Radicals and Rational Exponents

Rational Exponents and Nth Roots

Consider for a moment the possibility of rational exponents -- that is to say, exponents that can take on rational (i.e., fractional) values. Here again, it might be hard to imagine what we might mean by this, if we restrict ourselves to interpreting $a^n$ only as the product of $n$ factors of $a$. Certainly, having a product of a fractional number of factors seems problematic.

However, if all we require is consistency with the rules that must work for integer exponents, then at least $a^{1/n}$ has a very natural interpretation...

Consider $a^{1/2}$, and what happens if we square it and then apply the rule $(x^m)^n = x^{m\,n}$

$$\displaystyle{\left( a^{\frac{1}{2}} \right)^2 = a^{\frac{1}{2} \cdot 2} = a^1 = a}$$

As such, if we want consistency with the rule, $a^{1/2}$ must be defined to be a value that when squared equals $a$. Equivalently, $a^{1/2}$ must be a solution to $x^2 = a$.

In a like manner, $a^{1/3}$ must be a solution to $x^3 = a$; and $a^{1/4}$ must be a solution to $x^4 = a$; and so on...

As a matter of verbiage, let us call any value of $x$ that solves $x^2 = a$ a square root of $a$, and any value that solves $x^3 = a$ a cube root of $a$.

More generally, let us call any solution to $x^n = a$ an $\boldsymbol{n}^{\textrm{th}}$ root of $a$.

Recall that if a product of two or more factors is zero, one or more of those factors must be zero. Thus, if $n \ge 2$ and $x^n = 0$, it must be the case that $x=0$. As such, the only root that zero admits is zero itself -- whether we are talking about the square root, cube root, $4^{\textrm{th}}$ or $5^{\textrm{th}}$ root, etc...

However, roots of non-zero values are more interesting. For example, any positive value $a$ always admits two square roots -- one positive, the other negative. Take the square roots of $9$, for instance. There are two, $-3$ and $3$, since these both solve $a^2=9$. On the other hand, since the square of any real value must be non-negative, $a^{1/2}$ for negative values of $a$ must fail to exist.

To avoid any potential ambiguity, we define $a^{1/2}$ for $a \gt 0$ to be the positive square root of $a$, calling this the principle square root, and note that $a^{1/2}$ when $a \lt 0$ must be left undefined.

Fortunately, there is no ambiguity in writing $a^{1/3}$, assuming we wish this to be a real value. As an example, $x^3 = -8$ has only one real solution, so $-2$ is the clear choice for the principle cube root of $-8$.

Upon noting that the product of an odd number of negative factors is negative, and the product of an odd number of positive factors is positive, one can see that more generally when $n$ is odd there can never be more than one real solution to $x^n = a$.

On the other hand if $n$ is even (and non-zero), and recalling that both the product of an even number of positive factors, or an even number of negative factors, is positive, we can see that real solutions to $x^n = a$ for $a \gt 0$ must always occur in pairs, with one positive and the other negative. Further, when $a \lt 0$ solutions to this equation must fail to exist.

Given all the above, we make the following definition:

The principle $\boldsymbol{n^{\textrm{th}}}$ root of $a$, denoted $a^{1/n}$, is the real value that solves $x^n=a$, and is non-negative when $n$ is even. When $a$ is negative and $n$ is even, $a^{1/n}$ is undefined.

With the principle $n^{\textrm{th}}$ root defined, extending our definition of exponents to include more general fractions follows again naturally from the rule $(x^m)^n = x^{m\,n}$:

$$\boxed{\displaystyle{x^{m/n} = \left( x^{1/n} \right)^m \quad \textrm{ provided that } x^{1/n} \textrm{ is a real number}}}$$

The inclusion of a restriction that $x^{1/n}$ be a real number can be understood upon considering $(x^{1/n})^n$ when $n$ is even and $x$ is negative. If the expression inside the parentheses fails to exist, how can we possibly raise it to the $n^{\textrm{th}}$ power?

Radical Notation

The radical sign (or "radix"), "$\sqrt{\,\,}$", provides an alternative (older) way to write a principle root of a number. Specifically, we may write

The value of $x$ above, shown under the radical sign, is called the radicand. The value of $n$ is called the index.

Below are some simple examples:

We can convert between radicals and rational exponents by appealing to the following:

$$x^{\frac{m}{n}} = \left( x^{\frac{1}{n}} \right)^m = \left( \sqrt[n]{x} \right)^m$$

We can think of taking the $n^{\textrm{th}}$ root of something as the inverse of raising something to the $n^{\textrm{th}}$ power in many cases -- but not all of them. Often, taking the $n^{\textrm{th}}$ immediately after raising something to the $n^{\textrm{th}}$ power leaves the original expression unchanged. For example, $\sqrt[3]{2^3} = 2$.

However, when $n$ is even, the presence of multiple roots and our choice to define the principle root as the only non-negative real value among them requires that sometimes this is not the case. Consider for example: $\sqrt{(-3)^2} = \sqrt{9} = 3$. In this case, squaring and then taking a square root preserved the magnitude of the $-3$ with which we started, but changed it's sign.

In general,

$$\boxed{\begin{array}{l} \textrm{If } n \textrm{ is even, then } \sqrt[n]{x^n} = \left| x \right|\\ \textrm{If } n \textrm{ is odd, then } \sqrt[n]{x^n} = x \end{array}}$$

Simplifying Radical Expressions

Generally, we consider radical expression to be written in simplest form when:

Having a radical expression in this form generally makes approximating its value easier.

Recall, every radical expression can always be rewritten using rational exponents. As such, we can deduce some convenient rules for simplifying radical expressions directly from some corresponding rules for dealing with exponents .

For example, as $(xy)^{1/n} = x^{1/n} y^{1/n}$, we know

$$\boxed{\sqrt[n]{xy^{\phantom{1}}} = \sqrt[n]{x^{\phantom{1}}} \sqrt[n]{y^{\phantom{1}}}}$$

Similarly, given that $\displaystyle{ \left( \frac{x}{y} \right)^{1/n} = \frac{x^{1/n}}{y^{1/n}}}$, we have

$$\boxed{\displaystyle{\sqrt[n]{\frac{x}{y}} = \frac{\sqrt[n]{x}}{\sqrt[n]{y}}}}$$

Lastly, given that $\displaystyle{ \left( x^{\frac{1}{n}} \right)^{\frac{1}{m}} = x^{\frac{1}{n} \cdot \frac{1}{m}} = x^{\frac{1}{mn}}}$ we see that $$\boxed{\displaystyle{\sqrt[m]{\sqrt[n]{x}} = \sqrt[m n]{x}}}$$