# Is $(-1)^{2.16}$ a real number?

A lot of calculators actually agree with me saying that it is defined and the result equals 1, which makes sense to me because:

$$(-1)^{2.16} = (-1)^2 \cdot (-1)^{0.16} = (-1)^2\cdot\sqrt[100]{(-1)^{16}}\\ = (-1)^2 \cdot \sqrt[100]{1} = (-1)^2 \cdot 1 = 1$$

However, there are certain calculators (WolframAlpha among them) which contest this answer, and instead claim it is equal to:

Graphing this as an exponential function was not possible.

What's going on?

• – J.G.
Sep 14 '19 at 18:51
• Wolfram is taking the root before raising it to the 16th power. The principle 100th root of $-1$ has an angle of $180^o/100 = 1.8^o$. Then sixteen of these angles angles added together gives you $28.8^o$ Sep 14 '19 at 18:52
• For most people (who know about the nuances involved), the meaning of $(-1)^{2.16}$ depends on whether $2.16$ is interpreted as a real number or as a specific fraction (presumably $\frac{54}{25},$ and not $\frac{216}{200}$ or $\frac{108}{50}).$ Sep 14 '19 at 18:54
• How do you define a non-integer power of a negative number? For positive reals and rational powers there is a unique positive real root, and a real function can be defined. $0^0$ is notoriously problematic and you might want to think about negative powers of zero . Where the power is $\frac pq\gt 0$ and $q$ is odd, the function can be extended to the reals. Sep 14 '19 at 19:01
• It's a matter of definition. If $q = \frac ab; a\in \mathbb Z; b\in \mathbb N; \gcd(a,b) = 1$ and if $k < 0$ then is it a valid definition to define $k^q$ as $\sqrt[b]{k^a}$? It can be argued that is valid if $a$ is even but I'd say it has too many issues and handwaving and is inconsistant when we do bring in complex analysis to be acceptable. I'd say that definition holds if $k > 0$ but if $k < 0$ we will need to consider principal complex roots. By squaring $(-1)$ you are deliberating and artificially avoiding them. I'd say that is no good. Sep 14 '19 at 19:01

$$(-1)^{2.16}=(-1)^{\frac{54}{25}}=\exp({\frac{54}{25}(2k+1)i\pi})=\exp({\frac{4}{25}(2k+1)i\pi})$$

is a set of $$25$$ numbers corresponding to $$k=0,...24$$ as the exponential above has period $$25$$.

Choosing $$k=12$$ shows that $$1$$ is indeed in this set, though it doesn't correspond to the usual "principal" value which is for $$k=0$$ and which in this case, gives $$\exp({\frac{4}{25}i\pi})$$ and this is what Wolfram Alpha gave

Edit later - just to make it clear, here $$\exp(z)=\sum{\frac{z^n}{n!}}$$ is the uniquely defined usual entire exponential function

If we stay strictly in the real domain, then exponentiation is defined by the formula $$x^y=\exp(y\ln(x))$$.

Since the logarithm is defined only on positive numbers, exponentiation is also well defined only on positive bases $$x$$.

With this definition $$(-1)^{2.16}$$ is undefined.

However, since it also comes from rational exponentiation by continuity, which is merely an extension of integer exponentiation where things like $$(-1)^n$$ are perfectly defined, it is possible to extend exponentiation to negative real numbers as well, but with limitation to rational exponents.

If we call $$\mathbb Q_{odd}=\{\frac pq\in\mathbb Q\mid p\in\mathbb Z, q\in\mathbb N^*, \gcd(p,q)=1\text{ AND } q\text{ odd}\}$$

Then for $$x<0$$ we have $$x^r$$ defined for any $$r\in\mathbb Q_{odd}$$ by $$x^r=x^{\frac pq}=\sqrt[q]{x^p}=\left({\sqrt[q]{x}}\right)^p$$

Since the $$q-$$root is an odd function, it is perfectly defined for negative numbers.

Note that we explicitly require that $$\gcd(p,q)=1$$ to avoid contradictions like $$-1=(-1)^{\frac 13}=(-1)^{\frac 26}=\sqrt[6]{(-1)^2}=1$$

Considering this $$(-1)^{2.16}=(-1)^{\frac {54}{25}}=1$$ would be a reasonable answer.

Now it is also possible to take some high, and examine the situation on the complex domain.

$$z_1^{z_2}=\exp(z_2\ln(z_1))$$ is defined by the same formula that in the real case, but this time the logarithm has an extended definition.

The price to pay is that since $$\exp$$ is a $$2i\pi-$$periodic function, its inverse, the logarithm becomes multi-valued, meaning there will be more than one value per arithmetic expression.

We define $$\ln(z)=\ln|z|+i(\arg(z)+2k\pi)\quad k\in\mathbb Z$$

Let's examine the consequences for $$(-1)^{2.16}$$

$$\ln(-1)=\underbrace{\ln(1)}_0+i(\underbrace{\arg(-1)}_{\pi\text{ from }e^{i\pi}=-1}+2k\pi)=(2k+1)i\pi$$

Also since $$2.16$$ is in fact rational $$\frac {54}{25}$$ as we already seen, we won't get an infinite number of values for the expression, yet still $$25$$ different ones.

Let's call them for $$k=0\cdots24$$ $$z_{[k]}=\exp(\frac{54(2k+1)}{25}i\pi)=\exp(\frac{108k}{25}i\pi+\frac{54}{25}i\pi)=z_{[0]}\,\phi^k$$

With $$\begin{cases}z_{[0]}=e^{(2.16i\pi)} & \text{is called the principal value at }k=0\\\phi=\exp(\frac{108}{25}i\pi) & \text{is a cyclic multiplicative factor (in this case a rotation)}\end{cases}$$

As noted by Conrad, when $$k=12$$ then $$z_{[12]}=1$$ the answer found in the rational exponentiation case is effectively among the results, although it is not the principal value.

The problem is that the exponential function: $$f:\mathbb{R}\rightarrow\mathbb{R}, f(x) = ab^x$$ is only defined for $$b\in(0,\infty)$$ and $$a\neq0$$ so $$(-1)^x$$ is not an exponential function, thus it doesn't have its proprieties.

On the other hand $$(-1)^{2.16} = e^{2.16\log(-1)} = e^{2.16\pi i} = \cos(2.16\pi) + i \sin(2.16\pi)$$ which gives the same value as WolframAlpha.

The graphing is not working because the function $$(-1)^x = \cos(x\pi) + i \sin(x\pi)$$ takes real values $$\iff \sin(x\pi) = 0 \iff x\in\mathbb{Z}$$.

I don't think you are allowed to split negative numbers up like that. Since $$-1=e^{i\pi}$$ then $$(-1)^{2.16}=e^{i\pi\cdot 2.16}$$ which is not a real number.