Question

Is the statement \(\left(\frac{2}{3}\right)^{1 / 2} \leq\left(\frac{1}{2}\right)^{2 / 3}\) true or false?

Short answer

Answer: False

Step-by-step solution

Calculate Expression 1

Let's start by calculating the first expression, \(\left(\frac{2}{3}\right)^{1 / 2}\). Taking the square root of \(\frac{2}{3}\), we get: \(\sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}\)

Calculate Expression 2

Now, let's calculate the second expression, \(\left(\frac{1}{2}\right)^{2 / 3}\). Taking the cube root of \(\frac{1}{2}\), and then squaring it, we get: \(\left(\sqrt[3]{\frac{1}{2}}\right)^2 = \left(\frac{1}{\sqrt[3]{2}}\right)^2 = \frac{1}{\sqrt[3]{4}}\)

Compare the Expressions

Now, let's compare the two expressions: \(\frac{\sqrt{2}}{\sqrt{3}}\) and \(\frac{1}{\sqrt[3]{4}}\) Calculating a crude approximation of each expression, we have: \(\frac{\sqrt{2}}{\sqrt{3}} \approx \frac{1.41}{1.73} \approx 0.815\) \(\frac{1}{\sqrt[3]{4}} \approx \frac{1}{1.5874} \approx 0.630\)

Determine the Truth Value of the Inequality

Finally, we must determine if the inequality is true or false: \(\left(\frac{2}{3}\right)^{1 / 2} \leq\left(\frac{1}{2}\right)^{2 / 3}\) Using our crude approximations, we have: \(0.815 \leq 0.630\) This inequality is false. Therefore, the original statement is false.

Key concepts

Fractional Exponentiation

Understanding fractional exponentiation is vital for comparing expressions involving roots. This concept combines exponentiation with the operation of taking roots. When you see an exponent written as a fraction, for instance, \( a^{\frac{m}{n}} \) where \( n \) is not zero, this represents the \( n \)th root of \( a \) raised to the \( m \)th power. The process can be separated into two distinct steps: first taking the root and then applying the exponent.

Using the exercise as an example, \( \left(\frac{2}{3}\right)^{1/2} \) can be broken down into taking the square root of 2/3. Similarly, \( \left(\frac{1}{2}\right)^{2/3} \) involves first finding the cube root of 1/2, which is then squared. It's key to follow the correct order of operations: root before the exponent. This sequence affects the result and thus is critical for accurate calculation and comparison.

Square Root Calculation

Calculating the square root of a number means finding the value that, when multiplied by itself, gives the original number. For our exercise, the square root of 2/3 requires us to consider the numerator and denominator separately. We calculate \( \sqrt{2} \) and \( \sqrt{3} \) individually.

It's useful to remember that square roots of fractions can be simplified by taking the square roots of the numerator and denominator, as such: \( \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} \). Knowing how to compute square roots both exactly and approximately is an essential skill in various areas of math and science.

Cube Root Calculation

Cube roots are slightly more complex. The cube root of a number \( x \) is a value that, when raised to the power of three, returns the original number \( x \). In fractional exponentiation terms, it is expressed as \( x^{1/3} \).

For our exercise \( \left(\sqrt[3]{\frac{1}{2}}\right)^2 = \left(\frac{1}{\sqrt[3]{2}}\right)^2 \), the cube root of 1/2 is calculated before squaring the result. Cube root calculations can come in handy when dealing with three-dimensional geometrical problems or real-world situations such as finding the volume of a cube.

Inequality Comparison

The final piece of the puzzle lies in correctly comparing two quantities to validate an inequality. The exercise requires us to compare \( \frac{\sqrt{2}}{\sqrt{3}} \) with \( \frac{1}{\sqrt[3]{4}} \).

To effectively compare the expressions, we can either find the exact values or use approximations. In the provided solution, approximations were used to determine that \( \frac{\sqrt{2}}{\sqrt{3}} \) is approximately 0.815 and \( \frac{1}{\sqrt[3]{4}} \) is approximately 0.630. Since 0.815 is greater than 0.630, the inequality \( \left(\frac{2}{3}\right)^{1/2} \leq \left(\frac{1}{2}\right)^{2/3} \) is false. Inequality comparisons are not just mathematical exercises; they help us make decisions based on the relative size of values in various aspects of everyday life.