1}{(x - 1)}$
= $\frac{\frac{(1-x)}{x}}{(x - 1)}$
= $\frac{(1 - x)}{x}$ x $\frac{1}{(x - 1)}$
= $\frac{(1-x)}{(-x)(1-x)}$
= -$\frac{1}{x}$
Therefore, the required quotient = -$\frac{1}{x}$

5. Which is larger among $\sqrt[4]{6}$, $\sqrt[3]{4}$, $\sqrt[2]{5}$ ?
Solution:
Given surds are $\sqrt[4]{6}$, $\sqrt[3]{4}$, $\sqrt[2]{5}$
Surds are in the order of 4, 3, 2 respectively.
L.C.M. of 4, 3, 2 is 12
Now, change the each given surd in the order of 12
$\sqrt[4]{6}$ = ${6}^{(\frac{1}{4})}$ = ${6}^{(\frac{1}{4} * \frac{3}{3})}$ = ${6}^{(\frac{3}{12})}$ = ${(6^3)}^{(\frac{1}{12})}$ = ${(216)}^{(\frac{1}{12})}$
$\sqrt[3]{4}$ =${4}^{(\frac{1}{3})}$ = ${4}^{(\frac{1}{3} * \frac{4}{4})}$ = ${4}^{(\frac{4}{12})}$ = ${(4^4)}^{(\frac{1}{12})}$ = ${(256)}^{(\frac{1}{12})}$
$\sqrt[2]{5}$ =${5}^{(\frac{1}{2})}$ = ${5}^{(\frac{1}{2} * \frac{6}{6})}$ = ${5}^{(\frac{6}{12})}$ = ${(5^6)}^{(\frac{1}{12})}$ = ${(3125)}^{(\frac{1}{12})}$
Hence, ${(3125)}^{(\frac{1}{12})}$ > ${(256)}^{(\frac{1}{12})}$ > ${(216)}^{(\frac{1}{12})}$
Therefore, $\sqrt[2]{5}$ > $\sqrt[3]{4}$ > $\sqrt[4]{6}$