## number system questions

For the discussion of math. Duh.

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zenten wrote:j = 1-1

j has the rather interesting property where j = -j

However, 2j != j, which is important.

Maybe I'm missing something, but can you prove that 2j != j? Because, 2*(1-1) = 2*(0) = 0 = (1-1)

zenten wrote:S = for all x {x is an element of Complex numbers union Joshua numbers | x !=0} (this would be much easier with all the proper math notation symbols, but I think you get the picture).

Then S = Complex numbers. The restriction that x != 0 excludes j.

BTW electrical engineers already claimed j . They write j instead of i for complex numbers so as not to confuse the i representing current
aguacate

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buttons wrote:EDIT: Oh, okay. I guess you want S to not contain 0, so we can't use closure. Before we go any further, which of the ring properties (particularly the ones enumerated by jestingrabbit) do you want your system to exhibit?

Well, - a*j doesn't have a negation would kind of fit, since j=-j

j+b doesn't equal b for b = j or for b = a*j would definitely fit.

The rest don't apply though.

jestingrabbit wrote:What you have to do to describe a well formed system is to tell us what elements you want (J={a+bj | a≠0 or b>0}, I think), operations you are going to define (*, + as well as - and / too I think), where you are going to define them (a/0 is ruled out for the reals, but there are different rules here, so ...?) and what exactly the definitions are that you are going to use.

Yeah, that's part of the reason that I'm posting this here, since I have never developed something like this, I've just (briefly) read descriptions of the common number systems.

jestingrabbit wrote:(a+bj) + (c+dj) = (a+c) + (|b|+|d|)j unless a = -c and then...???

If a = -c, you would end up with
(a+bj) + (c+dj) = (a + |b| + |d|)j

However, that's assuming we're talking about b and d being real numbers. I'm not sure about what to do if they are imaginary/complex.

jestingrabbit wrote:edit: What's wrong with 0 ≠ 0+0? Quite a bit really, it means that you no longer have an additive identity, so that you can't really cancel anymore. That is, you can't prove things like

a+b = a+c if and only if b=c.

How does the proof work again for complex numbers?
zenten

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Maybe I'm missing something, but can you prove that 2j != j? Because, 2*(1-1) = 2*(0) = 0 = (1-1)

No, because your proof involves zero.

Assume 2j = j

2(1-1) = (1-1)

(2-2) = (1-1)

(1-1) = j, by definition

But, by defintion (2-2) != j

Therefore, 2j != j.

Then S = Complex numbers. The restriction that x != 0 excludes j.

No, because j is not the same thing as 0.

BTW electrical engineers already claimed j. They write j instead of i for complex numbers so as not to confuse the i representing current

I think every letter is claimed by someone. They can use a different letter for j if they want.
zenten

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*possibly unclear argument deleted*

EDIT: Alright, what I'm trying to say is that it would be helpful to define j a little better. I'm assuming that you meant 0 is not an element of the Joshua system, but this was not made clear.

Even if this system turns out to be consistent and what not, while interesting, I don't see how it pertains to the original question if you are going to exclude 0.
aguacate

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Okay, zenten, confirm or deny the following properties of the joshua system.

1. a + b = b + a
2. a + (b + c) = (a + b) + c
3. 5 - 3 = 2
Buttons

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I thought about the joshua number system a bit more and here is what I've come up with. The definition of j requires one of these two cases:

1) 1 - 1 != 0

2) 1 - 1 = 0 but the j system doesn't include 0.

Please note that in the examples the op gave of extending number systems, at no point in history did, say, 5 - 6 != -1. It's just that -1 was not part of that number system. It was still well defined and what not, it's just that the system was not closed under subtraction Also, you can think of extension of number systems as solutions to equations. For example, extending the naturals to the integers lets you solve x + 5 = 3. The integers to the rationals gives you 5x = 3. Going to the reals gives you x*x = 2. going to the complex numbers gives you x*x = -1.

1) If the j system is such that 1 - 1 != 0, the j system is going to need a new set of axioms. Also, redefining addition/subtraction operations? I honestly don't know enough algebra to say anything more, but this is a huge change.

2) If, on the other hand, the j system is case #2, then j is a number outside of the set of numbers the Joshua system defines. I'm assuming this was meant to be defined as all x st {x = aj for some a in [Reals - {0}]} (or Complexes instead of Reals, I suppose). In this case the set of all joshua numbers is empty. And thus the Silly numbers are merely the complex numbers, minus {0}.

EDIT:
zenten wrote:Assume 2j = j

2(1-1) = (1-1)

(2-2) = (1-1)

(1-1) = j, by definition

But, by defintion (2-2) != j

Therefore, 2j != j.

You cannot say that by definition (2-2) != j. Watch how this backfires in my face in the following proof that 1/2 != 2*(1/4)

Assume 1/2 = 2*(1/4)

1/2 = 2/4

But by definition 2/4 != 1/2.

Therefore 1/2 != 2*(1/4)
aguacate

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aguacate wrote:You cannot say that by definition (2-2) != j.

Sure you can. You can define whatever you want axiomatically, as long as your axioms don't contradict each other. Your "counterexample" points out that such definitions are often useless or absurd, but there's nothing fundamentally wrong with either definition.

Now, there might very well be some contradictions in the joshua system (especially if the answers to my earlier three questions are all 'yes'), but we haven't seen them yet. We need some more definitions. As has been said a bunch of times, we're not using the regular axioms here, so it's time to accept that they don't necessarily hold.
Buttons

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Buttons wrote:
aguacate wrote:You cannot say that by definition (2-2) != j.

Sure you can. You can define whatever you want axiomatically, as long as your axioms don't contradict each other. Your "counterexample" points out that such definitions are often useless or absurd, but there's nothing fundamentally wrong with either definition.

Now, there might very well be some contradictions in the joshua system (especially if the answers to my earlier three questions are all 'yes'), but we haven't seen them yet. We need some more definitions. As has been said a bunch of times, we're not using the regular axioms here, so it's time to accept that they don't necessarily hold.

Thrown in as an axiom is OK. The problem is proving 2j != j by throwing in an axiom. First come the axioms, next come the proofs. If you are modifying the number system as you go, fine, but I got the impression that it was a defense of the current system rather than a reworking.
aguacate

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aguacate wrote:If you are modifying the number system as you go, fine, but I got the impression that it was a defense of the current system rather than a reworking.

Well, it's quite definitely a reworking. As has been pointed out, there's no way to define x/0 without removing a whole bunch of axioms.
Buttons

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So how are you defining... everything? You don't even have a neutral element for addition, this is going to cause a lot of trouble. Addition isn't well defined, you don't even have a Monoid in (S,+). And you want to define multiplication with an inverse for every element? With the distributive law?

I'm not qualified to judge this, I haven't study enough algebra or logic, but I think that you're creating a really weak structure here. And only if it doesn't have any contradictions...
Sorry for my bad grammar and my wrong ortography. I'm Brazilian, never left my country and I am too lazy to go to English classes. They bore me

Spivak

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Woxor wrote:
jestingrabbit wrote:Whatever a*0 is, it has an additive inverse

Why? Clearly if you retain the axiom that all elements of your number system must have additive inverses, then you cannot divide by zero. But I can't think of a proof (of 1/0 being impossible) that doesn't somehow require that 1/0 has an additive inverse. Might be one, but I can't think of it.

There can't be a proof that doesn't rely on that, because there are systems, specifically the real projective line, where 1/0=∞. However, it is a nice property and one you would hope a collection of numbers would have, mostly because you want to be able to cancel things from both sides of an equation.

zenten wrote:
jestingrabbit wrote:(a+bj) + (c+dj) = (a+c) + (|b|+|d|)j unless a = -c and then...???

If a = -c, you would end up with
(a+bj) + (c+dj) = (a + |b| + |d|)j

However, that's assuming we're talking about b and d being real numbers. I'm not sure about what to do if they are imaginary/complex.

I think you'd probably actually want (a+bj) + (c+dj) = (|a| + |b| + |d|)j, or I could prove that j = 2j I think. And then you could use that for a, b, c and d complex too.

zenten wrote:
jestingrabbit wrote:edit: What's wrong with 0 ≠ 0+0? Quite a bit really, it means that you no longer have an additive identity, so that you can't really cancel anymore. That is, you can't prove things like

a+b = a+c if and only if b=c.

How does the proof work again for complex numbers?

Well, if you have that b=c, then, even in your system, that implies that a+b=a+c.

However, if you start with a+b=a+c, then to get where you want to, you want to add the additive inverse of a to both sides ie

a+b=a+c
-a+a+b=-a+a+c
0+b=0+c
b=c

But this is out in your system. For instance, say we want to look at the equation x+2 = 5. Then

x + 2 = 5
(x + 2) + (-2) = 5 + (-2)
x+2j = 3

but then we're stuck, there's no way we can get x by itself. x=3 is still a solution using the equation for addition you agreed with earlier, but... I think there are serious problems with what you want, but there's little that we actually gain by adopting it.

You ought to read up on the field axioms, think about how we manipulate formulae and consider the pros and cons of not working with a given axiom or collection of axioms.

Edit: and btw, we haven't even got up to multiplication yet, and when we do, J will be getting a whole lot bigger. What is j*j?

jestingrabbit

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aguacate wrote:EDIT: Alright, what I'm trying to say is that it would be helpful to define j a little better. I'm assuming that you meant 0 is not an element of the Joshua system, but this was not made clear.

Yeah, that's what I meant. Basically, j fills in the role of zero, but isn't exactly the same.

aguacate wrote:Even if this system turns out to be consistent and what not, while interesting, I don't see how it pertains to the original question if you are going to exclude 0.

Well, it's not fully what I'm asking for, since I realized from the posts and working it out that the reason I can never have a proper 0/0 is that using zero removes information, that can't be regained. So if I replace 0 with something that does retain the information, then I can do stuff like (2-2)/(1-1). Which would equal 2.

buttons wrote:1. a + b = b + a
2. a + (b + c) = (a + b) + c
3. 5 - 3 = 2

Confirm 1 and 2, deny 3 (I think), as 5-3 = 2 +3j.

Thrown in as an axiom is OK. The problem is proving 2j != j by throwing in an axiom. First come the axioms, next come the proofs. If you are modifying the number system as you go, fine, but I got the impression that it was a defense of the current system rather than a reworking.

I haven't written out anything formally yet, since I haven't done this type of math in at least five years, and I'm a bit rusty. Plus as I said I never had defined a number system before.

jestingrabbit wrote:I think you'd probably actually want (a+bj) + (c+dj) = (|a| + |b| + |d|)j, or I could prove that j = 2j I think. And then you could use that for a, b, c and d complex too.

You're right, I do need that.

jestingrabbit wrote:However, if you start with a+b=a+c, then to get where you want to, you want to add the additive inverse of a to both sides ie

a+b=a+c
-a+a+b=-a+a+c
0+b=0+c
b=c

But this is out in your system. For instance, say we want to look at the equation x+2 = 5. Then

x + 2 = 5
(x + 2) + (-2) = 5 + (-2)
x+2j = 3

So you end up with b + aj = c + aj. What's wrong with that?

jestingrabbit wrote:but then we're stuck, there's no way we can get x by itself. x=3 is still a solution using the equation for addition you agreed with earlier, but... I think there are serious problems with what you want, but there's little that we actually gain by adopting it.

Hey, I'm still working on defining it, and finding out (and proving if so) that it is internally consistent. Convincing others to use it, or even finding out if there's any use for it comes later

You ought to read up on the field axioms, think about how we manipulate formulae and consider the pros and cons of not working with a given axiom or collection of axioms.

I'm finding it difficult to grok field axioms from the online sources. Do you have any online or in print advice on things to learn from?

Edit: and btw, we haven't even got up to multiplication yet, and when we do, J will be getting a whole lot bigger. What is j*j?

I'm gonna go with this for now:

J*j = (1-1)*(1-1)
= 1 -1 -1 +1
= 2 - 2
= 2(1-1)
= 2j
zenten

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zenten wrote:Confirm 1 and 2, deny 3 (I think), as 5-3 = 2 +3j.
Good catch. (3) would've ruined a lot of your claims.

Interesting definition of multiplication. I'm not sure how this helps your original goal, though, because as far as I can tell, our multiplication is now

which, unless I'm mistaken, implies that 1/j is still undefined, since (a+bj)*j cannot be 1.
Buttons

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Interesting definition of multiplication. I'm not sure how this helps your original goal, though, because as far as I can tell, our multiplication is now

which, unless I'm mistaken, implies that 1/j is still undefined, since (a+bj)*j cannot be 1.

Don't we just need ((aj)^-1)*j = 1?
zenten

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zenten wrote:Don't we just need ((aj)^-1)*j = 1?

Well, sure, but what's (aj)^-1? That's hardly defined given what we've seen so far, and assuming multiplication works as above (that is, the product of anything with a j is still going to have a j in it), I don't see how you'd construct it.
Buttons

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Ok, I figured it out.

J*J = 2j

Therefore, 2j/j = j

Which seems to imply that j/j = 1

And of course j/j = j*(j^-1) = 1

Does that make sense?
zenten

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zenten wrote:Ok, I figured it out.

J*J = 2j

Therefore, 2j/j = j

Which seems to imply that j/j = 1

And of course j/j = j*(j^-1) = 1

Does that make sense?

No, that makes no sense at all. Here's how it breaks down. If there is such a thing as 1/j such that j*(1/j) = 1 then

j = j*1 = j*(j*(1/j)) = (j*j)*(1/j) = (2*j)*(1/j) = 2*(j*(1/j)) = 2*1 = 2

Seriously man, I think you need to take a few steps backwards and think about what's going on here. j must be a very small thing. Yet, when I multiply it by itself I don't get something extremely small, I get something that's bigger. This is a problem.

Also, I have absolutely no idea how you come to the conclusion "Which seems to imply that j/j = 1" from the equation 2j/j = j.

jestingrabbit

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Ok, does j*j = j work?
zenten

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Only if you want j=1. Either way though, it seems right to me that j*j = 2j is correct.

And when you have an element such that e*e=e you call it an idempotent. You always want 1 to be an idempotent, and your 1 is. With j*j=2j, you get that (1/2)j is an idempotent. Except for 1, they're not the sort of things that you want to have a multiplicative inverse ie 1/e shouldn't make sense except for e=1. If it does make sense then you can pretty quickly prove that e=1.

This idea of yours reminds me of the split complex numbers to an extent,
http://en.wikipedia.org/wiki/Split_complex
and also the dual numbers
http://en.wikipedia.org/wiki/Dual_numbers
though in both those places, you have a 0. Zero is a popular number for people to include. Most people like their zero. Even people who want to define 1/0 (which is where your query started)
http://en.wikipedia.org/wiki/Real_projective_line
like 0.

Seriously man, I just don't think you know enough about zeroes to decide whether you want them around or not. If you educate yourself about zeroes, get used to living with them on a day to day basis, the few things that zeroes aren't good for, like division, will be dwarfed by all the things that zeroes do that enrich our day to day lives. Reconsider your zeroist tendancies, for your sake, and for the sake of all the innocent zeroes around you everyday.

jestingrabbit

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What this thread is obviously getting at is: it's really easy to define division by 0 to mean something. What's difficult is not trashing the rest of your number system in the process.
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Strilanc

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A little late, but...

Yeah, the others are all correct. The essential difference between, say, extending your number system so that x^2 = -1 has a solution for x and extending it so that 1/0 = x has a solution for x is that, in the first one, you can create a structure that defines it and still obeys the field axioms. The second doesn't allow that (and it can be proven). I'm not sure of exactly which axioms are broken (more than likely, it's different ones depending on how you define it), but you'll never be able to get both addition and multiplication to work 'like they should' again.

Xanthir
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I agree with alky. If we do manage to come up with an internally consistent algebra on the complex numbers, or some subset of the complex numbers, normal operations between non-zero numbers would very likely not resemble what we are used to working with. The thought of such a number system is theoretically interesting; however, I doubt any use in application (though, who says we need to be able to apply math?) certainly nowhere near as useful as the integers, rationals, reals, or complex numbers, or any suitably chosen subset.

To stray slightly off topic for a second, i think everyone is forgetting that Bruce Schneier can divide by zero...

bonder

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