Problem 110 - Brief Applied Calculus

Question

When defining \(x^{m / n}\), why did we require that the exponent \(\frac{m}{n}\) be fully reduced? [Hint: \((-1)^{2 / 3}=(\sqrt[3]{-1})^{2}=1, \quad\) but with an equal but unreduced exponent you get \((-1)^{4 / 6}=(\sqrt[6]{-1})^{4}\). Is this defined?]

Step-by-Step Solution

Answer

Fractional exponents need reduction to ensure well-formed roots and powers.

Step 1: Understanding Fractional Exponents

Fractional exponents are a way of expressing roots using exponents. The exponent \( \frac{m}{n} \) means the \( n \)-th root of a number raised to the \( m \)-th power, i.e., \( x^{m/n} = (\sqrt[n]{x})^m \).

Step 2: Evaluating Exponent \((-1)^{2/3}\)

To solve \((-1)^{2/3}\), first find the cube root of -1, which is \( \sqrt[3]{-1} = -1 \). Next, raise this result to the power of 2: \((-1)^2 = 1\). Hence, \((-1)^{2/3} = 1\).

Step 3: Evaluating Exponent \((-1)^{4/6}\)

First, simplify the fraction \( \frac{4}{6} \) to \( \frac{2}{3} \). When attempted without reducing, compute \((-1)^{4/6}\) as follows. Find the sixth root of -1, \( \sqrt[6]{-1} \), which is not well-defined in the real numbers because it involves even roots of negative numbers. Thus, \((-1)^{4/6}\) is undefined.

Step 4: Reason for Fully Reducing Fractional Exponents

By reducing the fraction \( \frac{m}{n} \) to lowest terms, we ensure that each step in the exponentiation process (root and power) is well-defined. As seen in the example, \((-1)^{2/3} = 1\), but \((-1)^{4/6}\) is not meaningful in the real number system because it requires taking an even root of a negative number.

Key Concepts

Roots and Radicals

Roots and radicals go hand in hand with fractional exponents, helping us express roots more clearly using exponents. When we say a number is raised to the power of \( \frac{m}{n} \), it means two things happen. First, take the \( n \)-th root of the number, and then, raise the result to the "m" power: \( x^{\frac{m}{n}} = (\sqrt[n]{x})^m \).The process:

• Finding a square root is finding which number, multiplied by itself, gives the original number.
• Cubic roots find which number raised to the third power equals the original.

This concept of roots helps square, cube, and other roots become easier to understand. For example, in our exercise, the cube root of \(-1\) is part of solving \((-1)^{2/3}\), where \( \sqrt[3]{-1} = -1 \).

Exponentiation Rules

Exponentiation rules are fundamental when dealing with powers and fractional exponents. These rules help simplify expressions and make calculations consistent.Key rules to remember:

• Reducing fractions: Always reduce \( \frac{m}{n} \) to its simplest form to avoid unnecessary complications, especially with roots.
• Multiplying powers with the same base: \( x^a \cdot x^b = x^{a + b} \).
• Power of a power: \( (x^a)^b = x^{a \cdot b} \).

Reducing fractional exponents as seen in the solution ensures results remain defined. For instance, \((-1)^{4/6}\) appears complex until reduced to \((-1)^{2/3}\), which results in a real number.

Complex Numbers

Complex numbers extend real numbers to include solutions to equations that don't have real solutions alone. They consist of a real part and an imaginary part, defined as \(a + bi\), where \( i \) is the imaginary unit satisfying \(i^2 = -1\).In our exercise, understanding roots of negative numbers is where complex numbers come into play. With \( (-1)^{4/6} \), trying to find the sixth root of \(-1\) isn't possible in real numbers because even roots of negatives don't exist without imaginary numbers.Here's why the complexity:

• Imaginary numbers provide a way to work with negatives under even roots.
• In complex terms, even roots have complex results, like \(\sqrt[6]{-1} \).

Thus, complex numbers ensure calculations with roots of negatives remain meaningful and feed our exploration beyond real numbers.