Interpreting Rational (Fraction) Exponents

To know how rational exponents work, first we have to be able to take roots of an integer.

We will start simple, with the square root. The square root of $x$ is denoted $\sqrt{x}$, and denotes a number whose square is $x$. So the square root is the inverse operation of a square.

The square root operation is considerably trickier to define than the squaring operation, which is just multiplication with particular arguments. There are two problems. Firstly, the square root defined as above is not unique: both $2$ and $-2$ would be square roots of $4$. Secondly, the square root is not defined for all reals: $-4$ does not have a (real) square root.

The resolution for the first issue is simple: we just require that the square root of a number be nonnegative. The second issue cannot really be resolved with our current tools, so we have to bite the bullet and admit that the square root is not defined for negative arguments. (Even in complex analysis, the square root has to behave somewhat strangely on negative numbers.)

Now, we have a definition for the square root. But there's no reason to limit ourselves to squares. What about higher powers?

The cube root of $x$ is denoted $\sqrt[3]{x}$, and denotes a number whose cube (third power) is $x$.

Possibly surprisingly, cube roots are easier to define than square roots. The reason is that the power, $3$, is odd and turns negative numbers into negative numbers, whereas squaring, or taking any even exponent, indiscriminately makes all numbers nonnegative.

In general, the $n$th root of a number is denoted $\sqrt[n]{x}$ and is a number whose $n$th power is $x$. If $n$ is even, we have to deal with the problems above and state that the root is only defined for nonnegative arguments and its value is always nonnegative. If $n$ is odd, the simple definition is enough.

Another way to write $\sqrt[n]{x}$ is $x^{1/n}$. In general, rational exponents in the form $a^{x/y}$ are equal to $(\sqrt[y]{a})^x$. For instance, $8^{1/3} = \sqrt[3]{8} = 2$; and $8^{5/3} = (\sqrt[3]{8})^5 = 2^5 = 32$.

Hopefully you have learned something about Rational Exponents. $:)$

Rational exponents are obviously exponents with rational powers. This can take form of $a^{m/n}$ where $a,m,n$ are integers and $m,n$ are relatively prime.

Take a look at this simple example: $a^{1/2}$. This is equivalent to $\pm \sqrt{a}$. We notice that $a^{1/2}\ne \sqrt{a}$! Why? A square root must be positive, but $a^{1/2}$ doesn't have to be. Say $a^2=4$. We can't necessarily square root both sides, because a has 2 solutions! Those 2 solutions are positive $2$ and negative $2$.

Speaking of that equation, we see that if $f(x)=x^2$, $f^{-1}(x)=x^{1/2}$, or $x^{1/2}$ is the inverse of $x^2$.

Also, we can not square root a negative integer. If $a$ is positive, $a^2$ is also positive. If $a$ is negative, $a^2$ is positive. $a^2$ is never negative. Therefore, if $a^{1/2}$ or $\sqrt{a}$ is real, then $a$ must be positive.

Try this example.

Now let's move on to a different exponent: $f(x)=x^{1/3}$.

Here, if $x$ is negative, $x^3$ is also negative. Also, When we have a value $x^3$, it has only $1$ solution, because if a positive value works, it's respective negative will not work. Their respective cubes will also have a common ratio of $-1$ (one is positive, the other is the same number except it's negative). So we don't need to worry about negatives or anything.

So far, we've only covered exponents to a power that is a unit fraction (can be expressed as $\dfrac{1}{n}$ where $n$ is positive). How about for other fractions as a power?

Let's try $9^{3/2}$. First, we know about this logarithmic property $(a^b)(a^c)=a^{bc}$. What we want to do is to separate the numerator and the denominator. Plug the values in to get $9^{3/2}=(9^3)(9^{1/2})$. The first term is $9^3=729$ and the second is $9^{1/2}=\pm 3$. Multiplying gives $\pm 729\cdot3=\pm 2187$.

Great!

The best way to think about roots of higher order is to turn them into fractional exponents. Here is what I mean: $$\sqrt x=x^{1/2}\;\;\;\sqrt[3]x=x^{1/3}\;\;\;\sqrt[n]x=x^{1/n}$$ Once we have done this, we can apply our properties of exponents and deal with them in that fashion.