The solution discussion to Problem of the Week 11 may have got you thinking, what does it mean to raise a number to an irrational power? This will be the focus of this article.

### Recap: Rational powers

First, let’s take a detour through rational powers. As a reminder, rational numbers are those of the form $m/n$, where $m, n$ are integers. So what do we mean when we say:

$x^{m/n}$

For the numerator, we just mean multiply $x$ by itself $m$ times. Fair enough. What about the denominator? We mean “take the $n$th root of $x^m$” i.e. find the number that when multiplied by itself $n$ times equals $x^m$ (of course, you could take the root of $x$ first and then exponentiate too).

As far as calculation goes, it is possible to just approximate the $n$th root by ‘guessing’ values. However, there is a far more efficient approach: the Fractional Binomial Theorem. I won’t go into much depth here, but in short, the normal binomial theorem tells you how $(x+y)^n$ expands i.e. what it looks like when you multiply $(x+y)$ by itself $n$ times. Note that this is true when $n$ is a positive integer. The Fractional Binomial Theorem provides a way of extending this to negative and/or rational values of $n$. Considering negative/rational exponents of binomial expressions shouldn’t be too unfamiliar e.g. $\sqrt{1+x}$ is an example of something you could expand with the Fractional Binomial Theorem. The ‘expansion’ in this case just means finding an infinite sum of terms that are numerically equal to the expression.

### Irrational powers

$4^2 = 16$

$4^{0.5} = 2$

$4^{\sqrt{2}} = \, ?$

There are two ways you can think about irrational powers. The first involves something like approximation. The second involves the function $e^x$ and logarithms.

#### Approach 1

Using the example above as a stimulus, we know that $\sqrt{2}$ is 1.414213562… . Let’s take a step back now. We know what irrational numbers are, and we know that their decimal value is made up of an infinite string of digits. If we ‘cut’ the decimal representation at any point, discarding all digits beyond that point, we are left with a rational number, which we can have as a power. With this in mind, we can construct an infinite sequence of rational numbers:

$y_0, y_1, y_2, ...$

where $y_i$ includes all digits up to and including the $i$th digit after the decimal point (so when $i=0$, we only include the part of the number before the decimal point).

What we now do is set the value of $x^y$ for any irrational $y$ equal to $\lim_{i \to \infty} x^{y_i}$. This just means ‘whatever $x^{y_i}$ tends to’ i.e. is becoming closer and closer to as the value of $i$ increases. This value might seem difficult to pinpoint with this loose reasoning, especially when the actual value is irrational. Don’t worry: there is a way to formalize this process. If you’re interested in finding out more, check out this playlist on Khan Academy.

#### Approach 2

Another approach involves defining what exponentiation (raising to a power) means for a single base (the number being raised to a power), when raised to any real power, and then extending this so we can use it for any base.

The base mathematicians like to choose is $e = 2.71828...$, an irrational number associated with growth processes, among other things. One possible reason for this is that all powers of it are reasonably straightforward to define:

$e^a = \lim_{n \to \infty} (1+\frac{a}{n})^n$

What this definition means, similar to before, is that $e^a$ takes on the value that $(1+\frac{a}{n})^n$ tends to as you increase $n$.

Once you define this function, which works for any real number $a$, you can plug in whatever you want for $a$. What you would do to obtain $x^y$ is plug in $a = y \times ln(x)$, where $ln$ is the natural logarithm function. This function outputs the number that you would need to raise $e$ to in order to get $x$.

### Summary

There are two ways to think about irrational powers. The first is:

$x^y = x^{y_i}$

where $y_i$ is rational and approaches the value of $y$. Intuitively, this is kind of like we are approximating the irrational power with closer and closer rational powers.

The second is:

$x^y = (1+\frac{y \times ln(x)}{n})^n$

where $n$ approaches infinity.

Hopefully the notion of irrational powers isn’t so outlandish for your intuition anymore.