This seems to be similar to (I'd venture to say as hard as) a problem of Erdős open since 1979, that the base-3 representation of $2^n$ contains a 2 for all $n>8$.
Here is a paper by Lagarias that addresses the ternary problem, and for the most part I think would generalize to the question at hand (we're also looking for the intersection of iterates of $x\rightarrow 2x$ with a Cantor set). Unfortunately it does not resolve the problem.
But Conjecture 2' (from Furstenberg 1970) in the linked paper suggests a stronger result, that every $2^n$ for $n$ large enough will have a 1 in the decimal representation. Though it doesn't quantify "large enough" (so even if proved wouldn't promise that 2048 is the largest all-even decimal), it looks like it might be true for all $n>91$ (I checked up to $n=10^6$).
A negative base is a point of conflict between the three commonly used meanings of exponentiation.
- For the continuous real exponentiation operator, you're not allowed to have a negative base.
- For the discrete real exponentiation operator, we allow fractional exponents with odd denominators, and
$$(-a)^{b/c} = \sqrt[c]{(-a)^b}= \left( \sqrt[c]{-a} \right)^b = (-1)^b a^{b/c} $$
(and this is allowed because every real number has a unique $c$-th root)
- For the complex exponentiation operator, exponentiation is multivalued. An exponentiation with denominator $n$ generally takes on $n$ distinct values, although one is generally chosen as the "principal" value.
For $(-5)^{2/3}$, these three exponentiation operators give
- Undefined
- $\sqrt[3]{25}$
- $\omega \sqrt[3]{25}$ is the principal value. The other two are $\sqrt[3]{25}$ and $\omega^2 \sqrt[3]{25}$, where $\omega = -\frac{1}{2} + \mathbf{i} \frac{\sqrt{3}}{2}$ is a cube root of $1$.
Unfortunately, which meaning of exponentiation is meant is rarely ever stated explicitly, and has to be guessed from context.
I'm guessing that the second one is meant.
In case you're curious, here is part of the rationale for the first and third conventions.
In the first convention, 'continuity' is important. If two exponents are 'near' each other, then they should produce 'nearby' values when used to exponentiate. However, despite the fact $2/3$, $3/5$, and $\pi/5$ are all similarish in size, $(-5)^{2/3}$ and $(-5)^{3/5}$ are widely separated by the fact one 'should' be positive and the other negative. And it's not even clear that $(-5)^{\pi/5}$ should be meaningful!
For the third convention, the whole thing is like the idea of $\pm 2$ being the 'square root of 4', but for the fact the complexes cannot be cleanly separated into "negative" and "positive" to let us choose a specific one nicely.
A method is chosen for the principal value, based trying to get positive bases 'right' and trying to keep continuity as much as possible, but alas this convention gets the negative bases 'wrong'.
In some sense, this can be viewed as the principal value of $(-5)^{2/3}$ chosen to be "two-thirds of the way" from positive to negative.
Best Answer
Suppose $N\in\mathbb{N}$ odd. We can rewrite $N=2n+1$ for some $n\in\mathbb{N}$. Since any polynomial function $\mathbb{R}\to\mathbb{R}:x\mapsto x^{2n+1}$ is strictly increasing (the derivative is $(2n+1)x^{2n}>0\,\,\forall x\neq 0$ and equals $0$ only when $x=0$), if $a<b$, then $a^{2n+1}<b^{2n+1}$ for all $a,b\in\mathbb{R}$.
Suppose $N\in\mathbb{N}$ even. We can rewrite $N=2n$ for some $n\in\mathbb{N}$. If $N=n=0$, then $a^{n}<b^{n}$ is false for any $a,b\neq 0$ (since $x^{0}=1\,\forall x\in\mathbb{R}_{0}$) and not defined at $a=0$ or $b=0$. Suppose then $N\neq 0$. As $\mathbb{R}\to\mathbb{R}:x\mapsto x^{2n}$ is symmetric along the axis $x=0$ and strictly increasing on $\mathbb{R}^{+}$, so that $a^{2n}<b^{2n}$ if $|a|<|b|$.
Your rules are correct.