Why is a fractional exponent a root?

Question

You learn very early that $\sqrt2 = 2^{\frac12}$ and $\sqrt[3]{8} = 8^{\frac13}$ but why is this? Usually when I ask this I get the answer, "Because it is defined that way" but is there any logical reason why fractional exponents are roots?

Answer 1

Suppose we want $\sqrt 2$ to be $2^a$ for some value of $a$. We'd also like the exponent 'addition rule' for products to hold. Thus we would like

$$2 = \sqrt 2 \sqrt 2 = 2^a \cdot 2^a$$

or in other words

$$2^{a + a} =2^{2a} = 2^{1}$$

Hence $a = 1/2$ would make the notation consistent with the behavior of exponents for integers and therefore we are motivated to say that $\sqrt 2 = 2^{1/2}$.

Similarly, $\sqrt[3]{2} = 2^{1/3}$.

Answer 2

Let $x^a=\sqrt[b]x$ for some values of $a$ and $b$

By the definition of rooting and exponentiation, we can also say that.

$$x=(\sqrt[b]x)^b$$

Since we would like the Power of Powers Rule (see section on Five Laws of Exponents in Exponentiation) to hold true, let's take this equation and substitute in our initial identity for $\sqrt[b]x$ and simplify according to the Power of Powers Rule.

$$x=(\sqrt[b]x)^b\longrightarrow{x=(x^a)^b}=x^{ab}$$

and so

$$x=x^{ab}\longrightarrow{x^1=x^{ab}}$$

For this equation to logically hold, the exponents must be equal, and so we can say that

$$x^1=x^{ab}\longrightarrow1=ab$$

By the Multiplicative Inverse Property (see section on Reintroducing Arithmetic), we know that if $ab=1$ then $a$ and $b$ must be multiplicative inverses, and so Here is a general proof for all rational numbers.

$$a=\frac{1}{b}$$

so, we can substitute this identity for $a$ back into our initial equation and say that

$$x^a=\sqrt[b]x\longrightarrow{x^{\frac{1}{b}}}=\sqrt[b]x$$

Thus, the $a$ root of some number $x$ can be thought of as $x^{\frac{1}{a}}$ for vice-versa.

Technically, we can take the irrational root of a number (i.e. $\sqrt[\pi]x$); however, this proof above only works for rational numbers. We will not be able to prove the irrational root of a number is possible until we have some tools from calculus.