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?

  • 6
    $\begingroup$ Generally, an exponent between 0 and 1 is a "decimal root", of which the most commonly known are the square and cubed root. So your equation is correct. When you get to calculus, you'll learn that the equation $f(x) = x^a$, where $a$ is any real constant, has a bunch of ways to define it, usually using infinite polynomials. $\endgroup$ Commented Oct 6, 2016 at 19:29
  • $\begingroup$ Yes, when $y$ is a natural number and $x$ is positive, the equality you wrote is right. If $x<0$ it's still true but $y$ has to be odd. $\endgroup$
    – Augustin
    Commented Oct 6, 2016 at 19:31
  • 11
    $\begingroup$ Wait until you find out that $1.5!$ is defined as well... $\endgroup$
    – chepner
    Commented Oct 6, 2016 at 22:48
  • 1
    $\begingroup$ @chepner Bah, Python doesn't think so. ValueError: factorial() only accepts integral values (gamma(2.5) works, though.) $\endgroup$
    – JAB
    Commented Oct 7, 2016 at 0:55
  • 3
    $\begingroup$ Now... Look into negatives... Then negative fractions. Mind blown :) $\endgroup$
    – WernerCD
    Commented Oct 7, 2016 at 4:00

7 Answers 7


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.

  • 44
    $\begingroup$ +1 It's amazing how many of our unusual laws in mathematics can be traced back to "We tried to make things consistent, and the result turned out to be useful... so we kept it." $\endgroup$
    – Cort Ammon
    Commented Oct 7, 2016 at 0:41
  • 4
    $\begingroup$ @CortAmmon I think pretty much every definition or theorem in some sense is of that form. Definitions are attempts at drawing our attention to phenomena. Theorems are explanations of phenomena. $\endgroup$ Commented Oct 7, 2016 at 0:56
  • 1
    $\begingroup$ @MaskedMan: See my linked post. Alfred's point is exactly on target. It is not meaningful to simply lay down rules, but as mixedmath implicitly said we need to check that if we want to extend some operation to a larger domain they had better not contradict the previous rules! That is why a lot of definitions need to be justified by theorems. In this case we extend the old structure (reals with integer exponentiation) to a new one (reals with rational exponentiation), and have to prove that the new structure satisfies the desired properties. $\endgroup$
    – user21820
    Commented Oct 7, 2016 at 14:15
  • 3
    $\begingroup$ @MaskedMan Same reasoning... $x^1 \cdot x^{-1} = x^0 = 1$, so we should have $x^{-1} = \frac1x$. $\endgroup$
    – Barry
    Commented Oct 7, 2016 at 14:20
  • 2
    $\begingroup$ @MaskedMan Exact same thing. $x^{-\frac12} = x^{-1 \cdot \frac12} = (x^{-1})^{\frac12} = {\frac1x}^{\frac12}=\sqrt{\frac1x}$. $\endgroup$
    – Barry
    Commented Oct 7, 2016 at 15:24

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 $\sqrt[n]{b^m} = \sqrt[q]{\sqrt[n]{b^m}^q}=\sqrt[nq]{b^{mq}} = \sqrt[nq]{b^{np}} = \sqrt[q]{\sqrt[n]{(b^p)^n}} = \sqrt[q]{b^p}$

  • $\begingroup$ It amazes me that although this question is asked constantly, amazing answers such as this kept getting posted. On the one hand, extreme redundancy....but on the other hand, I somehow continue to learn new things from answers such as this one! As an unrelated tangent, newer versions of MathJax make root symbols so much more beautiful than they were in the past. And that makes me happy :D $\endgroup$ Commented Oct 7, 2016 at 3:13

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.


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

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).

  • 2
    $\begingroup$ OP: Note that this requires the definition of $e^x$ where $x$ is a real, not necessarily an integer. This can be done as, for instance, $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$. (This is in turn shorthand for $e^x = \lim_{n \to \infty} \sum_{k=0}^n \frac{x^k}{k!}$.) $\endgroup$
    – Brian Tung
    Commented Oct 6, 2016 at 19:35
  • 3
    $\begingroup$ @BrianTung It does get tricky though. You don't want $x^k$ to use the definition in this answer, since that leads to a recursive definition. At the same time, you don't want different definitions for $y^x$ depending on whether $x$ is an integer. Luckily, there are other ways of expressing $e^x$ that do not rely on exponentiation, and they can be as simple as spelling out $\frac{x^k}{k!}$ in your definition with repeated multiplication (e.g. $\prod_{i=1}^k{\frac xi}$). $\endgroup$
    – hvd
    Commented Oct 7, 2016 at 11:25
  • $\begingroup$ @hvd: Furthermore, note that complex exponentiation (as defined here) is incompatible with real exponentiation (that defines $(-1)^{1/3} = -1$). One has to be clear what one is dealing with! $\endgroup$
    – user21820
    Commented Oct 7, 2016 at 14:23
  • $\begingroup$ @hvd: Oops! Yup. I totally forgot what the original question was! $\endgroup$
    – Brian Tung
    Commented Oct 7, 2016 at 16:57

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 $b=0$).


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.


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

Now we can answer questions such as


But we can also ask


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."


And the number -4 is defined into existence Similarly


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.)


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .