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1: Discovering of negative numbers.

Assume a and b are positive integers

$x+a=b$ ----> if $b>a$ then $x$ is positive integer

$x+a=b$ ----> if $b=a$ then $x=0$

$x+a=b$ ----> if $b<a$ then $x=b-a$ is negative integer

2: Discovering of rational numbers.

$x+x+....+x=a.x=b$ ----> $b \equiv k\pmod a$ if $k$ is not zer0, $x=\frac{b}{a}$ is not integer .

3: Discovering of irrational numbers

$x+x+x......+x=x.x=x^2=2$ ----> $x=\sqrt{2}$ is not rational number

4: Discovering of complex numbers

$x^2=-1$ ----> $x$ is not irrational number

I wonder what 5th step can be for next generation numbers. Is there any known operator or equation to find next generation numbers?

or in other words, Are the complex numbers end of story for numbers to be found as an equation via an operator?

Could you please tell me your ideas and share your knowledge about this subject?

NOTE: I know the quaternions that are a number system that extends the complex numbers. Actually I wonder if possible or not to define next generation numbers via known operators or new operator such as previous numbers (negative numbers,rational numbers,irrational numbers, complex numbers) were defined as equation of $x$.

Thank you very much for answers and links.

Mathlover
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    The picture is not nearly as simple as this. For example at point 3 you can branch out and include some irrationals but not all of them while keeping the propety that the set is closed under the basic arithmetic operations. After step 4 you can continue in various ways, but you end up giving up on some of the properties (commutativity of multiplication and/or the ability to divide with every non-zero element). Already at step 4 you gave up on the ability to compare all the numbers. You don't see inequalities between complex numbers. And questions like: "Is $i>0$ ?" are meaningless. – Jyrki Lahtonen Jul 04 '12 at 07:52
  • Steps were just to demostrate simply to how jump next numbers as summary. I know there can some another equations to discover some kind of irrational numbers such as Transcendental number ($e$,$\pi$). The question is about next numbers system. – Mathlover Jul 04 '12 at 07:58
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    This question is perhaps a bit too broad. While the possibilities for "new" number systems are well understood, the answers branch out somewhat uncontrollably depending on which properties you want to have, AND at which point will you branch out! For example, if you are willing to give up only commutativity of multiplication, and extend the reals, your alternatives are limited to the complex numbers and quaternions. If you do the same starting from the rationals, there are infinitely many 4-, 9-, 16-, 25- dimensional "number systems" to choose from. There is no common "cover it all umbrella". – Jyrki Lahtonen Jul 04 '12 at 08:48
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    The fact of interest here would be Hurwitz's theorem. There are other ways to "extend" the complex numbers too - like $\Bbb C$-algebras. Also, there are some interesting numbery systems already that are missing, namely $p$-adics and perhaps adeles. It's a mistake to think that mathematical facts generalize in only one way: typically, generalizations and analogues branch out in many different ways. – anon Jul 04 '12 at 08:51
  • @anon: Thanks for link. I try to see an equation includes operators and $x$ that we want to find it in new number system. Is it possible?. I noticed this finish in next number system definations after complex numbers. – Mathlover Jul 04 '12 at 08:57
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    You don't need all the complex numbers to have solutions to all the polynomial equations, unless you include all the reals. Equations won't give you all the reals. You need topological properties as well (include limits of sequences according to some metric and such). – Jyrki Lahtonen Jul 04 '12 at 09:01
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    This looks like a duplicate of either Is there is a number system which is extension of complex number system? or The largest number system, or both. –  Jul 04 '12 at 09:04
  • @JyrkiLahtonen I've read the reals can be constructed purely algebraically as a wreath product, which of course is designed to capture the familiar digital expansions and their carrying properties. (Just tangential trivia.) – anon Jul 04 '12 at 09:06

1 Answers1

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Several extensions can be named. One of them is the set of "hypercomplex" numbers, see mathworld. They are somewhat like the quaternions in that you have additional elementary numbers $j$ and $k$, except the relations are chosen such that all numbers commute.

There are also "hyperreals", which is like a refining of real numbers. If you've heard about infinitesimals and infinities, this is how they're made rigorous. I'd recommend you read up on wikipedia. I'm confident the hyperreals can be extended to a 2D version, thus yielding a different "hypercomplex": this is not Davenport's algebra, but a 2D version which is, in a sense, more "dense" than the regular reals.

Though, as noted in the comments to the question, the real problem is Hurwitz' theorem, which says we cannot find very many algebras with interesting properties.

akkkk
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  • I do not see an equation of $x$ in hyper complex numbers. I wonder exactly this why we cannot write an equation includes $x$ and operators and equality as we wrote for complex numbers $x^2+1=0$. Is there any proof that such equation is impossible to write for hypercomplex numbers? thanks for your answers – Mathlover Jul 04 '12 at 10:38
  • Hi Mathlover, what do you mean by this? – akkkk Jul 04 '12 at 10:39
  • When I check hyper complex number links just some number definantions such as $j$,$k$ but no any equation that defines the new number system in an equation of $x$ as we can define for negative numbers,rational numbers,irrational numbers, complex numbers. – Mathlover Jul 04 '12 at 10:41
  • In case of the complex numbers, $i$ is defined as the solution of $i^2+1=0$. In quaternions, $i$, $j$ and $k$ are defined as the respective and axiomatically unequal solutions of $i^2=j^2=k^2=ijk=-1$ (notice that this is actually 3 equations with 3 unknowns). There is no natural way to "generate" them out of what you already have, because then they wouldn't be new numbers in any sense. – akkkk Jul 04 '12 at 11:43
  • I understand the definition of quaternions. Please consider $i^2+1=0$ thus $i^2=-1$ and then put it in $i^2=j^2=k^2=ijk=-1$. you will get $-1=j^2=k^2=ijk=-1$, and from that if we solve via algebra rules $j=i=\sqrt{-1}$ and $k=i=\sqrt{-1}$ and from last $\sqrt{-1}=-1$. Please explain how to solve that paradox. Is the known algebra wrong? where is the problem? – Mathlover Jul 04 '12 at 12:11
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    @Mathlover, no, we /define/ $i$ such that $i^2=-1$, but that does not mean that $i$ is unique - in the complex plane, we just only define one such number. In fact, quaternions make use of this by assuming there are more such "$i$", which, as I said, are defined to be different. But all this is no longer relevant to your original question. Also, if you understand the definition of quaternions, why do you think there is a paradox? – akkkk Jul 04 '12 at 12:16
  • If it is true , Fundamental theorem of algebra should be wrong. According to it that Gauss proved $x^2+1=0$ must have 2 roots. ${+\sqrt{-1},-\sqrt{-1}}$ you mention more than 2 roots can be. Is not it a contradiction? Thanks for answers.http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra – Mathlover Jul 04 '12 at 12:24
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    @Mathlover, firstly, the fundamental theorem of algebra is about the complex plane, not about quaternions. Secondly, it gives a minimum amount of roots, so it's not even a contradiction. – akkkk Jul 04 '12 at 12:30
  • :If so Quaternions cannot have any relation with algebra. Right?? we just write $x^2+1=0$ and the equation does not have any knowledge about complex plane we define everything while searching solution of it. $(x.x+1=0)$ has just with operators $(. , + )$. I believe it must be more clear definitions of quaternions. Thank you very much for your answers. Really very helpful – Mathlover Jul 04 '12 at 12:41
  • Yes, strictly speaking, whenever you give equations or anything, you need to specify what algebra you are using, because you are right, there are several. Mathematicians usually specify this by saying, for example "$x^2+1=0$ has no solution in $\mathbb{R}$," while indeed it does have solutions "in $\mathbb{C}$" – akkkk Jul 04 '12 at 12:51
  • Now more clear general picture of algebra. If we define different rules for some operators, we have different algebra. I read the remarks in the link below and it mention my question's answer above that $x^2+1=0$ has infinitely many quaternion solutions. Now I feel more stable. I understood how important to define clearly which algebra we use to solve an equation. http://en.wikipedia.org/wiki/Quaternion#Remarks – Mathlover Jul 04 '12 at 13:23