What's the difference between 0 and -0?

[qemist]

Just to give an example, consider non-commuting addition. In that case, we might want to introduce a left identity and a right identity. Specifically: 0_L + a = a = a + 0_R for all a that are not identities. If that is the case, and we want to preserve at least associativity, then we define inverses thusly. a + (-a) = 0_R, (-a) + a = 0_L. That way, a + b + (- = a. Note that (-(-a)) is not a.

This brings up the question, what is 0_R + 0_L? I can't find any reason not to define it to be 0_R. Consequently, 0_L = (-0_R). This leads to an algebra where -(a + = (- + (-a), which may look odd, but is entirely self-consistent.

Finally, as means of simplifying notation, I call 0_R simply 0, and 0_L = -0.

Lets look at your group-like structure then with distinct left and right identities, and associativity. I feel it's a bit of an abuse of notation to use (-) for both the right and left inverses, because with two distinct identities a can have distinct left and right inverses. So to make things simpler to write out, I'll use juxtaposition as the operator and call R the right identity and L left identity. Then from what you used to define the structure:

La = a = aR

aa^-R = R

a^-La = L

So far this is all the same as what you wrote. We can't know what RL is from the properties we have here, it could be R or L or any other element since neither are acting as an identity in this case. (We could define it to be R as you did, but that's redundant because the rest of the proof shows that RL = R = L) So lets look at LR.

LR = (a^-La)R = a^-L(aR) = a^-La = L

LR = L(aa^-R) = (La)a^-R = aa^-R = R

Therefore L = R, which is a contradiction.

So a group-like structure (I'm assuming you mean any structure with a set and a closed binary operation on that set) with inverses, associativity, a left identity, and a right identity (doesn't matter whether the left is the same as the right) as described above by you is actually a group, and a group has a unique identity.

[K^2]

Good catch. And that doesn't even require all inverses to exist. So long as there is a single element with both the left and the right inverses, the identities (and inverses) are identical.

It leaves me wondering how much more I need to strip down a group before I can get it to support multiple distinct identities. But it's probably moot. I can certainly strip it down to magma with identities, but then calling second identity -0 is absolutely arbitrary.

I concede. There is no good reason to have a -0 notation.

[ZetaX]

0_L + 0_R = 0_L, naturally. It follows from all the other definition. And no, identities don't have to act as identity on other identities. Not in algebras that do not have commutative property. Having only left/right identity (or both) and having special rules for these is pretty common in math. Consult this table.

You really only need the definition of "identity" to conclude that left ones are equal to right ones if both exist.

For e to be a left identity means to satisfy e°x=x for all x. For e' to be a right identity means to satisfy x°e'=x for all x. You can't just exclude this properties in the cases where x is an "identity"; how the heck would that definition even work¿ You would define identity using the word identity, making a circular definition.

The list you gave has not a single such example, too; only cases where only identities on one side exist. It even says: "But if there is both a right identity and a left identity, then they are equal and there is just a single two-sided identity".

Proof: If you have such e and e' then automatically e = e°e' = e', both equalities by definition of the respective identities. And if there is another one on any side, the same shows it's equal to e.

Edited October 28, 2014 by ZetaX

[K^2]

You can't just exclude this properties in the cases where x is an "identity"

Sure you can. Every element in a field has a multiplicative inverse, unless it's additive identity. When definitions don't work for one particular element, it doesn't hurt the generalization. We make one special exception and move on.

And it's not a problem that exception is itself, either. For example, I can point out that, "There exists a real number, such that any number, except itself, to the power of that number is 1." You can tell me right away what that number is, and the fact that definition includes itself isn't a problem.

The problem here is specifically with inverses, as qemist pointed out. I can have a magma with any number of distinct identities. It's not useful, but it's not a contradiction by itself. Contradictions happen when we try to assign some useful properties to the binary operation in question.

[Beowolf]

I like how this question had so many different correct answers.

[ZetaX]

Sure you can. Every element in a field has a multiplicative inverse, unless it's additive identity. When definitions don't work for one particular element, it doesn't hurt the generalization. We make one special exception and move on.

That one is completely different. 1 is a multiplicative identitiy, even for 0. But even if not, that would not be a problem, as you would have defined 0 (additive identity) before you define 1 (multiplicative identity; only field without 0 should be a group); and we are talking about two structures here, not just one. As I said, an identity is one for all elements in question; that's the definition.

All you could do is make a definition such as "a left-neutral element is one such that e°x=x for all x" and "a right-neutral is one such that x°e'=x for all x except left-neutrals". That would be a correct definition, but it would a) not be the accepted one, be asymetrical (calling the latter a "almost-right-neutral element" would make more sense, for example).

Or you go with "a set of identities is a set S of elements such that s°x=x for all s in S and all x not in S". Then every single real number is an identity if you just wish so: set S = IR [the reals]; then clearly s°x=x for all x in IR-IR aka the empty set.

And it's not a problem that exception is itself, either. For example, I can point out that, "There exists a real number, such that any number, except itself, to the power of that number is 1." You can tell me right away what that number is, and the fact that definition includes itself isn't a problem.

That definition would indeed not be problematic, but it is again of a different type. And by the way, 0^0 = 1; that's not just some random convention, but the one accepted by mathematicians.

It works very well, for example: a^b is the cardinality (i.e. the number/amount of elements) of the set A^B := {f:B->A} of maps f from a set B with b elements to a set A with a elements. And there is indeed exactly one map from the empty set to itself.

And when was the last time you wrote a polynomial/power series not as sum a_i x^i, but as a_0 + sum a_i x^i instead¿

The problem here is specifically with inverses, as qemist pointed out.

He still uses that a°e=a for a right identity e (similiar for left); the same reasoning I used. I used the defining propterty of identity; he used that, plus associativity plus inverses of some kind. Yet you accept his, which uses everything I use, but not mine, which works in a more general case by the very same argument.

Do you really agree with his proof more because he wrote a (general element) instead of e (neutral one of some kind)¿

I can have a magma with any number of distinct identities. It's not useful, but it's not a contradiction by itself. Contradictions happen when we try to assign some useful properties to the binary operation in question.

So where exactly do you disagree with both my proof and Wikipedia¿ That magma can only have distinct identities if they are all on the same side. unless you now also doubt transitivity of "=".

Edited October 29, 2014 by ZetaX

[K^2]

only field without 0 should be a group

A field with a zero is also a group. Two groups, actually. I'm not sure where you are going with this whole thing.

That one is completely different.

You keep saying that. Yet you aren't providing any reasons for why it's different.

And by the way, 0^0 = 1; that's not just some random convention, but the one accepted by mathematicians.

It is not. It is useful as a shorthand in a lot of places, just like sqrt(-1) = i is a useful shorthand. Neither is actually true. Map x^y has an undefined point at (0, 0), and you cannot avoid that.

And when was the last time you wrote a polynomial/power series not as sum a_i x^i, but as a_0 + sum a_i x^i instead¿

How about every time I needed to write code that doesn't crash because of a numerical singularity?

Yet you accept his, which uses everything I use, but not mine, which works in a more general case by the very same argument.

I accept his proof, because it is a proof. A strict mathematical proof. You have not constructed one. You simply keep insisting that identity acting on identity must give identity, when the whole conversation is, "What happens if it is not?" You seriously don't understand why it is silly? qemist used all of the definitions set and has derived a contradiction. You introduce contradiction by bringing in a contradicting assumption. I can prove anything the way you "prove" it.

So where exactly do you disagree with both my proof and Wikipedia¿ That magma can only have distinct identities if they are all on the same side. unless you now also doubt transitivity of "=".

Erm... I'm not sure which Wikipedia you are reading, but the version I'm getting states, quite correctly, that Magma is simply a map S x S -> S with no other requirements. It doesn't even have to have identity elements. But if it does, it can have any number of distinct elements that act as identity on anything other than each other. There are no axioms this can contradict, because there are literally no other axioms.

[rtxoff]

Dang. (10chars)

You must spread some Reputation around before giving it to K^2 again.

[sal_vager]

I don't know if anyone here's spotted this yet, but when you reach your node in KSP, there's a pause when it reaches zero.

KSP is counting -0

[ZetaX]

A field with a zero is also a group. Two groups, actually. I'm not sure where you are going with this whole thing.

I was talking about the multipicative one, obviously. Please stop trying to "contradict" things that way, it is just tedious to repeat the whole terms every sentence.

You keep saying that. Yet you aren't providing any reasons for why it's different.

I did. I gave both examples and proofs. You instead gave only examples that are flawed (the 1 is an identity even for 0; and the linked wikipedia list has not a single example supporting what you say).

It is not. It is useful as a shorthand in a lot of places, just like sqrt(-1) = i is a useful shorthand. Neither is actually true. Map x^y has an undefined point at (0, 0), and you cannot avoid that.

Name a definition of a^b that does not simply extend to 0^0 without any problem. The map x^y is simply 1 at (0,0); what you wan to say is that it is not continuous there. But that does not make the definition (it is one) invalid.

How about every time I needed to write code that doesn't crash because of a numerical singularity?

Then you are obviously not using a power series; so that's not an answer at all. Or you seriously think that x^0 will evaluate wrongly if x is close to 0, which it won't; at least if you set 0^0=1. Clearly it fails if you consider 0^x instead, but that's again just the lack of continuity (and smoothness), but even more, it is not the example I gave, thus: stop using strawmen.

I accept his proof, because it is a proof. A strict mathematical proof. You have not constructed one. You simply keep insisting that identity acting on identity must give identity, when the whole conversation is, "What happens if it is not?" You seriously don't understand why it is silly? qemist used all of the definitions set and has derived a contradiction. You introduce contradiction by bringing in a contradicting assumption. I can prove anything the way you "prove" it.

What. I used the definition of "identity" (look up any math page of wikipedia). I even repeated what that definition means.

And I will repeat qemist's (he did by the way, like you, not state all his axioms in full, thus no, it is not as precise as mine; but I don't even claim his proof is wrong, just that it uses more properties than necessary) proof for you, and if you think there is an error or he did different:

A: Assume we have a left neutral L: LA = a for all a.

B: Assume we have a left neutral R: aR = a for all a.

C: Assume the structure is associative.

D: Assume we have left-inverses a^-L (in regard to L): a^-Ra = L for all a.

E: Assume we have right-inverses a^-R (in regard to R): aa^-R = R

F: Assume equality is transitive (I really don't think I need to mention this...).

Idea: evaluate LR in two ways (I will supress mentioning associativity, and instead write no bracket at all):

qemist's version:

1) LR = Laa^-R = aa^-R = R

[using axioms D, A, D in that order]

2) LR = a^-LaR = a^-La = L

[using axioms E, B, E in that order]

Thus L = LR = R, so by F we have L=R.

Now in comparision, my argument; poin out any presumed errors explicitely!

1) LR = R [use axiom A, with the special case of a = R]

2) LR = L [use axiom B, with the special case of a = L]

Again get L = LR = R and conclude by F that L=R.

Note that C,D,E were not used at all. I even only used A and B for a being the neutral elements.

Also, it's your turn to at least give SOME definition. You act like mine is "wrong" despite it being the accepted one in mathematics, and without you ever demonstrating any problems with that. You just calculate as you want, sometimes following a rule/axiom and sometimes claiming one is not allowed to use it. That's definitely no how it works.

You are, as gpisics behaviour alraedy demonstrates (*sigh*), on a road to an argument like "what if I can transmit information instantly", but ignore that relativity does make that "instantly" very observer dependent. Your construct needs to be self-consistent; or, and it's not even that; at least have a definition. You can change the rules, but you a) can't break logics itself, need to propery say what it means. You did neither.

Erm... I'm not sure which Wikipedia you are reading, but the version I'm getting states, quite correctly, that Magma is simply a map S x S -> S with no other requirements. It doesn't even have to have identity elements. But if it does, it can have any number of distinct elements that act as identity on anything other than each other. There are no axioms this can contradict, because there are literally no other axioms.

And my proof uses only that and the definition of "identity". Point out an error or stop acting that way.

I never said every magma has one; I only said that the existence of an identity on each side implies they are equal; and then every other identity is equal to that, too. You claimed you can have distinct identities on both sides in a magma, and I proved you wrong; just accept it.

I don't get what is so difficult to understand here. If the proof above still is no explicit enough, answer me the following:

a) Give the definition of left and right identity.

Are we at least using a magma¿

c) Do you accept that = is transitive¿

d) What is not clear or what do you consider faulty.

In all seriousity, the last time I needed to expand down to that level for formality for such a triviality was when I explained groups to 10th graders not making this up; giving lectures to interested students all he time). I knon you are a physicist and would have expected a better mathematical understanding from such (despite the jokes about them).

[rtxoff]

Also, it's your turn to at least give SOME definition. You act like mine is "wrong" despite it being the accepted one in mathematics, and without you ever demonstrating any problems with that. You just calculate as you want, sometimes following a rule/axiom and sometimes claiming one is not allowed to use it. That's definitely no how it works.

You are, as gpisics behaviour alraedy demonstrates (*sigh*), on a road to an argument like "what if I can transmit information instantly", but ignore that relativity does make that "instantly" very observer dependent. Your construct needs to be self-consistent; or, and it's not even that; at least have a definition. You can change the rules, but you a) can't break logics itself, need to propery say what it means. You did neither.

Leave out my forumname in the future, this thread is not about me. Also this thread is not about relativity or FTL comms. As far as breaks of logic are involved i see very clearly who is constantly breaking the logic here.

[ZetaX]

Make me.

Seriously, I will continue to use you as a bad example as long as you continue your behaviour. Also, the example I gave is very close to the error K^2 makes.

You are also still invited to point out actual errors. You just choose not to, or you can't find any; "your argument is invalid" is not a proper argument, so make better ones in the future.

[rtxoff]

So the thread is now about me beeing a bad example and about my behaviour? Seriously, are you taking drugs?

[tetryds]

At this point I know that everyone here is aware of all the issues regarding this damned number, just remember to not divide your manners by it

-0 is not something meaningful on our physical universe, and showing maths about it will not change this fact.

Just remember to keep the conversation civil when discussing about the rest of the implications of such number.

[ZetaX]

I really want to keep it civil, but I would want to add a word for this to be possible: be rational and reasonable. If a argument, especially such a solid one as mathematical proof, is given, you can't just apply your version of wishful thinking.

To make my point clear: I gave a proof in my second to last post (the same I already gave 3 times). Either accept it or point out errors; other things are not accepteble. This is mathematics, where decisions are not made by popular vote (not even a closed circle votes as in the dwarf planet demotion), but by the power of logical conclusions only.

[tetryds]

You forgot that even maths have it's contradictions.

And even though it exists being metaphorical and independent even from the existence of the universe itself, there are many things about it that we may never find out.

But -0 is just a number, like any other.

As this topic is clearly not leading to any useful and interesting conclusions anymore, other than debating on whether one's methodology is correct or biased, closed.

Edited October 29, 2014 by tetryds