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Section 6.5 Going Deeper: Radicals and Fractional Exponents

At some point in the past, you might have remembered learning that \(x^{\frac{1}{2}} = \sqrt{x}\text{.}\) Most likely, you learned this as another rule. However, you have enough knowledge at this point to learn why this is the most (and perhaps only) sensible thing that the fractional exponent could represent.

Before we take a look at fractional exponents, we will do a quick review of radicals. The first radical that students encounter is the square root. The symbol represents the non-negative number that has the property that \(\left( \sqrt{x} \right)^2 = x\text{.}\) Certain integers are considered to be perfect squares because their square roots are themselves integers. For example, we have \(\sqrt{4} = 2\) (since \(2^2 = 4\)) and \(\sqrt{25} = 5\) (since \(5^2 = 25\)). Square roots of other numbers exist, and we can either represent the exact values symbolically (such as being the number such that \(\left( \sqrt{2} \right)^2 = 2\)) or we can use a decimal approximation (\(\sqrt{2} \approx 1.41421\) because \(1.41421^2 = 1.9999899241 \approx 2\)).

An important feature of square roots is that we always take the value to be non-negative. When we think about numbers that have the property that squaring them gives you the value 4 we quickly realize that there are two possible values: 2 and -2. and This creates a potential ambiguity in our notation, because could theoretically be one of two values. However, mathematicians have adopted the convenient convention that the square root function is always non-negative.

This framework clashes with what some students have learned in their previous algebra experiences. Some students have been taught that \(\sqrt{4} = \pm 2\text{.}\) Unfortunately, this is not a proper understanding of the symbols. When we write the only value this can represent is and it never represents the value The distinction comes down to understanding what question is being asked:

  • What number does \(\sqrt{4}\) represent? The value of \(\sqrt{4}\) is 2, so that \(\sqrt{4} = 2\text{.}\)

  • What numbers have the property that \(x^2 = 4\text{?}\) The value can be either 2 or -2, which we can denote as \(x = \pm 2\text{.}\)

In the first case, we're given a specific mathematical expression and are asked to determine its value. In the second case, we are given a mathematical equation and are asked to solve it. These are two different questions, which is why we end up with two different answers.

The reason that there are two solutions to \(x^2 = 4\) is because squaring a negative number results in a positive number. In fact, raising a negative number to any even power leads to a positive result. But a negative number raised to an odd power remains odd. This observation is helpful for defining higher roots.

Definition 6.5.1. The \(n\)-th root.

The \(n\)-th root of \(x\), denoted \(\sqrt[n]{x}\) is the number that has property \(\left( \sqrt[n]{x} \right)^n = x\text{.}\) If \(n\) is even, we pick this value to be a nonnegative number (by convention). If \(n = 2\text{,}\) we often omit the \(n\) in the notation, so that \(\sqrt{x} = \sqrt[2]{x}\text{.}\)

We can take odd roots of both positive and negative numbers, so that \(\sqrt[3]{8} = 2\) and \(\sqrt[3]{-8} = -2\text{.}\) However, we can only take even roots of nonnegative numbers and the roots are only ever nonnegative values. So \(\sqrt[4]{81} = 3\) even though both \(3^4 = 81\) and \((-3)^4 = 81\text{.}\) And it turns out that doesn't exist because it's impossible for a number raised to the fourth power to be negative.

This takes us back to the concept of fractional exponents. The key is to think about them in terms of properties. If we expect the properties of exponents to be consistent, then we must have the following:

\begin{equation*} \left( x^{\frac{1}{2}} \right)^2 = x^{\frac{1}{2} \cdot 2} = x^1 = x \end{equation*}

This means that whatever \(x^{\frac{1}{2}}\) is, it really ought to have the property that when you square it, you get \(x\text{.}\) And we have already seen that this is the definition of \(\sqrt{x}\text{,}\) which tells us that \(x^{\frac{1}{2}} = \sqrt{x}\text{.}\)

We can use the exact same logic to determine the value of \(x^{\frac{1}{3}}\text{.}\) Since \(\left( x^{\frac{1}{3}} \right)^3 = x\text{,}\) we see that \(x^{\frac{1}{3}} = \sqrt[3]{x}\text{.}\) In fact, we can see that \(x^{\frac{1}{n}} = \sqrt[n]{x}\text{.}\)

What can we say about \(x^{\frac{m}{n}}\text{?}\) Once again, we're going to look at the properties of exponents. Notice that

\begin{equation*} x^{\frac{m}{n}} = \left( x^{\frac{1}{n}} \right)^m = \left( \sqrt[n]{x} \right)^m \end{equation*}

and

\begin{equation*} x^{\frac{m}{n}} = \left( x^m \right)^\frac{1}{n} = \sqrt[n]{x^m}. \end{equation*}

These calculations give us the ability to evaluate every fractional exponent.

Definition 6.5.2. Fractional Exponents.

For any integer \(m\) and positive integer \(n\text{,}\) we define \(x^{\frac{m}{n}}\) to be \(\left( \sqrt[n]{x} \right)^m = \sqrt[n]{x^m}\text{.}\)

A key observation about this definition is that we have set it up so that it would be consistent with the product rule and power rule (Theorem 6.1.2). It is also consistent with negative exponents (Definition 6.1.3). This consistency means that there are no new "rules" to learn. This is how mathematicians like to generalize results. It would be very confusing if we had one set of rules that only worked with integer exponents, and then an entirely different set of rules for working with fractional exponents. It makes much more sense to work towards consistent notation and consistent structures.