New types of numbers

[tomtom2357]

I have an idea, that there might be different types of numbers out there. First we (ancient humans) thought that only rational numbers exist, then someone proved them wrong, then we thought that only real numbers exist, but again they were proven wrong (imaginary numbers), so my question is, is there any other form of number out there (no, I don't want infinitesimals or anything like that). The new numbers were only found as solutions to a problem (the first number to be proven irrational was the square root of 2, it arises as the length of the diagonal of the unit square, i was found as the solution to x²+1=0, but is there any other equation (that isn't contradictory) that doesn't have any solutions on the complex plane? Such an equation would have to be a transcendental equation, because it has been shown that any polynomial has solutions in the complex plane.

[Qaanol]

Define the word “number” and you’ll have your answer. /notkidding

There are hypercomplex numbers. Analogous to the complex numbers (which compared to the reals gain algebraic closure but lose order), there are the quaternions (which lose commutativity), octonions (which lose associativity), and the sedenions (which have zero-divisors).

[tomtom2357]

I'm sorry, I should have been more clear, numbers that satisfy an equation that is not satisfied on the complex plane, the hyper-complex numbers are just a useful extension, they only provide more solutions, they don't fill in a gap in the number system (like the complex numbers do). As always, feel free to prove me wrong.

[Proginoskes]

In On Numbers and Games, John Conway mentions that there are gaps in the surreal numbers, and that someone should investigate them.

[tomtom2357]

Thanks, I'll try to get a copy of that!
Also, is there a complex solution to x^x=1/2?

[Sizik]

Thanks, I'll try to get a copy of that!
Also, is there a complex solution to x^x=1/2?

Many

[tomtom2357]

Okay, obviously x^x won't work (Lambert W), we may need a more complicated equation.

[Talith]

tomtom2357, you seem to be very interested in many areas of mathematics and you want to prove novel results (normally famous unsolved problems which might be a problem). That's great, and it's a good thing that you're so curious, but might I suggest you try concentrating your efforts on one small area first, instead of spreading them across a wide range of topics and making a new post for each one... You're asking good questions, I just worry that you're asking them for the sake of asking, when you can easily Google a lot of these answers and generally do independent research on the topics. My advice would be, become a master of one trade, before trying to become a jack of all trades.

[Xanthir]

I have an idea, that there might be different types of numbers out there. First we (ancient humans) thought that only rational numbers exist, then someone proved them wrong, then we thought that only real numbers exist, but again they were proven wrong (imaginary numbers), so my question is, is there any other form of number out there (no, I don't want infinitesimals or anything like that). The new numbers were only found as solutions to a problem (the first number to be proven irrational was the square root of 2, it arises as the length of the diagonal of the unit square, i was found as the solution to x²+1=0, but is there any other equation (that isn't contradictory) that doesn't have any solutions on the complex plane? Such an equation would have to be a transcendental equation, because it has been shown that any polynomial has solutions in the complex plane.

An equation that doesn't have a solution in the complex plane is "x*0 = 1". However, you can't give it an answer without breaking the complex numbers. Other number systems *can* give it a solution, but they have a different structure than the complex numbers.

[Qaanol]

I'm sorry, I should have been more clear, numbers that satisfy an equation that is not satisfied on the complex plane, the hyper-complex numbers are just a useful extension, they only provide more solutions, they don't fill in a gap in the number system (like the complex numbers do). As always, feel free to prove me wrong.

|x²+1| + |y²+1| + |(xy)²+1| = 0

This equation has a solution in quaternions but not in complex numbers.

[jestingrabbit]

I'm sorry, I should have been more clear, numbers that satisfy an equation that is not satisfied on the complex plane, the hyper-complex numbers are just a useful extension, they only provide more solutions, they don't fill in a gap in the number system (like the complex numbers do). As always, feel free to prove me wrong.

|x²+1| + |y²+1| + |(xy)²+1| = 0

This equation has a solution in quaternions but not in complex numbers.

Where by "a" you mean at least one. For any x with Re(x)=0 and x^2 + 1 = 0, there's a circle of ys that satisfy this equation. I'd also have a preference for |z|^2 over just |z|. |z|^2 just needs conjugates, which is a lot less hassle than square roots.

[tomtom2357]

So now x²,y²,(xy)²=-1? Okay, that is an equation that requires quarternions. Thanks, I would never have thought of using two variables to form the equation!

[Afif_D]

http://forums.xkcd.com/viewtopic.php?f=17&t=66816&hilit=crazy+numbers

This guy invented crazy numbers such that |x|<1

[Dason]

This guy invented crazy numbers such that |x|<1

I'm not sure I can believe in numbers like 0, 1/2, or .252. Too crazy for me.

(ok so I followed the link and it's supposed to |x|<0)

[WarDaft]

In On Numbers and Games, John Conway mentions that there are gaps in the surreal numbers, and that someone should investigate them.

Gaps? I thought they were the largest possible ordered field. How can you have gaps in that?

[jestingrabbit]

In On Numbers and Games, John Conway mentions that there are gaps in the surreal numbers, and that someone should investigate them.

Gaps? I thought they were the largest possible ordered field. How can you have gaps in that?

They're not a set so they're not a field. Some people write that that they are a Field.

I'm not sure that I understand what you mean when you say that they're the largest possible. You could use an ultraproduct construction to get something that would be larger in that it would contain a subset that was the surreals (though it would, of course, have the same "cardinality") and it would still be ordered (because of the nature of the ultraproduct construction).

[WarDaft]

I don't actually know anything about them that it doesn't say here.

[tomtom2357]

In On Numbers and Games, John Conway mentions that there are gaps in the surreal numbers, and that someone should investigate them.

Gaps? I thought they were the largest possible ordered field. How can you have gaps in that?

They're not a set so they're not a field. Some people write that that they are a Field.

I'm not sure that I understand what you mean when you say that they're the largest possible. You could use an ultraproduct construction to get something that would be larger in that it would contain a subset that was the surreals (though it would, of course, have the same "cardinality") and it would still be ordered (because of the nature of the ultraproduct construction).

So, you couldn't make a number system (that was a set) that had cardinality greater than that of the real numbers?
Also, if the surreal numbers are not a set, then how can it have cardinality?

[Deedlit]

No, there are fields of every infinite cardinality. For example, take an arbitrary subset of the surreals, and close it under the field operations.

Yes, the surreals are not a set so they technically do not have a cardinality. However, all proper classes (classes that are not sets) can be said to be of the same size, since there is a class bijection between any two proper classes, and this size can be said to be larger than the size of any set, since any set can be injected into any proper class.

Gaps? I thought they were the largest possible ordered field. How can you have gaps in that?

For example, there is a "gap" between the finite surreals (surreals bounded by two real numbers) and the positive infiinite surreals. Or, for example, between the surreals smaller than any real > 0, and the remaining surreals. Note that while the reals do not seem to have any gaps, enlarging the set can introduce gaps. So you can have a "largest possible field" with gaps in it.

[Qaanol]

So now x²,y²,(xy)²=-1? Okay, that is an equation that requires quarternions.

No, you are wrong.

[tomtom2357]

I'm sorry, I should have been more clear, numbers that satisfy an equation that is not satisfied on the complex plane, the hyper-complex numbers are just a useful extension, they only provide more solutions, they don't fill in a gap in the number system (like the complex numbers do). As always, feel free to prove me wrong.

|x²+1| + |y²+1| + |(xy)²+1| = 0

This equation has a solution in quaternions but not in complex numbers.

So now x²,y²,(xy)²=-1? Okay, that is an equation that requires quarternions.

No, you are wrong.

Wait, what? I am just confirming what you first said.

[Qaanol]

Wait, what? I am just confirming what you first said.

No you are not. The equation I gave—as well as the equivalent system of equations you gave—has a solution in the quaternions. It does not require the quaternions to solve. The quaternions are sufficient, but not necessary, to solve it. There are plenty of other number systems besides the quaternions in which that equation has a solution. The complex numbers, however, are not among them.

[Proginoskes]

Gaps? I thought [ the surreals ] were the largest possible ordered field. How can you have gaps in that?

For example, there is a "gap" between the finite surreals (surreals bounded by two real numbers) and the positive infiinite surreals. Or, for example, between the surreals smaller than any real > 0, and the remaining surreals. Note that while the reals do not seem to have any gaps, enlarging the set can introduce gaps. So you can have a "largest possible field" with gaps in it.

What I refered to is on page 37 of ONAG (either edition).

This comment also reminded me of a similar structure in nonstandard number theory (the Peano axioms minus the induction axiom). In this case, numbers come in "blocks": the block B(n) containing n is \{ n + i : i {\rm~~is~a~"finite~integer"} \} (except for the block containing 0, which is just \{ 0, 1, 2, 3, ... \}. Between any two blocks B(a) and B(b), if b is not in B(a), there is another block B(c).

[tomtom2357]

Wait, what? I am just confirming what you first said.

No you are not. The equation I gave—as well as the equivalent system of equations you gave—has a solution in the quaternions. It does not require the quaternions to solve. The quaternions are sufficient, but not necessary, to solve it. There are plenty of other number systems besides the quaternions in which that equation has a solution. The complex numbers, however, are not among them.

Sorry, that's what I was trying to say, but I'm curious, what other systems allow a solution to the suggested equation?

[Afif_D]

This guy invented crazy numbers such that |x|<1

I'm not sure I can believe in numbers like 0, 1/2, or .252. Too crazy for me.

(ok so I followed the link and it's supposed to |x|<0)

Sorry. That was a typing mistake.. of mine..

[tomtom2357]

Any other equations to make new types of numbers? Also, if anyone's interested, I figured out how to circumvent the blackout on wikipedia!!!

[undecim]

Any other equations to make new types of numbers? Also, if anyone's interested, I figured out how to circumvent the blackout on wikipedia!!!

http://mohamedmansour.com/wikipedia-sopa.html

On Topic:
Due to Gödel's incompleteness theorem, any sufficiently powerful formal system of numbers (e.g. anything that includes the integers) has equations that cannot be solved. So if you define "number" as the solution to an equation, then you can always have more types of numbers.

[tomtom2357]

Good point!

[Desiato]

Due to Gödel's incompleteness theorem, any sufficiently powerful formal system of numbers (e.g. anything that includes the integers) has equations that cannot be solved. So if you define "number" as the solution to an equation, then you can always have more types of numbers.

One needs to be careful here.

For instance, the theory of real closed fields is complete, yet the reals obviously include the integers. Similarly for the theory of algebraically closed fields of given characteristic.

The difference here is that Gödel really does not refer to the integers, but their first-order theory (as given e.g. by Peano). And there's a few things that can complete the theory of integers in some sense (for instance, if we go beyond recursively enumerable axioms - the theory of "all statements true in the natural numbers" is complete). Things also look different if we go beyond first-order logic. Now, second-order logic has other issues (lack of a proof system), but there are a lot of variations such as second-order arithmetic.

You could also take the model-theoretic perspective and consider that, via Löwenheim-Skolem, any first-order theory that has a model at all will have models of any cardinality. But which cardinalities - and thus, in a certain sense, which "numbers" or, perhaps better, "objects" - there are is defined by the set-theoretic universe in which we operate. In that sense, Gödel cannot really be relevant to the question, as we already need to "define" what numbers we have before we can check what results we get from Gödel.

And for day-to-day mathematics, we freely speak about (and quantify over) the power set of natural numbers, so we technically operate in some form of second order logic already.

So, ultimately, Gödel is not really relevant here at all. If you go down that path and really deduce some "new numbers" from there, you need to be clear about the underlying systems and universe you're discussing.

[tomtom2357]

How about countable sets that are contained in the real numbers?

[MartianInvader]

If you're willing to go to statements and not just equations, you could have statements like "x has no square root", which is untrue of any complex number but is true about elements of, say, the field of rational functions of one complex variable. Whether that's a nicer field or not is up to you.

[tomtom2357]

How about this: I define an extension of the algebraic numbers, a quasibraic number is defined as a root of a quasinomial function. A normal polynomial is of the form axⁿ+bx^n-1+...+yx+z, where a,b,...,y,z,n are integers. Quasinomials allow the exponents (and the coefficient) to be quasibraic numbers or other quasinomials.

[mfb]

Your definition is circular: Imagine the equation x-a=0, which clearly has the solution x=a. Now, x is a quasibraic number if a is a quasibraic number. Wait, what?
For the definition of algebraic numbers, only rational numbers are allowed as coefficients (and integers for the exponents) to avoid this problem.

Maybe you get more numbers if you look at solutions for equations

0=\sum\limits_{i=0}^n a_i\, x^{b_i}

with algebraic a_i and b_i.
Maybe.
This set is still countable.

[mike-l]

You don't get any more solutions by allowing algebraic coefficients, as the set of all algebraic numbers is the algebraic closure of Q.

Allowing different exponents is hairy, as for non integers there isn't a unique value, so what does it even mean for something to be a solution? That 0 is among is (often infinite) possible values? Why is such a thing useful in the first place?

[tomtom2357]

Your definition is circular: Imagine the equation x-a=0, which clearly has the solution x=a. Now, x is a quasibraic number if a is a quasibraic number. Wait, what?
For the definition of algebraic numbers, only rational numbers are allowed as coefficients (and integers for the exponents) to avoid this problem.

Maybe you get more numbers if you look at solutions for equations

0=\sum\limits_{i=0}^n a_i\, x^{b_i}

with algebraic a_i and b_i.
Maybe.
This set is still countable.

Oops, sorry. How about this: a quasibraic number cannot have itself in the definition, and the integers are quasibraic numbers. That should work. This way things like x^x=a can happen.

[Desiato]

Maybe you get more numbers if you look at solutions for equations

0=\sum\limits_{i=0}^n a_i\, x^{b_i}

with algebraic a_i and b_i.
Maybe.
This set is still countable.

I don't see this, are you able to prove it? Certainly there are only countably many equations, but I don't see how you'll prove a countable number of solutions?

I also assume you mean b_i != b_j for i != j. Otherwise, the set is trivially the entire base set, so R or C, and thus uncountable.

[url=http://en.wikipedia.org/wiki/Gelfond–Schneider_theorem]Gelfond–Schneider[/url] proves that this set does extend the algebraic numbers (e.g. sqr(2)^sqr(2) is a solution to x^sqr(2) - 2 = 0, but is itself transcendent per Gelfond-Schneider); however, I don't see trivial ways to show even that this would be a group with addition or multiplication, let alone a ring or field or whatever.

[Proginoskes]

How about this (definition): a quasibraic number cannot have itself in the definition,

I think we're heading for Epimenides's ground here ...

[tomtom2357]

How about this (definition): a quasibraic number cannot have itself in the definition,

I think we're heading for Epimenides's ground here ...

Fine then, and no circular definitions of quasibraic numbers allowed. So no definitions like ax-1=0 defining b and and bx-1=0 defining a.

[mfb]

Maybe you get more numbers if you look at solutions for equations

0=\sum\limits_{i=0}^n a_i\, x^{b_i}

with algebraic a_i and b_i.
Maybe.
This set is still countable.

I don't see this, are you able to prove it? Certainly there are only countably many equations, but I don't see how you'll prove a countable number of solutions?

Well, I forgot to say "non-trivial sum". Of course, 0=0*x has all real numbers as solution.
To avoid issues with complex numbers, require x>0. Let -x be a "quasibraic number" if x is one, so this does not harm. In that case, the highest b_i defines the behavior for x->infinity, so there is a solution with the largest x (if it has any). Although a proof might be tricky, I see no way to get an uncountable number of solutions with a non-constant function of the type given above. Similar to polynomials, I would expect at most n (maybe 1 more or less, but even n more would not harm) solutions.

>> I also assume you mean b_i != b_j for i != j. Otherwise, the set is trivially the entire base set, so R or C, and thus uncountable.
This is not necessary, as a_i x^{b_i} + a_j x^{b_i} = (a_i+a_j) x^{b_i}. Again, the trivial sum has to be excluded.