Problem 86 - Algebra and Trigonometry

Question

\(83-88=\) Rationalize the denominator. $$ \begin{array}{llll}{\text { (a) } \frac{1}{\sqrt[5]{2^{3}}}} & {\text { (b) } \frac{2}{\sqrt{3}}} & {\text { (c) } \frac{3}{\sqrt{2^{3}}}}\end{array} $$

Step-by-Step Solution

Answer

(a) \( \frac{\sqrt[5]{4}}{8} \); (b) \( \frac{2\sqrt{3}}{3} \); (c) \( \frac{3\sqrt{2}}{4} \).

Step 1: Understand Rationalization

Rationalizing the denominator involves eliminating any irrational numbers like square roots or cube roots from the denominator. This is achieved by multiplying the numerator and the denominator by a value that will remove the irrational component from the denominator.

Step 2: Rationalize (a)

Given \( \frac{1}{\sqrt[5]{2^3}} \). To rationalize the denominator, multiply both the numerator and denominator by \( \sqrt[5]{2^2} \) which will make the denominator \( \left(\sqrt[5]{2^3} \right)^5 = 2^3 \):\[\frac{1}{\sqrt[5]{2^3}} \times \frac{\sqrt[5]{2^2}}{\sqrt[5]{2^2}} = \frac{\sqrt[5]{2^2}}{2^3} = \frac{\sqrt[5]{4}}{8}\].

Step 3: Rationalize (b)

Given \( \frac{2}{\sqrt{3}} \). Multiply both the numerator and the denominator by \( \sqrt{3} \) to remove the square root in the denominator: \[\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\].

Step 4: Rationalize (c)

Given \( \frac{3}{\sqrt{2^3}} \) which simplifies to \( \frac{3}{2\sqrt{2}} \). Multiply both the numerator and the denominator by \( \sqrt{2} \): \[\frac{3}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{4}\].

Key Concepts

Understanding Irrational Numbers

Irrational numbers are numbers that cannot be expressed as a simple fraction. They often appear whenever you see square roots or cube roots that don’t result in a clean whole number or a repeating, terminating decimal. Examples include \(\sqrt{2}\), \(\pi\), and cube roots such as \(\sqrt[3]{5}\). This makes these numbers unique. It is this non-repeating and non-terminating nature that makes them irrational. Our main focus here is the expression of these numbers in fractions, especially when they are in the denominator. As you'll see, rationalizing these denominators involves converting these expressions so that the denominator is a rational number. This makes it easier to read and work with calculations.

Exploring Square Roots

Square roots are one of the most common irrational numbers encountered in math. The square root of a number \(x\) is a number \(y\) such that \(y^2 = x\). For other numbers, like \( \sqrt{4} \), which equals 2, the square root is a whole number making it rational. However, most square roots are irrational, like \( \sqrt{3} \). When such irrational numbers appear in the denominator of a fraction, as with \( \frac{2}{\sqrt{3}} \), the goal is often to eliminate the square root from the denominator. This involves multiplying the numerator and the denominator by the square root itself (in this case, \(\sqrt{3} \)), resulting in \( \frac{2\sqrt{3}}{3} \). Doing this transforms the expression into a form that is easier to use.

Unpacking Cube Roots

Cube roots work similarly to square roots, but instead of finding a number that multiplies by itself twice to reach the original number, you multiply three times. If \( y^3 = x \), then \( y \) is the cube root of \( x \). Just like with square roots, cube roots of most numbers are irrational, such as \( \sqrt[3]{2} \).In our exercises, we rationalized an expression involving \( \sqrt[5]{2^3} \). To remove the cube root from the denominator, it’s necessary to use a manipulation that turns it into a higher power that is rational. We achieve this by multiplying it by a suitable form of 1 (a fraction like \( \frac{\sqrt[5]{2^2}}{\sqrt[5]{2^2}} \)) to make the expression manageable and straightforward.

Numerator and Denominator Manipulation

Manipulating numerators and denominators is a useful technique in solving equations involving irrational numbers. When rationalizing, multiplying the numerator and the denominator by the same value is essentially multiplying by 1, which doesn’t change the value of the expression.In the exercises, for example, we used this technique to eliminate square and cube roots from denominators. By calculating something like \( \frac{\sqrt{3}}{\sqrt{3}} \) or \( \frac{\sqrt[5]{2^2}}{\sqrt[5]{2^2}} \), we are creatively using the property of 1 to transform an expression into a simpler and more useful form without altering its value. This practice helps in achieving a rational number in the denominator, making the expression easier to comprehend and work with in subsequent steps.