2

It is said that the first numbers we used were natural numbers like $0$, $1$ ,$2$... in $\mathbb{N}$.

Then we discovered negative numbers $-1$,$-2$... , and classified them all in $\mathbb{Z}$.

Then the rational like $1/2$, $3/5$ and classified them in $\mathbb{Q}$.

Then we discovered irrational numbers like $\sqrt2$ and $\pi$ and classified everything in the real numbers $\mathbb{R}$.

Then, after a lot of debates, we introduced imaginary numbers thanks to the imaginary unit $i$ and we classified them along with real numbers in the complex set $\mathbb{C}$.

But if you're reading this, you certainly already now all that. My question is :

Have we proved the existence of numbers that don't suit in these sets ? or is $\mathbb{C}$ the 'ultimate' set of numbers ? If not, is there a set of numbers which contains them all ?

Pyrofoux
  • 1,758
  • 1
    I think that the extension of each set into the "next one in line" was motivated by the fact that certain operations on certain elements in the set resulted with an element that was not in the set. For example: $2,5\in\mathbb{N}$ but $2-5\not\in\mathbb{N}$, so we "invented" $\mathbb{Z}$. Another example: $-2\in\mathbb{R}$ but $\sqrt{-2}\not\in\mathbb{R}$, so we "invented" $\mathbb{C}$. Therefore, you should ask yourself if there are any operations and elements $\in\mathbb{C}$, such that the result is $\not\in\mathbb{C}$. – barak manos Dec 21 '14 at 15:23
  • Some previous discussions on the topic of alternative number systems. Also, why is this tagged as elementary set theory? – Asaf Karagila Dec 21 '14 at 15:42
  • @barakmanos I find your hypothesis very clever. Do you happen to know the answer ? – Pyrofoux Dec 22 '14 at 12:31
  • @Pyrofoux: Thank you. As far as I know, $\mathbb{C}$ is complete in the sense that every operation applied on any elements $\in\mathbb{C}$ results with an element $\in\mathbb{C}$ (with the exception of dividing by zero and raising zero to the power of zero). But perhaps there are some operations that we haven't thought of... – barak manos Dec 22 '14 at 12:52

4 Answers4

1

From what I've read you can keep going with quaternions and octonions.

http://en.wikipedia.org/wiki/Octonion

I'd imagine if you wanted to you could make a 16-tuple equivalent, and keep going, but I don't think there's much use for them or they would have a name.

Ssarutto
  • 110
1

The term number really doesn't make too much sense and exists mostly for historical reasons (imo). It vaguely describes some element of some algebraic structure, so we can't really speak of an 'ultimate' set of numbers. But since you are looking for something 'bigger': Consider quarternions, hyperreals, ordinals, surreals and surcomplex numbers (use Wikipedia). For example, the surreal numbers, vaguely speaking, extend the reals by adding an infinite number (ha) of different "infinities" and their multiplicative inverses, which are usually called "infinitesimals". The surcomplex numbers are to the surreals, what the complex numbers are to the reals. But there really is no real "limit", on how many "numbers" we can have.

I should mention though, that ordinals, surreals and surcomplex numbers are proper classes and therefore not exactly sets of numbers.

Stefan Perko
  • 12,467
1

A good thing to note is that if you have a polynomial with all of its coefficients in $\mathbb C$, then all roots of this polynomial lie in $\mathbb C$. This does not hold for something like $\mathbb R$ or even $\mathbb Z$ (see $x^2 + 1$). This explains why complex numbers are so important.

Now on the other hand, you may ask why not use quaternions, octonions, or even sedenions. These algebras get more "weird" as you continue:

  1. Quaternion multiplication is not commutative ($ab \neq ba$ for all $a, b$).
  2. Octonion multiplication is not commutative or associative $a(bc) \neq (ab)c$ for all $a,b,c$).
  3. Sedenion multiplication isn't even alternative (which is weaker than associativity). In addition, sedenions do not have a division algorithm (you can have two sedenions $a, b \in S$ with $a, b \neq 0$ but $ab = 0$).
  4. You can keep constructing these algebras (see Cayley-Dickenson construction). However the only $\mathbb R$ algebras with division have $n=1,2,4$ (the reals, complex numbers, and quaternions).
George V. Williams
  • 5,213
1

The issue here is that you need to define what a 'number' actually is. According to wikipedia - not the best of sources, but should be enough here -'A number is a mathematical object used to count, measure, and label'. Basically, all you need to have some kind of 'numbers' is a set where you have defined operations, giving you - depending on the exact properties - a group, a ring or something else.

Now, if you are talking about extensions of our usual number system with $\mathbb{N},\mathbb{Z},\mathbb{Q}, \mathbb{R},\mathbb{C}$, then for most practical purposes we simply do not need anything more than $\mathbb{C}$.

However, this does in no way mean that there are no further extensions. An instance of those would be the quaternions or octonions mentioned in the answer above. And, indeed, you can keep going indefinitely. However, with each extension while you gain some properties - eg that every polynomial has roots in $\mathbb{C}$ - you also lose some, for example that the expression $z_1<z_2$ does not make sense for complex numbers $z_1,z_2$ anymore.

Still, to go back to the beginning, depending on how you define a number, you really don't need much to get a number. To give you a rough idea, if you say a number is 'a mathematical object used to count, measure and label', then you could simply take two objects - say, bananas and apples -, consider these some weird kind of values, and assign these values to elements of a set. Then you'd already have the property of your 'numbers' labeling something. And, as for being mathematical objects, there's not really much standing in the way of bananas and appleas being such.

As long as you do not explicitly require your numeral system to fulfill some properties - in case of our usual numbers these are given by eg the Peano axioms for naturals - you are fairly free to choose anything as potential numbers.

So, to answer your question: No, we have in no way proven that there are no numbers that don't fit into $\mathbb{C}$. There are even extensions of $\mathbb{C}$ that can be considered numbers. And unless you give a set of properties all numbers have to fulfill, you also won't be able to find a set containing them all - given such a set $A$, you could find a larger set (eg $P(A)$), and just interpret those as numbers. However, for practical purposes, as far as I know the numbers we 'usually' have are wholly sufficient.

Some Math Student
  • 1,502