Question

x^n is not real if x≤0, and n∈T, where T is the set of transcendental numbers, right?

Answer 1

First rule of proofs: Find counterexamples.

Example 1: \((-1)^{\pi}\). Unfortunately, \((-1)^{\pi}\) is not real.
Example 2: \((-2)^{\phi}.\) Unfortunately, \((-2)^{\phi}\) is not real.
Example 3: \((-3)^{e}.\) Unfortunately, \((-3)^{e}\) is not real.

So, it would seem that this whole "find counterexamples" technique is not working. Perhaps, then, it is true that \(x^n\) is not real if \(x\) is negative and \(n\) is transcendental.

Let \(\mathcal{T}_{\mathbb{C}}\) denote the set of transcendental numbers of the form \(t=s+ri\) where \(s\) and \(r\) are real transcendentals. Likewise, let \(\mathcal{T}_{\mathbb{R}}\) denote the set of transcendental numbers of the form \(t=s+ri\) where \(s\) is a real transcendental and \(r=0.\)

Let \(y \in \mathcal{T}_{\mathbb{R}}\). Then, \(y=\lfloor y \rfloor + y'\) for some \(y': 0 < y' < 1.\) Thus, \(x^{y}=x^{\lfloor y \rfloor}x^{y'}.\) \(x^{\lfloor y \rfloor} \in \mathbb{R},\) hence the only issue is proving \(x^{y'} \notin \mathbb{R}.\)

Observe that by definition, \(x^{y'}=(-1)^{y'}|x|^{y'}.\) \(|x|^{y'}\) is real, since \(b^{p}\) is real for all positive \(b\) and all real numbers \(p.\) Via the wonderful identity of Euler, \(e^{\pi i}=-1\) and hence \((-1)^{y'}=e^{\pi i y'}.\) By de Moivre's theorem, \((-1)^{y'}=e^{\pi iy'}=\cos(\pi y')+\sin(\pi y')i.\)

If \((-1)^{y'}\in \mathbb{R},\) then \(\sin(\pi y')=0.\) However, \(\sin(x)=0\) if and only if \(x=\pi n\) with \(n \in \mathbb{Z}.\) \(y' \notin \mathbb{Z};\) therefore, \((-1)^{y'} \notin \mathbb{R}.\) \(\square\)

Let \(z \in \mathcal{T}_{\mathbb{C}}.\) Then, \(z=u+vi.\) Since \(x^{z}=x^{u}x^{vi},\) we can conclude that \(x^z\) is not real by applying the same reasoning as before on \(u.\) \(\square\)

...But it doesn't end here.

There is a very strange thing I discovered. There is a counterexample! There's a lot, actually. An infinite amount: \((-1)^{n}\) with \(\{n: n \in \mathcal{T}_{\mathbb{C}} \wedge s=0\}.\)

Here's the proof: \((-1)^n=e^{\pi i n}.\) If \(n=ri\) for some real transcendental \(r\), then \(e^{\pi i n}=e^{\pi i r i}=e^{-\pi r} \in \mathbb{R}.\) \(\square\)

Interestingly, \(x=-1\) seems to be the only \(x\) such that \(x^{n} \in \mathbb{R}\) when \(n\) is of the form \(n=ri\). Why could this be so, I wonder... Let's look at it like this: \(x^{ki}=(-1)^{ki}|x|^{ki}=e^{-\pi k}|x|^{ki}.\) Interesting. \(-1\) is the only \(x\) such that \(x^n \in \mathbb{R}\) because it is the only \(x\) such that \(|x|^{ki}\) is real. (Fun fact: \(x=0\) and \(x=1\) also work.)

So, what is the conclusion here?
\[\left(n \in \mathcal{T}_{\mathbb{R}} \cup \mathcal{T}_{\mathbb{C}} \right )\wedge \{x: x < 0 \wedge x \ne -1\} \Rightarrow x^n \notin \mathbb{R}.\]