Up until now we have been dealing with \emph{whole} numbers in our powers. We have been manipulating expressions such as \(a^n\) where $n \in \mathbb{Z}$. But truthfully, there is no reason why we must confine ourselves to $\mathbb{Z}$, we have the tools to do arithmetic with $\mathbb{Q}$, \fillin[the set of all quotients][2in]. Later, if you fall in the black hole of mathematics, you will learn to grasp $a$ to the power of \emph{all real} numbers $\mathbb{R}$, things such as \(2^\pi\) and later even complex numbers (denoted $\mathbb{C}$) to understand the most beautiful equation in mathematics (by votes): \[e^{\pi i} = -1\] But for now we will just understand the \emph{quotient} powers thoroughly. There are 3 main types: \begin{enumerate} \begin{centering} \item numerator of one: $2^\frac{1}{3}$ \item numerator of not one: $2^\frac{2}{3}$ \item negative versions of the above cases $\uparrow$ \end{centering} \end{enumerate} \begin{derivebox}%TODO: update the boxstyle You could just remember that $\sqrt{a} = a^\frac{1}{2}$. Or you could \emph{understand} that this is obviously true from the arithmetic laws you learnt last lesson. Recall that $(a^2)^2 = a^4$, then $a^1 = (a^\frac{1}{2})^2$ which we could say is the same as $a^\frac{1}{2} \times a^\frac{1}{2}$. But then which mathematical entity becomes itself when squared? \begin{solutionorbox} $\sqrt{a}$ \end{solutionorbox} Thus, obviously $a^\frac{1}{2} = \fillin[$\sqrt{a}$]$ \end{derivebox} \begin{examplebox} \subsection{Examples:} \begin{questions} \Question[3]$\,$ \begin{parts}\begin{multicols}{3} \part $100^\frac{1}{2} = \fillin[10]$ \part $4^\frac{1}{2} = \fillin[2]$ \part $256^\frac{1}{2} = \fillin[16]$ \end{multicols}\end{parts} \end{questions} \end{examplebox} \begin{derivebox} We can extend this more generally, where the denominator does not need to be 2. Recall that $\sqrt[3]{8} = \fillin[2]$.\\ And so $8^\frac{1}{3} = 2$ and $a^\frac{1}{n} = \sqrt[n]{a}$. \end{derivebox} \begin{examplebox} \subsection{Examples} \begin{questions} \Question[6]$\,$ \begin{multicols}{3} \begin{parts} \part $\sqrt[3]{27} = \fillin[3]$ \part $\sqrt[4]{16} = \fillin[2]$ \part $\sqrt[3]{125} = \fillin[2]$ \end{parts} \begin{parts} \part $9^\frac{1}{2} = \fillin[3]$ \part $27^\frac{1}{3} = \fillin[3]$ \part $16^\frac{1}{4} = \fillin[2]$ \end{parts} \end{multicols} \end{questions} \end{examplebox} \begin{derivebox} Notice that up until now all of the numerators have been the number 1. It is easy to equip ourselves to handle bigger numbers though. Just recall the law from last lesson: \[(a^m)^n = a^mn\] Which we can apply to the \(a^\frac{1}{n}\)'s we have by raising this to whatever number we need. Thus \(a^\frac{2}{3} = (a^\frac{1}{3})^2\). We can even change the order around to be $(a^2)^\frac{1}{3}$ which then looks nicer as $\sqrt[3]{a^2}$. \begin{theorembox} \subsection*{Positive fractional indices} \[a^\frac{p}{q} = (a^\frac{1}{q})^p = (\sqrt[q]{a})^p\] \end{theorembox} \end{derivebox} \begin{examplebox} \subsection{Examples:} \begin{questions} \Question[3]$\,$ \begin{multicols}{3}\begin{parts} \part $4^\frac{3}{2} = \fillin[8]$ \part $8^\frac{2}{3} = \fillin[4]$ \part $81^\frac{3}{4} = \fillin[27]$ \end{parts}\end{multicols} \end{questions} \end{examplebox} \begin{derivebox} The final thing in this section to learn is the \emph{negative fractional indices}. Recall that \[a^{-m} = \frac{1}{a^m}\]. Then in exactly the same fashion, \[a^{-\frac{p}{q}} = \frac{1}{a^\frac{p}{q}}\] \end{derivebox} \begin{examplebox} \subsection{Examples:} \begin{questions} \Question[2]$\,$ \begin{multicols}{2}\begin{parts} \part $4^{-\frac{3}{2}}=$ \begin{solutionordottedlines} $\frac{1}{8}$ \end{solutionordottedlines} \part $(\frac{8}{27})^{-\frac{1}{3}}=$ \begin{solutionordottedlines} $\frac{3}{2}$ \end{solutionordottedlines} \end{parts}\end{multicols} \end{questions} \end{examplebox} \begin{exercisebox} \subsection{Exercises:} \begin{questions} \Question[3] Evaluate: \begin{parts} \begin{multicols}{3} \part \(\sqrt[3]{8}\) \begin{solutionordottedlines}[2cm] 2 \end{solutionordottedlines} \part \(\sqrt[5]{32}\) \begin{solutionordottedlines}[2cm] 2 \end{solutionordottedlines} \part \(\sqrt[3]{216}\) \begin{solutionordottedlines}[2cm] 6 \end{solutionordottedlines} \end{multicols} \end{parts} \Question[3] Write using fractional indices. Evaluate, correct to 4 decimal places. \begin{parts} \begin{multicols}{3} \part \(\sqrt{14}\) \begin{solutionordottedlines}[2cm] \(14^{\frac{1}{2}} \approxeq 3.7417\) \end{solutionordottedlines} \part \(\sqrt[4]{64}\) \begin{solutionordottedlines}[2cm] \(64^{\frac{1}{4}} = 2.8284\) \end{solutionordottedlines} \part \(\sqrt[5]{7}\) \begin{solutionordottedlines}[2cm] \(7^{\frac{1}{5}} \approxeq 1.4758\) \end{solutionordottedlines} \end{multicols} \end{parts} \Question[6] Evaluate: \begin{parts} \begin{multicols}{3} \part \(4^{\frac{1}{2}}\) \begin{solutionordottedlines}[2cm] 2 \end{solutionordottedlines} \part \(27^{\frac{1}{3}}\) \begin{solutionordottedlines}[2cm] 3 \end{solutionordottedlines} \part \(64^{\frac{1}{2}}\) \begin{solutionordottedlines}[2cm] 8 \end{solutionordottedlines} \part \(25^{\frac{1}{2}}\) \begin{solutionordottedlines}[2cm] 5 \end{solutionordottedlines} \part \(32^{\frac{1}{5}}\) \begin{solutionordottedlines}[2cm] 2 \end{solutionordottedlines} \part \(625^{\frac{1}{4}}\) \begin{solutionordottedlines}[2cm] 5 \end{solutionordottedlines} \end{multicols} \end{parts} \Question[6] Evaluate: \begin{parts} \begin{multicols}{3} \part \(4^{\frac{5}{2}}\) \begin{solutionordottedlines}[2cm] 32 \end{solutionordottedlines} \part \(25^{\frac{3}{2}}\) \begin{solutionordottedlines}[2cm] 125 \end{solutionordottedlines} \part \(32^{\frac{2}{5}}\) \begin{solutionordottedlines}[2cm] 4 \end{solutionordottedlines} \part \(81^{\frac{3}{4}}\) \begin{solutionordottedlines}[2cm] 27 \end{solutionordottedlines} \part \((\sqrt[4]{16})^{3}\) \begin{solutionordottedlines}[2cm] 8 \end{solutionordottedlines} \part \((\sqrt[3]{27})^{2}\) \begin{solutionordottedlines}[2cm] 9 \end{solutionordottedlines} \end{multicols} \end{parts} \Question[8] Simplify: \begin{parts} \begin{multicols}{2} \part \(\left(a^{\frac{1}{2}}\right)^{2}\) \begin{solutionordottedlines}[2cm] a \end{solutionordottedlines} \part \(\left(b^{\frac{1}{3}}\right)^{6}\) \begin{solutionordottedlines}[2cm] \(b^2\) \end{solutionordottedlines} \part \(x^{\frac{1}{2}} \times x^{\frac{3}{2}}\) \begin{solutionordottedlines}[2cm] \(x^2\) \end{solutionordottedlines} \part \(y^{\frac{1}{3}} \times y^{\frac{2}{3}}\) \begin{solutionordottedlines}[2cm] \(y\) \end{solutionordottedlines} \part \(x^{\frac{3}{2}} \div x^{\frac{1}{2}}\) \begin{solutionordottedlines}[2cm] \(x\) \end{solutionordottedlines} \part \(y^{\frac{2}{3}} \div y^{\frac{1}{3}}\) \begin{solutionordottedlines}[2cm] \(y^{\frac{1}{3}}\) \end{solutionordottedlines} \part \(\left(4 m^{6}\right)^{\frac{1}{2}}\) \begin{solutionordottedlines}[2cm] \(2m^3\) \end{solutionordottedlines} \part \(\left(27 n^{12}\right)^{\frac{1}{3}}\) \begin{solutionordottedlines}[2cm] \(3n^4\) \end{solutionordottedlines} \end{multicols} \end{parts} \Question[4] Evaluate: \begin{parts} \begin{multicols}{2} \part \(4^{-\frac{1}{2}}\) \begin{solutionordottedlines}[2cm] \(\frac{1}{2}\) \end{solutionordottedlines} \part \(25^{-\frac{1}{2}}\) \begin{solutionordottedlines}[2cm] \(\frac{1}{5}\) \end{solutionordottedlines} \part \(\left(\frac{1}{81}\right)^{-\frac{1}{4}}\) \begin{solutionordottedlines}[2cm] \(3\) \end{solutionordottedlines} \part \(81^{-\frac{1}{4}}\) \begin{solutionordottedlines}[2cm] \(\frac{1}{3}\) \end{solutionordottedlines} \end{multicols} \end{parts} \Question[6] Simplify, expressing the answer with positive indices. \begin{parts} \begin{multicols}{3} \part \(\left(a^{\frac{1}{2}}\right)^{-2}\) \begin{solutionordottedlines}[2cm] \(a^{-1}\) \end{solutionordottedlines} \part \(\left(b^{-\frac{2}{3}}\right)^{6}\) \begin{solutionordottedlines}[2cm] \(b^{-4}\) \end{solutionordottedlines} \part \(y^{\frac{1}{3}} \times y^{-\frac{2}{3}}\) \begin{solutionordottedlines}[2cm] \(y^{-\frac{1}{3}}\) \end{solutionordottedlines} \part \(p^{\frac{3}{4}} \times p^{-\frac{2}{5}}\) \begin{solutionordottedlines}[2cm] \(p^{\frac{7}{20}}\) \end{solutionordottedlines} \part \(p^{\frac{3}{4}} \div p^{-\frac{2}{5}}\) \begin{solutionordottedlines}[2cm] \(p^{\frac{19}{20}}\) \end{solutionordottedlines} \part \(q^{\frac{3}{2}} \div q^{-\frac{2}{3}}\) \begin{solutionordottedlines}[2cm] \(q^{\frac{13}{6}}\) \end{solutionordottedlines} \end{multicols} \end{parts} \end{questions} \end{exercisebox}