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

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


• Surds can be multiplied using [latex]\sqrt{x}[/latex] x [latex]\sqrt{x}[/latex]= [latex]x[/latex] and [latex]\sqrt{x}[/latex] x [latex]\sqrt{y}[/latex] = [latex]\sqrt{xy}[/latex]


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


• A surd can be expressed in index form as a fractional index. Example: [latex]\sqrt[n]{a}[/latex] = [latex]a^{(\frac{1}{n})}[/latex] here [latex]\sqrt[n]{a}[/latex] is in surd form and [latex]a^{(\frac{1}{n})}[/latex] 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:[latex]4^6[/latex]  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: [latex]2^6[/latex], [latex]2^4[/latex], [latex]2^8, etc[/latex]

Example 1: Show that for any positive real number p, the expression [latex]a^{-p}[/latex] is equivalent to [latex]\frac{1}{a^{p}}[/latex].
Solution:

We proceed with the following manipulation – [latex]a^{-p}[/latex] = [latex]a^{0 - p}[/latex]
Using Law 2 i.e. [latex]\frac{a^{m}}{a^{n}}[/latex] = [latex]a^{(m - n)}[/latex], we can rewrite the above expression as -
[latex]\frac{a^{0}}{a^{p}}[/latex]
= [latex]\frac{1}{a^{p}}[/latex], which is the required result.

Example 2: Simplify and evaluate [latex](\frac{16}{81})^{-(\frac{3}{4})}[/latex]
Solution:

Using the laws of indices and some manipulation –
[latex](\frac{16}{81})^{-(\frac{3}{4})}[/latex] = [latex]\frac{1}{(\frac{16}{81})^{\frac{3}{4}}}[/latex]
= [latex](\frac{81}{16})^{\frac{3}{4}}[/latex]
= [latex]((\frac{81}{16})^{\frac{1}{4}})^{3}[/latex]
= [latex](\frac{81^{\frac{1}{4}}}{16^{\frac{1}{4}}})^{3}[/latex]
= [latex](\frac{3}{2})^{3}[/latex]
= [latex]\frac{3^{3}}{2^{3}}[/latex]
= [latex]\frac{27}{8}[/latex]

Example 3: Simplify the expression [latex]y = x^{a - b} \times x^{b - c} \times x^{c - a} \times x^{-a - b}[/latex]
Solution:

Using the laws of indices:
[latex]y = x^{a - b} \times x^{b - c} \times x^{c - a} \times x^{-a - b}[/latex]
[latex]y = x^{(a - b) + (b - c) + (c - a) + (-a - b)}[/latex]
[latex]y = x^{-a - b}[/latex]
[latex]y = \frac{1}{x^{a + b}}[/latex]

Laws of Surds/ Rules of Surds:
Example: Simplify [latex]\sqrt{18}[/latex]
Solution:

Since 18 = 9 x 2 = [latex]3^{2} \times 2[/latex], as 9 is the largest perfect square factor of 18.
[latex]\sqrt{18}[/latex] = [latex]\sqrt{3^{2} \times 2}[/latex]
= [latex]\sqrt{3^{2}}[/latex] x [latex]\sqrt{2}[/latex] (Using the rule [latex]\sqrt{(a \times b)}[/latex] = [latex]\sqrt{a}[/latex] x [latex]\sqrt{b}[/latex])
= 3[latex]\sqrt{2}[/latex]

Example: Simplify [latex]\sqrt{\frac{12}{121}}[/latex]
Solution:

[latex]\sqrt{\frac{12}{121}}[/latex] = [latex]\frac{\sqrt{12}}{\sqrt{121}}[/latex] (Using the rule [latex]\sqrt{\frac{a}{b}}[/latex] = [latex]\frac{\sqrt{a}}{\sqrt{b}}[/latex])
= [latex]\frac{\sqrt{2^{2} \times 3}}{11}[/latex] (Since 4 is the largest perfect square factor of 12)
= [latex]\frac{\sqrt{2^{2} \times \sqrt{3}}}{11}[/latex] (Using the rule [latex]\sqrt{(a \times b)}[/latex] = [latex]\sqrt{a}[/latex] x [latex]\sqrt{b}[/latex])
= [latex]\frac{2\sqrt{3}}{11}[/latex]

Example: Rationalise [latex]\frac{5}{\sqrt{7}}[/latex]
Solution:

[latex]\frac{5}{\sqrt{7}}[/latex] = [latex]\frac{5}{\sqrt{7}}[/latex] x [latex]\frac{\sqrt{7}}{\sqrt{7}}[/latex] (Multiply both numerator and denominator by [latex]\sqrt{7}[/latex])
= [latex]5\frac{\sqrt{7}}{7}[/latex]

Example: Simplify [latex]5\sqrt{6} + 4\sqrt{6}[/latex]
Solution:

[latex]5\sqrt{6} + 4\sqrt{6}[/latex] = (5 + 4)[latex]\sqrt{6}[/latex] (Using the rule [latex]a\sqrt{c} \pm b\sqrt{c} = (a \pm b)\sqrt{c}[/latex])
= [latex]9\sqrt{6}[/latex]

Following this rule enables to rationalise the denominator.
Example: Rationalise [latex]\frac{3}{2 + \sqrt{2}}[/latex]
Solution:

[latex]\frac{3}{2 + \sqrt{2}}[/latex] = [latex]\frac{3}{2 + \sqrt{2}}[/latex] x [latex]\frac{2 - \sqrt{2}}{2 - \sqrt{2}}[/latex] (Multiply the numerator and denominator by [latex]2 - \sqrt{2}[/latex])
= [latex]\frac{6 - 3\sqrt{2}}{4 - 2}[/latex]
= [latex]\frac{6 - 3\sqrt{2}}{2}[/latex]

Following this rule enables to rationalise the denominator.
Example: Rationalise [latex]\frac{3}{2 - \sqrt{2}}[/latex]
Solution:

[latex]\frac{3}{2 - \sqrt{2}}[/latex] = [latex]\frac{3}{2 - \sqrt{2}}[/latex] x [latex]\frac{2 + \sqrt{2}}{2 + \sqrt{2}}[/latex] (Multiply the numerator and denominator by [latex]2 + \sqrt{2}[/latex])
= [latex]\frac{6 + 3\sqrt{2}}{4 - 2}[/latex]
= [latex]\frac{6 + 3\sqrt{2}}{2}[/latex]

1. Laws of Indices/ Laws of Exponents:

• [latex]a^m[/latex] x [latex]a^n[/latex] = [latex]a^{m + n}[/latex]


• [latex]\frac{a^m}{a^n}[/latex] = [latex]a^{m - n}[/latex]


• [latex]({a^m})^{n}[/latex] = [latex]a^{mn}[/latex]


• [latex](ab)^m[/latex] = [latex]a^m[/latex] x [latex]b^m[/latex]


• [latex](\frac{a}{b})^n[/latex] = [latex]\frac{a^n}{b^n}[/latex]


• [latex]a^0[/latex] = 1

2. Laws of Surds:

• [latex]\sqrt[n]{a}[/latex] = [latex]a^{(\frac{1}{n})}[/latex]


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


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


• [latex](\sqrt[n]{a})^n[/latex] = a


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


• [latex](\sqrt[n]{a})^m[/latex] = [latex]\sqrt[n]{a}^m[/latex]

1 Find [latex]\sqrt{{8}\sqrt{{8}\sqrt{{8}\sqrt{8.....∝}}}}[/latex] = ?
Solution:

Assume [latex]\sqrt{{8}\sqrt{{8}{\sqrt{{8.....}}}}}[/latex] = [latex] x [/latex]
Now, consider [latex]\sqrt{{8}\sqrt{{8}\sqrt{{8}\sqrt{8.....∝}}}}[/latex]
⇒[latex]\sqrt{8 x}[/latex] = [latex] x [/latex]
⇒[latex] x^2 [/latex] = 8[latex] x [/latex]
⇒[latex] x [/latex] = 8
Therefore, the value of [latex]\sqrt{{8}\sqrt{{8}\sqrt{{8}\sqrt{8.....∝}}}}[/latex] = 8

2. Simplify the value of [latex](125)^{(\frac{2}{3})}[/latex]
Solution:

Consider[latex](125)^{(\frac{2}{3})}[/latex]
=[latex](5^3)^{(\frac{2}{3})}[/latex]
=[latex]5^2[/latex]
=25
Therefore, the value of [latex](125)^{(\frac{2}{3})}[/latex] = 25

3. Find the value of [[latex](8)^{\frac{5}{3}}[/latex] + [latex](8)^{\frac{-5}{3}}[/latex]]?
Solution:

Consider [[latex](8)^{\frac{5}{3}}[/latex] + [latex](8)^{\frac{-5}{3}}[/latex]]
= [[latex](2^3)^{\frac{5}{3}}[/latex] + [latex](2^3)^{\frac{-5}{3}}[/latex]]
= [[latex](2^5)[/latex] + [latex](2^{-5})[/latex]]
= [latex](2^5)[/latex] + [latex]\frac{1}{(2^5)}[/latex]
= [latex]\frac{2^5 + 1}{2^5}[/latex]
= [latex]\frac{1024 + 1}{32}[/latex]
= [latex]\frac{1025}{32}[/latex]
Therefore, [[latex](8)^{\frac{5}{3}}[/latex] + [latex](8)^{\frac{-5}{3}}[/latex]] = [latex]\frac{1025}{32}[/latex]

4. What will be the quotient if [latex](x^{-1}-1)[/latex] is divided by [latex] (x - 1)[/latex]?
Solution:

Given that
Dividend = [latex](x^{-1}-1)[/latex]
Divisor = [latex] (x - 1)[/latex]
Now, Consider [latex]\frac{(x^{-1}-1)}{(x - 1)}[/latex]
= [latex]\frac{\frac{1}{x} -1}{(x - 1)}[/latex]
= [latex]\frac{\frac{(1-x)}{x}}{(x - 1)}[/latex]
= [latex]\frac{(1 - x)}{x}[/latex] x [latex]\frac{1}{(x - 1)}[/latex]
= [latex]\frac{(1-x)}{(-x)(1-x)}[/latex]
= -[latex]\frac{1}{x}[/latex]
Therefore, the required quotient = -[latex]\frac{1}{x}[/latex]

5. Which is larger among [latex]\sqrt[4]{6}[/latex], [latex]\sqrt[3]{4}[/latex], [latex]\sqrt[2]{5}[/latex] ?
Solution:

Given surds are [latex]\sqrt[4]{6}[/latex], [latex]\sqrt[3]{4}[/latex], [latex]\sqrt[2]{5}[/latex]
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
[latex]\sqrt[4]{6}[/latex] = [latex]{6}^{(\frac{1}{4})}[/latex] = [latex]{6}^{(\frac{1}{4} * \frac{3}{3})}[/latex] = [latex]{6}^{(\frac{3}{12})}[/latex] = [latex]{(6^3)}^{(\frac{1}{12})}[/latex] = [latex]{(216)}^{(\frac{1}{12})}[/latex]
[latex]\sqrt[3]{4}[/latex] =[latex]{4}^{(\frac{1}{3})}[/latex] = [latex]{4}^{(\frac{1}{3} * \frac{4}{4})}[/latex] = [latex]{4}^{(\frac{4}{12})}[/latex] = [latex]{(4^4)}^{(\frac{1}{12})}[/latex] = [latex]{(256)}^{(\frac{1}{12})}[/latex]
[latex]\sqrt[2]{5}[/latex] =[latex]{5}^{(\frac{1}{2})}[/latex] = [latex]{5}^{(\frac{1}{2} * \frac{6}{6})}[/latex] = [latex]{5}^{(\frac{6}{12})}[/latex] = [latex]{(5^6)}^{(\frac{1}{12})}[/latex] = [latex]{(3125)}^{(\frac{1}{12})}[/latex]
Hence, [latex]{(3125)}^{(\frac{1}{12})}[/latex] > [latex]{(256)}^{(\frac{1}{12})}[/latex] > [latex]{(216)}^{(\frac{1}{12})}[/latex]
Therefore, [latex]\sqrt[2]{5}[/latex] > [latex]\sqrt[3]{4}[/latex] > [latex]\sqrt[4]{6}[/latex]