Are binary, octal, decimal and hexadecimal numbers natural, whole, integer, real numbers?

Question

I think that the title is clear in itself. But still i wanna ask whether can we consider these binary, octal, decimal and hexadecimal number formats at least as rational numbers or is it illogical to intersect these different topics? We know that the hexadecimal number uses A to F or a to f for the numbers 10 to 16; in this case, if accepted, will it still be considered as a rational number?

Edit: Ok just a small edit. Are the hex notations A to F or a to f numbers or numerals? I am also confused with these terms as different sources quote them as synonyms while the rest don't.

Answer 1

You should try to disambiguate between representations of numbers and types of numbers.

Natural (and 'whole' is usually a synonym for natural in this context, though whether zero is included or not is sometimes the difference), integer, rational, and real numbers (and the various other kinds) are all different sets of numbers, and they can be thought of as different 'kinds'; each one satisfies a different set of properties. For instance, in the set of natural numbers $\mathbb{N}$ there is no member $n$ satisfying $n+1=0$, but all the other sets have such a member (which we usually write as the number -1). In the set of integers $\mathbb{Z}$ there's no member $h$ satisfying $2\times h=1$; the 'bigger' sets have such a number, and we write it $\frac12$.

In fact, this process is often the motivating factor behind defining larger and larger sets of numbers; the lack of a member $r$ of the set of rational numbers $\mathbb{Q}$ such that $r\times r=2$ vexed the Pythagoreans to no end, but now we recognize that as a sign that we should consider extending the rationals in some way. There are a bunch of ways of doing this, but the one that turns out to be the most canonically useful is the one that 'completes' the rationals, so that every bounded sequence of elements of the set has an upper bound that's a member of the set. This leads to the classic construction of the reals $\mathbb{R}$ by 'cut sets'. Similarly, the lack of any solution to the 'simple' equation $x^2+1=0$ in $\mathbb{R}$ suggests that we might want to pass to some set of numbers that can actually solve all such equations; that would be the set $\mathbb{C}$. (This process doesn't come for free; we give up things at each step of the way, too, often in the form of 'simplicity' in one sense or another. But that starts to take things very far afield...)

By contrast, binary, decimal, etc. are all ways of representing numbers; in particular, the sets of numbers in each of these representations are the same. (Note that I'm deliberately glossing over what 'the same' means here to keep from going even further afield). There are some very thin exceptions to this rule; for instance, we might want to talk about the set of numbers that have a terminating representation in some base, for instance 0.1101_b or $0.2_{10}$. Now, every number with a terminating binary representation also has a terminating decimal representation, but the opposite isn't true; $0.2_{10}=\frac15$ has no terminating binary representation. But this means that we don't even have 'division' available in our sets of terminating-expansion numbers, so they're smaller than the rationals $\mathbb{Q}$, and this makes them less practical from a mathematical standpoint. Generally, when we talk about representations of numbers in bases, we stick to integers.

That said, there's no reason that it doesn't make sense to talk about, e.g., the rational number $\dfrac{\mathtt{6}_{16}}{\mathtt{1E}_{16}}$, using hexadecimal to represent the integers that are the numerator and denominator of the number; this can be useful for computer representation, for instance, of exact rational calculations (so that you always have $3\times \frac13=1$ exactly.) But there's no difference of kind here between $\mathtt{1E}_{16}$, $30_{10}$, 11110_b, etc; they're just different ways of naming the number thirty.

Answer 2

Numbers written in a different base are just as nice of integers as the ones you are accustomed to, and in fact, there is an obvious bijection between them and the normal integer types, e.g. a number written in octal with digits $n_m\ldots n_2n_1n_0$ is exactly $\sum_{k=0}^m n_k 8^k$ in decimal...

So these are regular integers, for sure they can be thought of as rational numbers as well since $\mathbb{Z} \subsetneq \mathbb{Q}$.