How does an exponent work when it's less than one?

Question

I'm rather familiar with exponents, I know that $y^x = y_1 \cdot y_2 \cdot y_3 .... y_x$, but what if the exponent is less than one, how would that work?

I put in my computer $25^{1/2}$ anyway, expecting it to give me an error, and I got an answer!! And even more surprising, when I did this with more numbers and a little research, I found that $$x^{1/y} = \sqrt[y] x$$ Is it just me, or can exponents take on the role of square roots?

Answer 1

You're right that there is something interesting going on. It's certainly true that $25^{1/2} = 5$ is a bit different than $3^2 = 3 \cdot 3 = 9$.

One of the reasons why we've chosen $25^{1/2}$ to mean $\sqrt{25}$ goes like this.

For normal, positive integer exponents, we have the really nice exponent laws $a^b \cdot a^c = a^{b+c}$. For instance, $3^3 \cdot 3^5 = 3^8$. What if we wanted this law to work even when we used numbers less than $1$, or maybe fractions between integers?

Then we would want $25^{1/2} \cdot 25^{1/2} = 25^{\frac{1}{2} + \frac{1}{2}} = 25^1 = 25$, and so we would want $25^{1/2} = \sqrt {25}$. This works with others, too. For instance, $27^{1/3}$ should satisfy $27^{1/3} \cdot 27^{1/3} \cdot 27^{1/3} = 27^1 = 27$, so that $27^{1/3} = \sqrt[3]{27} = 3$.

For this also works for fractions bigger than $1$. Thus $36^{3/2}$ should satisfy $36^{3/2} \cdot 36^{3/2} = 36^3$, so $36^{3/2} = \sqrt{36^3}$. Note that you could also rationalize this as thinking that $36^{3/2}$ should equal $(36^3)^{1/2}$, and indeed that works here!

More generally, choosing $25^{1/2}$ to mean $\sqrt {25}$ (and the related identities) agrees with all of our previous rules regarding exponents, so it seems like a very natural choice to make. And indeed, it is the choice we make.

As an aside: one can further extend exponents to decimal expansions instead of fractions, or go further still.

Answer 2

$$y^x = y\cdots y$$

only if $x$ is a positive integer.

In general:

$$y^x = \exp(x\log y)$$

provided that $y >0$ (the functions $\exp$ and $\log$ can be defined via power series).

Also one can prove:

$$\sqrt[x]{y} = y^{1/x}$$

as you noted (this can also be taken as the definition of roots).

Answer 3

For integer exponents we have the property that $(y^a)^b = y^{ab}$ (among others). In order to define $y^q$ for rational $q$, we want it to preserve this property. Writing $q = a/b$ with $a,b$ integers and $b$ positive, we would like $$ (y^q)^b = (y^{a/b})^b = y^{(a/b)b} = y^a. $$ So we would like $y^q$ to be a $b$th root of $y^a$. This always exists if $y$ is positive. So we define $$y^{a/b} = \sqrt[b]{y^a} \quad(y \ge 0).$$ Then it can be verified that the other properties of exponentiation are satisfied for rational exponents, so this is a good definition.

Answer 4

This isn't really a proof but just another way to look at it.

If you raise $a^b$ to the $n$th power, i.e. $(a^b)^n$, you can calculate this by multiplying the exponents:

$$(a^b)^n = a^{bn}.$$

Now consider the case of $n = 1/b$:

$$(a^b)^{1/b} = a^{b\cdot \frac1b} = a^1 = a.$$

So if I square a number ($b=2$), and then get the number back after I raised it to the $1/2$ power, it sure looks like I took the square root.

In more general terms, raising to the $1/b$th power is the inverse operation of raising to the $b$th power (as long as we stay away from $b=0$).

Answer 5

One thing we notice right away about $b^n; n \in N$ is that $b^nb^m = b^{n+m}$ (This is obvious because $b^n = \underbrace{b\cdot b\cdots b}_{n\text{ times}}$ and $b^m = \underbrace{b\cdot b\cdots b}_{m\text{ times}}$, so $b^nb^m = \underbrace{b\cdot b\cdots b}_{n\text{ times}} \cdot \underbrace{b\cdot b\cdots b}_{m\text{ times}} = \underbrace{b\cdot b\cdots b}_{n+m\text{ times}} = b^{n+m}$).

And that $(b^n)^m = \underbrace{b^n\cdot b^n\cdots b^n}_{m \text{ times}} = b^{n+n+...+n} = b^{n\cdot m}$.

So if we want to extend the definition of $b^n$ so that $n$ is not just a natural number but maybe $n = 0$ or $n < 0$ or $n \in \mathbb Z$ we realize that we want to define it so that $b^0b^n = b^{0+n} = b^n$. That means we must define $b^0 = b^n/b^n = 1$. (We really have not choice).

We also want it so that if $0 < n < m$ then $b^{m-n}=b^mb^{-n}$ so $b^{-n} = \frac{b^{m-n}}{b^m}= \frac{b^{m-n}}{b^{(m-n)+n}} = \frac {b^{m-n}}{b^{m-n}b^n} = \frac 1 {b^n}$.

So we must define $b^{-n} = 1/b^n$. (We really have no choice.)

Now we also have $(b^n)^m = b^{nm}$ this means $ \sqrt[m]{b^{nm}} = b^n$. This really isn't that surprising. After all $\sqrt[m]{b^{nm}} =\sqrt[m]{(b^{n})^m} = b^n$, after all.

But what if we aren't talking about whole integers? What if $\sqrt[m]{b^n} = \sqrt[m]{(b^{nm/m})}=\sqrt[m]{(b^{n/m})^m} = b^{n/m}$. Does that make any sense at all?

Well, it makes perfect sense if $m$ divides $n$ and $n/m$ is an integer.

But if $n/m$ then .... we haven't defined what $b^{n/m}$ means if $n/m$ isn't an integer.

But why shouldn't we define $b^{n/m}$ if $n/m$ isn't an integer? If we define $b^{n/m} = \sqrt[m]{b^n}$ that is a fine definition[@]. And because of our rules $(b^r)^n = b^{rn}$ we really have no choice. We must define it that way.

So $a^{1/2} = \sqrt{a}$. This is because if $a^{1/2} = x$ then $x^2 = (a^{1/2})^2 = a^{\frac 12 * 2} = a^1 = a$. So $x = \sqrt{a}$.

[@] Actually, we have to show that if $r = m/n = p/q$ then $\sqrt[n]{b^m} = \sqrt[q]{b^p}$ as it turns out that is true. $m/n = p/q$ mean $mq= np$ and $\sqrt[n]{b^m} = \sqrt[q]{\sqrt[n]{b^m}^q}=\sqrt[nq]{b^{mq}} = \sqrt[nq]{b^{np}} = \sqrt[q]{\sqrt[n]{(b^p)^n}} = \sqrt[q]{b^p}$

Answer 6

Also consider this: $25 = 5 \cdot 5$.

So if we take "half" of $25$, written as a multiplication, then it certainly makes sense that we get $5$ as the result. Other answers show the general extrapolation of this idea.

Answer 7

I'd like to weigh in with how we define numbers, to give a big picture of it all.

Start with the natural numbers, define a starting point and the notion of a successor.

0
0+1
0+1+1
...

Now define addition as repeated succession, and multiplication as repeated addition. Easy enough.

Now we can answer questions such as

2+6=?
2*6=?

But we can also ask

2+?=6
2*?=6

We now define 6-2 as "the number which when added to 2 gives 6 and similarly define 6/2 as "the number which when multiplied by 2 gives 6.

We have a problem when we try to do things such as 6+?=2 because we only have the natural numbers defined so far. But here we say "I define the negative numbers to exist such that they obey the rules of addition and subtraction as above."

6+?=2
?=2-6=0-4=-4

And the number -4 is defined into existence Similarly

6*?=2
?=2/3

We have now defined the fractions. They are imaginary numbers which are only "real" insofar as they are consistent with the rules.

Likewise we define surds: $\sqrt{2}$ is the number such that $\sqrt{2}\times\sqrt{2}=2$.

We can define exponentiation as repeated multiplication and solve the problem of "what is the nth root of something" by defining it as "it's whatever thing that, when you multiply it by itself n times, gives you that something." That's what your question is essentially about.

$x^{-3}=\frac{1}{x^3}$ because $x^{-3}=x^{-1\times 3}=(x^{-1})^3=(\frac{1}{x})^3=\frac{1}{x^3}$ because we defined $x^{-1}=\frac{1}{x}$ and then followed the established rules.

The natural conclusion to this is to define imaginary numbers such as $i=\sqrt{-1}$ and this gives us an algebraic closure which means we don't have to invent any other kinds of number in order to answer any algebraic question. (You can invent other kinds of number but they have their own algebras.)