How do we know $(-1)^\sqrt{2}$ is nonreal without using advanced concepts?

Question

Note: Please do not give answers with advanced material. I only know up to Algebra 2.

Whenever I put in $(-1)^\sqrt{2}$ into the google calculator, I get something like $-0.266255342 - 0.963902533i$. I know that $(-1)^\frac{p}{q}$ is nonreal if $q$ is even (given that $\frac{p}{q}$ is a common fraction) and that it is real whenever $q$ is odd. But how do I evaluate $(-1)^\sqrt{2}$? $\sqrt{2}$ is not a rational number, so it cannot be expressed as a common fraction.

Once again, I would like a simple answer without advanced concepts.

Answer 1

Before asking how to evaluate an expression, you should ask what it even means.

The calculator you are using has defined $a^b$ to mean $e^{b \log a}$ where $e^z$ is the exponential function with base $e$ and complex number inputs $z$ (depends by a power series or by $e^{x+iy} = e^x(\cos y + i\sin y)$ for real $x$ and $y$). In particular, $e^{it} = \cos t + i\sin t$ for real numbers $t$.

Since $e^{i\pi} = -1$, the calculator you use has chosen to define $\log(-1) = i\pi$, so it regards $(-1)^{\sqrt{2}}$ to mean $e^{\sqrt{2}\log(-1)} = e^{\sqrt{2}i\pi} = \cos(\sqrt{2}\pi) + i\sin(\sqrt{2}\pi)$. Using a calculator, $\cos(\sqrt{2}\pi) \approx -0.266255$ and $\sin(\sqrt{2}\pi) \approx - 0.963902$, which should look familiar to you.

There are other numbers $z \not= i\pi$ for which $e^z = -1$, such as $z = 3i\pi$, so one might also decide to define $\log(-1) = 3i\pi$, which leads to another value for $(-1)^{\sqrt{2}}$. The bottom line is that the expression $(-1)^b$ for real numbers $b$ can sometimes be ambiguous. We have universal conventions on this when $b$ is an integer, but other cases can be delicate.

I'm sorry for violating your request to avoid advanced concepts, but what you are asking about is beyond the level of Algebra 2 even if it doesn't look like it. In Algebra 2 you learn about $a^b$ for some $a$ and $b$, but that doesn't mean the definition of $a^b$ for all $a$ and $b$ has to be at the level of Algebra 2. For an analogue, consider factorials: $n!$ for $n = 1, 2, 3, \ldots$ is first defined in school as the product $1 \cdot 2 \cdot \ldots \cdot n$ (e.g., $5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120$), which is accessible at the level of Algebra 2, but you can plug other kinds of numbers into the factorial function on a calculator or in a computer algebra system: $(1/2)!$ turns out to be approximately $0.88622692$. This makes no sense based on the elementary factorial definition, so you can ask what it means and to keep the explanation at the level of Algebra 2. Well, that's not a reasonable request, because the extension of $n!$ to numbers like $n = 1/2$ uses math beyond the level of Algebra 2 (in fact, it uses calculus) and the exact value of $(1/2)!$ is $\sqrt{\pi}/2$; try to compute that and you'll see it is approximately $0.8862\ldots$.

Answer 2

This is an interesting question. There is no answer that does not involve "higher concepts" since you need higher concepts just to define what $(-1)^\sqrt{2}$ means, let alone compute it.

But you do seem to know enough about powers and roots and complex numbers to see a little deeper into what is going on.

You know that $1$ has two square roots, $\pm 1$. Although the expression $\sqrt{1}$ always refers to the positive root, it's important to keep both in mind.

If you allow complex numbers then $1$ has three cube roots. Clearly $1$ is one of them. The other two are $$ \frac{1 \pm i\sqrt{3}}{2}. $$ (You can check that with ordinary algebra, using the fact that $i^2 = -1$.)

It's even easier to see that $1$ has four fourth roots: $1, i, -1, -i$.

In general, $1$ has $n$ $nth$ roots, though you can't always write them nicely. If you plot them in the complex plane where $x+iy$ corresponds to the point with coordinates $(x,y)$ they show up on a circle and form a regular polygon with $n$ sides.

You can use these roots of $1$ to find all the $n$th roots of any (real) number. If $x$ is positive then you get its $n$ $n$th roots by multiplying its real $n$th root by the $n$th roots of $1$. If $x$ is negative you get its $n$ $n$th roots by multiplying the real $n$th root of $|x|$ first by $1$ and then by the $2n$th roots of $1$ that are not themselves $n$th roots. (That's a little tricky, but doable with just high school algebra).

So $(-1)^{p/q}$ will always have $q$ "answers" when you write the rational exponent $p/q$ in lowest terms. When you study more advanced mathematics you will see why $(-1)^{\sqrt{2}}$ has infinitely many answers, none of which are real. The Google calculator reported an approximation to one of them.

Answer 3

$p$ and $q$ are assumed to be in lowest terms, i.e. relatively prime. If one is even, the other is odd and vice versa.

If $q$ is odd, then we can first take $(-1)^p$ Which is either 1 or -1. Then we take the $q_{th}$ root. If that's odd it doesn't matter what $(-1)^p$ is, the answer is real. If $p$ is odd, then $(-1)^p=-1$. So $(-1)^{p/q}=(-1)^{1/q}$ which is complex if $q$ is even, which it must be.

$\sqrt{2}$ is not a rational number, however, one can choose an rational number as close to $\sqrt{2}$ as you like. After wall, $1.41421...$ is after all specifying a sequence of rational numbers that grow ever closer to $\sqrt{2}$.

For any approximation to the square root of two, there exists yet another closer possible approximation in the rational numbers with a choice of even or odd denominator. This is called the density of the rational numbers in the reals. It's proof might be out of scope. It essentially means between any two real numbers is a rational number. That statement uses Algebra 2 concepts, the proof doesn't. This principle can be iterated to get ever closer rational approximations to any real. We are free to choose whether to use an even or an odd denominator in our sequence of approximations. If $q$ is odd, $(-1)^{p/q}=1$.

If $q$ is even, then $q=2k$ and we have $((-1)^{1/2})^{1/k}=i^{1/k}=u$. Since any real number to an integer power is real and $u^k=i$, we know that $u$ must be complex.