# Exponent of a number is a square root?

Say you have a number $x^{\sqrt 2}$.

Is there any way to represent this number so that there's no root (or irrational) as the exponent (so that it's easier to understand for me)? I just can't wrap my head around this.

I was thinking something like $$x^{2^{1/2}}$$ and extending on that idea, but I don't know how.

Also, what about for $x^{\sqrt 3}$? Or $x^{\sqrt n}$?

• $e^{\sqrt{2}\cdot\log(x)}$. – Michael Hoppe May 18 '16 at 6:58
• @MichaelHoppe There's still a $\sqrt 2$ as the exponent, though. Sorry, should have worded the question better. – Qwerp-Derp May 18 '16 at 7:00
• It's the number that, when raised to the power of $\sqrt 2$ gives you $x^2$. It's a number that lies between $x^{141/100}$ and $x^{142/100}$, between $x^{1414/1000}$ and $x^{1415/1000}$, and so on. I don't think that there is any better way of understanding what it is; irrational exponents are not defined in any intuitive way like integer or rational exponents are. – Arthur May 18 '16 at 7:02
• @Arthur Ah, OK. I was thinking about that as well, just wondering if there was a better solution to this (i.e. the thing that you posted). – Qwerp-Derp May 18 '16 at 7:04
• There might be, but I don't know one. – Arthur May 18 '16 at 7:04

I guess you can't wrap your head around an irrational exponent because you don't know what it really is supposed to mean, and you're not just asking how to compute it. That's an excellent question, and here is the answer.

## Integer exponents

The basic idea is that exponentiation is repeated multiplication. That worked nicely for natural exponents, and we obviously get:

$x^{a+b} = x^a \times x^b$ for any real $x$ and natural $a,b$.

This extends immediately to integer exponents as long as we understand a negative exponent as denoting undoing repeated multiplication, which is of course repeated division of the base, which requires nonzero base. So we get:

$x^{a+b} = x^a \times x^b$ for any real $x \ne 0$ and integer $a,b$.

## Rational exponents

The first big conceptual jump is to rational exponents. Here we've to make some crucial decision. We would like $x^{1/2}$ to mean something like "do half of the operation of multiplying by $x$" What on earth would that mean?

Well, if we take real numbers as scaling factors (you can think of them as zoom factors), then $9$ is a $9$-scaling, which is the same as two $3$-scalings done one after the other. So we'd like $x^{1/2}$ to be $3$, don't we?

Seems fine, but also $9 = -3 \times -3$. $-3$ corresponds to a negative scaling (stretched the other way, or reflected through the origin and then scaled by $3$ times), and indeed if you perform a $-3$-scaling twice you are essentially performing a $9$-scaling. In general there are two possible operations that when done twice is the same as $x$-scaling, when $x > 0$. Which to pick? By convention we pick the positive one. Note that if $x < 0$, then there is no such operation and so we don't allow writing $x^{1/2}$.

We can generalize this motivating idea to define $x^{1/n}$ for any even $n$ and $x \ge 0$. For odd $n$ you can check that there is only one scaling that fits the requirement, for any real $x$ including negative ones.

Then we define $x^{m/n} = (x^{1/n})^m$ for any real $x$ and integers $m,n$ with no common factor such that $n > 0$, and of course we require $x > 0$ if $n$ is even. Note that this definition is valid because for every rational $q$ there is a unique pair of integers $m,n$ with the above properties such that $q = m/n$. This is important, because we must make sure we don't give two conflicting definitions of the same thing.

By such a definition, we expect that we again have:

$x^{a+b} = x^a \times x^b$ for any real $x > 0$ and rational $a,b$.

And indeed it is so, although it is tedious to prove.

## Real exponents

The next big jump is to real exponents. We would like to do the same thing, so $x^\sqrt{2}$ should be something like "perform $\sqrt{2}$ times of the multiplication by $x$". But before we can even grasp what it ought to mean, we need to go back to what the real numbers themselves are. Each real number can be approximated to arbitrary precision by rationals. (Some definitions of real numbers state exactly this!) So we hope that if we take a sequence of rationals $q_n \to \sqrt{2}$ as $n \to \infty$, then $x^{q_n}$ tends to some limit as $n \to \infty$ as well. Indeed, that is true for $x > 0$, but the proof is harder than before.

Here is an intuitive sketch of why it works. First note that given any real $x > 0$ and rationals $a,b,c$ such that $b$ is between $a$ and $c$, we can prove that $x^b$ is between $x^a$ and $x^c$. (The order may reverse, but the middle one remains the middle one.) As $n \to \infty$, the change in $q_n$ is smaller and smaller. Eventually given any natural $k$ there is a point past which all subsequent $q_n$ are pairwise less than $\frac1k$ apart. For any two such $q_m,q_n$ such that $q_m < q_n$, we have $q_m < q_n < q_m + \frac1k$ and hence $x^{q_n}$ is between $x^{q_m}$ and $x^{q_m+1/k} = x^{q_m} \times x^{1/k}$. But $x^{1/k}$ can be made as close to $1$ as we wish by choosing a sufficiently large $k$. (This is not trivial to prove but at least intuitive enough to believe.) Therefore $x^{q_m}$ can be forced (by the sandwiching) to be as close to $x^{q_n}$ as we wish, as long as we go far enough in the sequence. We can conclude that $x^{q_n}$ tends to some limit as $n \to \infty$. (This is called Cauchy convergence, but that is beyond the scope of this post.)

After everything it turns out that real exponentiation satisfies our dream:

$x^{a+b} = x^a \times x^b$ for any real $x > 0$ and real $a,b$.

There are other dream properties that also turn out to be true, which you would have been taught as 'rules of exponentiation', but which you can now see are more like 'miracles of exponentiation'.

## Back to $x^\sqrt{2}$

Okay so what is a concrete understanding of $x^\sqrt{2}$? We first need a rational sequence that tends to $\sqrt{2}$. One obvious one would be via its decimal expansion, but then the question arises of how to compute it. One such sequence would be the one recursively defined as follows:

Let $q_0 = 2$.

Let $q_{n+1} = q_n - \frac{{q_n}^2-2}{2 q_n} = \frac{q_n}{2} + \frac{1}{q_n}$ for every natural $n$.

This comes from Newton-Raphson approximation.

You can similarly use Newton-Raphson approximation to approximate $x^{1/m}$ for any integer $m > 0$:

Let $y_0$ be the integer with smallest absolute value such that $x$ is between $0$ and $y_0^m$.

Let $y_{n+1} = y_n - \frac{{y_n}^m-x}{m {y_n}^{m-1}} = \frac{m-1}{m} y_n + \frac{x}{m {y_n}^{m-1}}$ for every natural $n$.

So this is a concrete algorithm to compute $x^\sqrt{2}$ that you can implement in any programming language of your choice or even do by pencil and paper!

Anyway, if you just want purely mathematical intuition, here is another:

$x^\sqrt{2} = x^1 \times x^{0.4} \times x^{0.01} \times x^{0.004} \times \cdots$. (The digits can be computed via the above algorithm...)

Note that the terms in this infinite product tend to $1$ sufficiently fast so that the sequence of partial products actually converge. It isn't always true that this happens, which is why I said earlier that real exponentiation is really not a trivial self-justifying thing. For a simple infinite product with terms going to $1$ but not converging:

$(1+\frac11) \times (1+\frac12) \times (1+\frac12) \times (1+\frac13) \times (1+\frac13) \times (1+\frac13) \cdots \to \infty$

$$\sqrt2= 1.4142135623730951...$$ Let $(a_n):1;1.4;1.41;1.412; 1.4121,...$ Then $$x_1=x^{1}$$ $$x_2=x^{a_2}=x^{1.4}=x^{\frac{14}{10}}=x^{\frac{7}{5}}=\sqrt[5]{x^7}$$ $$...$$ $$x_n=x^{a_n}$$ $$...$$ Then $$x^{\sqrt2}=\lim_{n\rightarrow \infty}x_n$$

Perhaps you could consider it as the root of a function composition.

Suppose that $(f\circ f)(x) = x^2$ then what would $f(x)$ equal?

• How do you show there's a unique such function? (For that matter, how do you show such a function exists, without refering to $x^{\sqrt 2}$?) – Barry Cipra May 18 '16 at 14:35
• Showing existence and uniqueness wasn't my intent here. I was just trying to present something analogous to finding a multiplicative root. (e.g. the roots of $x^3$), using composition instead of multiplication as an operator. But to partially answer your question (for existence), assume that $a^x$ is a continuous function for Real $x$, $a\gt 0$ and that $(a^b)^c = a^{bc}$. – John Joy May 18 '16 at 22:25

I am not sure you will find this more to your liking, but you can use the fact that $$\cos\frac{\pi}{4}=\frac{\sqrt2}{2}$$

to obtain $$x^\sqrt2=x^{2\cos\frac{\pi}{4}}$$

Of course there is still an irrational hidden there, but a geometric interpretation might help your intuition.