For instance, the square root of 4 can be written as \(4^{\frac{1}{2}}\text{.}\) Squaring this gives us back 4:
\begin{equation*}
\left(4^{\frac{1}{2}}\right)^2 = 4
\end{equation*}
Section 8.4 Exponents and Irrational Numbers
Subsection 8.4.1 nth Root as Exponent 1/n
An \(n\)th root of a number can be represented as raising the number to the power of \(\frac{1}{n}\text{.}\) This means that for any positive real number \(a\) and any natural number \(n\text{,}\) the expression \(\sqrt[n]{a}\) is equivalent to \(a^{\frac{1}{n}}\text{.}\) To understand this, consider that raising \(a^{\frac{1}{n}}\) to the power of \(n\) gives us back the original number:
\begin{equation*}
\left(a^{\frac{1}{n}}\right)^n = a
\end{equation*}
Example 8.4.1.
Subsection 8.4.2 Multiplying and Dividing nth Roots
When working with roots, we can multiply or divide nth roots, but we cannot add or subtract them in a straightforward manner. The rules for multiplying and dividing roots come from the properties of exponents.
For example, to simplify the product of two square roots, we use the property that:
\begin{equation*}
\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}
\end{equation*}
Example 8.4.2.
Simplify \(\sqrt{2} \times \sqrt{8}\text{:}\)
\begin{equation*}
\sqrt{2} \times \sqrt{8} = \sqrt{2 \times 8} = \sqrt{16} = 4
\end{equation*}
The same applies to division:
\begin{equation*}
\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}
\end{equation*}
Example 8.4.3.
Simplify \(\frac{\sqrt{18}}{\sqrt{2}}\text{:}\)
\begin{equation*}
\frac{\sqrt{18}}{\sqrt{2}} = \sqrt{\frac{18}{2}} = \sqrt{9} = 3
\end{equation*}
However, adding or subtracting roots is not as straightforward. For example, \(\sqrt{a} + \sqrt{b}\) does not equal \(\sqrt{a + b}\text{.}\)
Example 8.4.4.
Simplify \(\sqrt{9} + \sqrt{16}\text{:}\)
\(\sqrt{9} + \sqrt{16} = 3 + 4 = 7\text{,}\) but \(\sqrt{9 + 16} = \sqrt{25} = 5\text{,}\) which is different.
Subsection 8.4.3 Simplifying \(n\)th Principal Roots by Factoring Out \(n\)th Powers
When simplifying \(n\)th principal roots, you can factor out any \(n\)th powers that appear in the prime factorization of the number you are taking the root of. Let’s do some examples to see the method first.
Example 8.4.5.
Consider the cube root of 54:
\begin{equation*}
\sqrt[3]{54}
\end{equation*}
First, we factor 54 into its prime factors:
\begin{equation*}
54 = 2 \times 3^3
\end{equation*}
Notice that \(3^3\) is a perfect cube. We can use the property of cube roots that states:
\begin{equation*}
\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}
\end{equation*}
Applying this property:
\begin{equation*}
\sqrt[3]{54} = \sqrt[3]{2 \times 3^3} = \sqrt[3]{2} \times \sqrt[3]{3^3}
\end{equation*}
Since \(\sqrt[3]{3^3} = 3\text{,}\) we simplify the expression to:
\begin{equation*}
\sqrt[3]{54} = 3 \times \sqrt[3]{2}
\end{equation*}
Example 8.4.6.
Now consider the cube root of 162 and that the prime factorization of 162 is \(2 \times 3^4\text{.}\) Now, we can realize that we can factor out a perfect cube from \(3^4\text{.}\) Indeed \(3^4=3^{3+1}=3^3 \times 3\text{.}\)
Using the same idea as the previous example we can say
\begin{equation*}
\sqrt[3]{162} = \sqrt[3]{2 \times 3^4} = \sqrt[3]{2 \times 3} \times \sqrt[3]{3^3}
\end{equation*}
Since \(\sqrt[3]{3^3} = 3\text{,}\) we simplify the expression to:
\begin{equation*}
\sqrt[3]{162} = 3 \times \sqrt[3]{2 \times 3}
\end{equation*}
In the examples above, we showed that we can indeed simplify an \(n\)th root as long as we can factor out an \(n\)th power from its factors. We say this more precisely below.
Subsubsection 8.4.1 Using the Quotient Remainder Theorem
The quotient remainder theorem states that for any integer \(a\) and a positive integer \(n\text{,}\) there exist unique integers \(q\) (quotient) and \(r\) (remainder) such that:
\begin{equation*}
a = n \times q + r \quad \text{with} \quad 0 \leq r < n
\end{equation*}
This theorem is particularly useful in identifying and extracting \(n\)th powers from the prime factorization. For instance, if we have a prime factor \(p\) raised to the power of \(k\text{,}\) we can express \(k\) as:
\begin{equation*}
k = nq + r
\end{equation*}
Here, \(nq\) represents the highest \(n\)th power within \(k\) that can be factored out, leaving \(r\) as the remainder.
Subsubsection 8.4.2 Additional Examples
Let’s explore three more examples of increasing difficulty.
Example 8.4.7.
Solution.
First, factor 162 into its prime factors:
\begin{equation*}
162 = 2 \times 3^4
\end{equation*}
Notice that \(3^4\) is a perfect fourth power. Applying the property of fourth roots:
\begin{equation*}
\sqrt[4]{162} = \sqrt[4]{2 \times 3^4} = \sqrt[4]{2} \times \sqrt[4]{3^4}
\end{equation*}
Since \(\sqrt[4]{3^4} = 3\text{,}\) we simplify the expression to:
\begin{equation*}
\sqrt[4]{162} = 3 \times \sqrt[4]{2}
\end{equation*}
Example 8.4.8.
Solution.
First, factor 2430 into its prime factors:
\begin{equation*}
2430 = 2 \times 3^5 \times 5
\end{equation*}
Notice that \(3^5\) is a perfect fifth power. Applying the property of fifth roots:
\begin{equation*}
\sqrt[5]{2430} = \sqrt[5]{2 \times 3^5 \times 5} = \sqrt[5]{2} \times \sqrt[5]{3^5} \times \sqrt[5]{5}
\end{equation*}
Since \(\sqrt[5]{3^5} = 3\text{,}\) we simplify the expression to:
\begin{equation*}
\sqrt[5]{2430} = 3 \times \sqrt[5]{2 \times 5} = 3 \times \sqrt[5]{10}
\end{equation*}
Example 8.4.9.
Solution.
First, factor 5184 into its prime factors:
\begin{equation*}
5184 = 2^6 \times 3^4
\end{equation*}
Notice that \(2^6\) is a perfect sixth power. Applying the property of sixth roots:
\begin{equation*}
\sqrt[6]{5184} = \sqrt[6]{2^6 \times 3^4} = \sqrt[6]{2^6} \times \sqrt[6]{3^4}
\end{equation*}
Since \(\sqrt[6]{2^6} = 2\text{,}\) we simplify the expression to:
\begin{equation*}
\sqrt[6]{5184} = 2 \times \sqrt[6]{3^4} = 2 \times \sqrt[6]{81}
\end{equation*}
Further simplification yields:
\begin{equation*}
\sqrt[6]{5184} = 2 \times \sqrt[6]{3^4} = 2 \times 3^{2/3}
\end{equation*}
Subsection 8.4.4 Simplifying Expressions with Fractional Powers
Simplifying expressions containing fractional powers often involves using the properties of exponents. Understanding these properties is crucial for algebraic manipulation and solving various mathematical problems efficiently. Also, it allows us to see the relationship between multiplication of numbers with the same base, and addition of their exponents.
When working with fractional powers, it’s important to remember that the numerator of the fraction represents the power, and the denominator represents the root. For example, \(a^{\frac{m}{n}}\) can be interpreted as the nth root of \(a\) raised to the mth power, as in \((a^{m})^{1/n}\) or as the nth root of mth power of \(a\text{,}\) as in \((a^{1/n})^m\text{.}\) This dual interpretation is helpful when simplifying or transforming expressions.
Example 8.4.10.
Simplify \(a^{\frac{1}{2}} \times a^{\frac{1}{2}}\text{:}\)
Using the property of exponents that \(a^m \times a^n = a^{m+n}\text{,}\) we get:
\begin{equation*}
a^{\frac{1}{2}} \times a^{\frac{1}{2}} = a^{\frac{1}{2} + \frac{1}{2}} = a^1 = a
\end{equation*}
This example demonstrates that multiplying two expressions with the same base and fractional exponents involves simply adding the exponents. Here, since \(\frac{1}{2} + \frac{1}{2} = 1\), the result is \(a^1\), which simplifies to \(a\).
Example 8.4.11.
Simplify \(\left(b^{\frac{1}{3}}\right)^3\text{:}\)
Using the property of exponents that \(\left(a^m\right)^n = a^{m \times n}\text{,}\) we get:
\begin{equation*}
\left(b^{\frac{1}{3}}\right)^3 = b^{\frac{1}{3} \times 3} = b^1 = b
\end{equation*}
This example shows how raising a power to another power results in multiplying the exponents. Here, \(\frac{1}{3} \times 3 = 1\), so the expression simplifies to \(b^1\), which is simply \(b\).
Example 8.4.12.
Simplify \(\left(\frac{c^{\frac{1}{4}}}{d^{\frac{1}{2}}}\right)^2\text{:}\)
Using the property of exponents that \(\left(\frac{a^m}{b^n}\right)^p = \frac{a^{m \times p}}{b^{n \times p}}\text{,}\) we get:
\begin{equation*}
\left(\frac{c^{\frac{1}{4}}}{d^{\frac{1}{2}}}\right)^2 = \frac{c^{\frac{1}{4} \times 2}}{d^{\frac{1}{2} \times 2}} = \frac{c^{\frac{2}{4}}}{d^1} = \frac{c^{\frac{1}{2}}}{d}
\end{equation*}
This example illustrates the application of the exponent rule to a fraction. By raising both the numerator and the denominator to the power of 2, we multiply their respective exponents by 2, allowing us to simplify the expression.
To further deepen our understanding, let’s consider a more complex example involving a sum and product of fractional powers:
Example 8.4.13.
Simplify \((x^{\frac{1}{2}} \times y^{\frac{1}{3}}) \times (x^{\frac{1}{2}} \times y^{\frac{2}{3}})\text{:}\)
Using the properties of exponents, we combine like terms:
\begin{equation*}
(x^{\frac{1}{2}} \times y^{\frac{1}{3}}) \times (x^{\frac{1}{2}} \times y^{\frac{2}{3}}) = x^{\frac{1}{2} + \frac{1}{2}} \times y^{\frac{1}{3} + \frac{2}{3}} = x^1 \times y^1 = xy
\end{equation*}
This example combines both addition of exponents and multiplication of terms with fractional powers. By adding the exponents of \(x\) and \(y\) separately, we simplify the expression to \(xy\).
Example 8.4.14.
Simplify \(\frac{(a^2b^{-1})^{\frac{1}{2}}}{b^3c^{-2}} \div \left(\frac{a^2b^2c^{-1}}{(a^{-1}c)^{-1}}\right)^2\text{:}\)
First, simplify each part of the expression separately:
\begin{equation*}
(a^2b^{-1})^{\frac{1}{2}} = a^{2 \times \frac{1}{2}} b^{-1 \times \frac{1}{2}} = a^1 b^{-\frac{1}{2}} = ab^{-\frac{1}{2}}
\end{equation*}
\(\frac{1}{b^3c^{-2}} = b^{-3}c^2\)
\(\left(\frac{a^2b^2c^{-1}}{(a^{-1}c)^{-1}}\right)^2\text{:}\)
Simplify the denominator:
\begin{equation*}
(a^{-1}c)^{-1} = a^1c^{-1} = ac^{-1}
\end{equation*}
Now the expression inside the parenthesis becomes:
\begin{equation*}
\frac{a^2b^2c^{-1}}{ac^{-1}} = a^{2-1}b^2c^{-1+1} = ab^2
\end{equation*}
Squaring this, we get:
\begin{equation*}
(ab^2)^2 = a^2b^4
\end{equation*}
Now combine all parts:
\begin{equation*}
\frac{ab^{-\frac{1}{2}}}{b^{-3}c^2} \div a^2b^4 = \frac{ab^{-\frac{1}{2}} \times c^2}{b^{-3}} \div a^2b^4
\end{equation*}
Simplify the multiplication:
\begin{equation*}
\frac{ab^{-\frac{1}{2}} \times b^3 \times c^2}{a^2b^4} = \frac{ab^{3-\frac{1}{2}}c^2}{a^2b^4} = \frac{ab^{\frac{5}{2}}c^2}{a^2b^4}
\end{equation*}
Now combine the exponents:
\begin{equation*}
\frac{ab^{\frac{5}{2}}c^2}{a^2b^4} = \frac{ac^2}{a^2b^{4-\frac{5}{2}}} = \frac{ac^2}{a^2b^{\frac{3}{2}}} = \frac{c^2}{ab^{\frac{3}{2}}}
\end{equation*}
