Quantitative Aptitude :: Surds and Indices

\( \mathbb{i.}\) \(a^{m} \times a^{n}\) = \(a^{m + n}\)

\( \mathbb{ii.}\) \(\frac{a^{m}}{a^{n}}\) = \(a^{m - n}\)

\( \mathbb{iii.}\) \(\left(a^{m}\right)^{n}\) = \(a^{mn}\)

\( \mathbb{iv.}\) \(\left(ab\right)^{n}\) = \(a^{n}b^{n}\)

\( \mathbb{v.}\) \(\left(\frac{a}{b}\right)^{n}\) = \(\frac{a^{n}}{b^{n}}\)

\( \mathbb{vi.}\) \(a^{0}\) = 1

Let a be rational number and n be a positive integer such that a^(1/n) = a

Then, \(\sqrt[n]{a}\) is called a surd of order n.

\(\mathbb{i.}\) \(\sqrt[n]{a}\) = \(a^{\frac{1}{n}}\)

\(\mathbb{ii.}\) \(\sqrt[n]{ab}\) = \(\sqrt[n]{a} \times \sqrt[n]{b}\)

\(\mathbb{iii.}\) \(\sqrt[n]{\frac{a}{b}}\) = \(\frac{\sqrt[n]{a}}{\sqrt[n]{b}}\)

\(\mathbb{iv.}\) \(\left(\sqrt[n]{a}\right)^{n}\) = a

\(\mathbb{v.}\) \(\sqrt[m]{\sqrt[n]{a}}\) = \(\sqrt[mn]{a}\)

\(\mathbb{vi.}\) \(\left(\sqrt[n]{a}\right)^{m}\) = \(\sqrt[n]{a^{m}}\)