The way that I was taught it in 8th grade algebra, a number raised to a fractional exponent, i.e. $a^\frac x y$ is equivalent to the denominatorth root of the number raised to the numerator, i.e. $\sqrt[y]{a^x}$. So what happens when you raise a number to an irrational number? Obviously it is not so simple to break it down like above. Does an irrational exponent still have a well formed meaning?

The only example that comes to mind is Euler's identity, but that seems likes a pretty exceptional case. What about in general?

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    $\begingroup$ I'm just gonna say this real quick as it's not really an answer, but there isn't just one value for $a^b$ where $b$ is irrational. $\endgroup$ Aug 2, 2011 at 6:42
  • $\begingroup$ @El'endia: Do you have an example for such case? (at least for a positive $a$) $\endgroup$
    – Asaf Karagila
    Aug 2, 2011 at 6:47
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    $\begingroup$ @El'endia: Why do you think so? Does $e^e$ have more than one value? $\endgroup$ Aug 2, 2011 at 6:48
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    $\begingroup$ @El'endia, I suppose you are talking about exponentiation for complex numbers, rather than for reals. Also, isn't it the case that $a^b$ is not unique for any $b$ that is not an integer (assuming we're working with complex numbers)? E.g.: $a^{1/2}$ has $2$ values... $\endgroup$
    – Srivatsan
    Aug 2, 2011 at 6:48
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    $\begingroup$ @Srivatsan: No, $a^{1/2} = \sqrt{a}$ has, by convention, a single value for positive $a$. It's the positive one. $\endgroup$ Aug 2, 2011 at 6:58

5 Answers 5


What do we mean when we write $a^b$, say for $a>0$? The question is a very good one. The answer, unfortunately, is fairly complicated, and the full details are quite lengthy.

We have a clear understanding of what we mean by $a^2$, or $a^5$. And from fairly early on, we learn to define $a^n$, where $n$ is negative, as $\frac{1}{a^{-n}}$.

After a while, we develop an understanding of what we mean by something like $a^{3/4}$. For (we are led to believe) there is a unique positive number $s$ such that $s^4=a$, and then we can define $a^{3/4}$ to be $s^3$. This idea can be used to define $a^{p/q}$, where $p$ and $q$ are integers.

After a while, we can show, more or less rigorously, that the laws of exponents that worked for integer powers also work for expressions of the form $x^{p/q}$, where $p$ and $q$ are integers.

However, what do we mean, for example, by $3^{\sqrt{2}}$? Certainly it is not $3$ multiplied by itself $\sqrt{2}$ times!

There are several ways to resolve the question. One way is to note that $\sqrt{2}=1.41421356\dots$ and consider the sequence $3^{1.4}$, $3^{1.41}$, $3^{1.414}$, $3^{1.4142}$, and so on. All these powers make sense, because the exponents $1.4$, $1.41$, $1.414$, and so on, can be expressed as fractions. But, intuitively, these numbers are getting closer and closer to something, and we define $3^{\sqrt{2}}$ to be that something. We can do a partial informal verification of the "getting closer and closer" part in this case, by using a calculator.

More formally, let $b$ be a real number, and let $b_1, b_2, b_3, \dots$ be an infinite sequence of rational numbers such that the sequence $(b_n)$ has limit $b$. It can be shown that the sequence $(a^{b_n})$ has a limit, which is independent of the particular sequence $(b_n)$ that we have chosen, as long as the sequence has $b$ as a limit.

Then we can define $a^b$ as the limit of the sequence $(a^{b_n})$. With quite a lot of effort, we can then show that the familiar laws of exponentiation hold.

The above approach, though intuitively very natural, is unwieldy. So in practice, we usually take another approach.

The standard way is to first define the function $\ln x$. Then we define the exponential function $\exp(x)$, also known as $e^x$, as the inverse function of $\ln x$. Or else, depending on taste, we first define the function $\exp(x)$, and then its inverse $\ln x$. There is a fair variety of (provably equivalent) definitions.

For example, we could define $\ln x$ by $$\ln x=\int_1^x \frac{1}{t}dt.$$ It is not terribly difficult to show that $\ln$ as defined above satisfies the usual basic "laws of logarithms," and that it is an increasing function, so has an inverse, that we call $\exp$.

Finally, after this background work, we define $a^b$ (for $a>0$) by $$a^b=\exp(b\ln a).$$ We can then easily verify that in the cases where we already "know" what $a^b$ should be, namely rational $b$, the above definition agrees with our intuition, and that the usual "laws of exponents" hold for this more general notion of power.

Warning: In the entire post, it is assumed that $a$ is a positive real number, and that all exponents are real numbers. Complex exponentials are a lot more--complex.

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    $\begingroup$ There is a warning about "laws of exponents" to put here. When negative numbers or complex numbers are allowed, it may fail that $(a^b)^c = a^{bc}$. $\endgroup$
    – GEdgar
    Aug 2, 2011 at 14:49
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    $\begingroup$ @GEdgar: Yes. At the beginning of the answer, I specify that $a>0$, and it is implicit that $b$ is real, but I should be much more emphatic. Will add prominent warning. $\endgroup$ Aug 2, 2011 at 15:53
  • $\begingroup$ Question: What happens in your formula for $ln x$ when x < 1? $\endgroup$
    – Marcos
    Jul 17, 2017 at 4:43
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    $\begingroup$ @MARXOS, we know that the area under $\frac{1}{t}$ is positive for $t>0$ ( the graph is above the x-axis and integration is done from left to right[ Riemann Integral ] ). Mathematically, $\int_{1}^{x}\frac{1}{t}dt>0$. When $0<x<1$, $\int_{1}^{x}\frac{1}{t}dt$ is done from right to left. By definition, we can interchange the limits by writing, $-\int_{x}^{1}\frac{1}{t}dt$ where $\int_{x}^{1}\frac{1}{t}dt>0$. If there was a problem understanding the above, I suggest you understand the concept of a Riemann Integral if you haven't already. $\endgroup$
    – R004
    Sep 15, 2017 at 3:27
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    $\begingroup$ What are the 'complications' in proving the laws fo exponents in the first definition? It seems that if $b_n \to b$ and $c_n \to c$ are (Cauchy) sequences of rationals, then $a^{b_n}a^{c_n} = a^{b_n+c_n}$ by the rational law of exponents, and, by def, this sequence goes now to $a^{b+c}$. The same approach (apparently) works for every other law of exponents I know of. So what's the deal? $\endgroup$
    – seldon
    Nov 12, 2019 at 16:41

Yes you can, when the number you're raising to a power is positive. There are (at least) two ways to define $a^x$, where $a > 0$ and $x$ is a real number that may not be rational:

  1. You can pick a sequence of rational numbers $x_n$ converging to $x$ (i.e., $\lim\limits_{n\to\infty} x_n = x$) and define $$a^x = \lim_{n\to\infty} a^{x_n}.$$

  2. You can use the exponential function $e^x$ (defined in many ways, say as $e^x = \lim_{n\to\infty} (1+\frac{x}n)^n$ or with a power series), and its inverse the logarithmic function that satifies $e^{\ln t} = t$ for all positive $t$, and since $a = e^{\ln a}$, define $$a^x = e^{x \ln a}.$$

Both definitions give the same answer. See also the Wikipedia article on exponentiation, section on real exponents. When $a$ is negative, it's hard to do this while staying in the reals, but see the section on powers of complex numbers.

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    $\begingroup$ '...and its inverse logarithmic function that satisfies... for all strictly positive $t$' (something you need to be careful about on the HS level). $\endgroup$
    – Gerben
    Aug 2, 2011 at 9:09
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    $\begingroup$ @Gerben: Right! Thanks, I've made the correction. Actually I'd originally written $\ln (e^t) = t$, which is true for all real $t$, but when I changed it I didn't think carefully. :-) $\endgroup$ Aug 2, 2011 at 14:40
  • $\begingroup$ @ShreevatsaR if we define in a way of sequence , then it must be independent from choice of sequence.and i think it is, but how can i prove it? $\endgroup$
    – Meet Patel
    Sep 19, 2022 at 0:48
  • $\begingroup$ @MeetPatel Indeed. Proving it is a bit of pain, but can be done with some effort (feel free to ask a separate question about it and maybe someone will answer or point to an existing duplicate): basically we'd want to prove that for every $\epsilon$ there exists a $\delta$ such that if $|y - x| < \delta$ and $|z - x| < \delta$ (where $y$ and $z$ are rational numbers), then $|a^y - a^z| < \epsilon$. $\endgroup$ Sep 19, 2022 at 1:42
  • $\begingroup$ @ShreevatsaR why we need $a$ to be positive? $\endgroup$
    – Meet Patel
    May 2, 2023 at 16:55

For a positive real number $a$ and any real number $b$, you can make the definition:

$$a^b \equiv \exp(b \log a)$$

having first defined the exponential function as an infinite power series (and showing that it converges for every real argument) and the logarithm as the inverse of the exponential function (after showing that the exponential function is monotonic and that its image is the positive real line).

More formally, you can define the function $f:\mathbb{R}^+ \times \mathbb{R} \to \mathbb{R}^+$ which satisfies $f(a,b)=a^b$ for rational $a,b$ and fills in the gaps continuously for irrational $a,b$.

  • $\begingroup$ AFAIK you don't have to define it that way, for positive $a$ it is $a^b=\exp(\log a^b)$, since $\exp$ and $\log$ are inverse, and then using $\log a^b=b\log a$ you get $a^b=\exp(b\log a)$. $\endgroup$
    – eudoxos
    Aug 2, 2011 at 7:09
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    $\begingroup$ @eudoxos To say anything like $\log (a^b) = b \log a$, we'll need a definition of $a^b$, which makes the problem circular. Defining things the way Chris did does not have this trouble... $\endgroup$
    – Srivatsan
    Aug 2, 2011 at 7:16
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    $\begingroup$ There are, of course, many other ways to define the exponential function (e.g. as the solution of $\frac{dy}{dx} = y$ with $y(0)=1$). Or you could define the logarithm first, and the exponential as the inverse of that. Or (although I have never seen it developed this way) you could try defining $x^b$ as the solution of $\frac{dy}{dx} = \frac{by}{x}$ with $y(1) = 1$. $\endgroup$ Aug 2, 2011 at 7:42
  • $\begingroup$ @Robert Israel all good points, thanks. $\endgroup$ Aug 3, 2011 at 8:58

The above answers are fine provided that you are only concerned with real numbers. However, just as $x^{a/b}$ has $b$ solutions in the complex plane, $x$ to the power of an irrational will have an infinite number of solutions.



We note that irrational numbers can be arbitrarily approximated by rational numbers,. For a positive real number $r$ irrational number $j$, integers $p_1$, $q_1$, $p_2$, and $q_2$, the $j$ power can be ordered as well as well-defined, by choosing the $p_i$ and $q_i$ values so that their quotients are arbitrarily close to $j$.


Going beyond the positive reals to arbitrary complex numbers, we can rely on Euler's formula. If $e$ is the base of the natural logarithm and $i^2=-1$, then


So, for all integers $z$, and real numbers $a$ and $b$


(I explains the arg function in a comment.)

Note that $\sqrt{a^2+b^2}$ is always a positive real. Also note that 2πzj is never an integer multiple of 2π except for $z=0$, because j is irrational. So the cosines and sines do not cycle, and each nonzero complex number has a countable infinitude of $j$th powers (and $j$th roots!)

  • $\begingroup$ arg function definition. Let ($a, b$)be an ordered pair of real numbers on a circle whose center is represented by ($0, 0$) Then $arg(a,b)$ is the number of radians in the arc connecting ($a,b$) and ($\sqrt{a^2+b^2}, 0$) $\endgroup$ Feb 12 at 18:33

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