Negative x Negative = Positive?

[Hypercube]

Does anyone know why a negative times a negative equals a positive? I have never been able to understand the reasoning behind that.

[Tom Mattson]

Why yes, I posted a complete proof of that just one month ago!

 

http://www.scienceforums.net/forum/showthread.php?t=26984

[Bignose]

Besides Tom's proof (which was very nice), I really, really liked the tree's visualization of the concept. Multiplying by a negative is a reflection, so multiplying a negative by a negative is in effect the reflection of a reflection which would be the original back again.

[phyti]

When numbers are directed (pos and neg, relative to zero), they are treated as vector. Historically they were not viewed as such, and a makeshift rule was used, "the product is pos if like signs and neg if unlike signs".

Technically the neg number was the first use of a complex number.

In that context, j is a 180 deg rotation, so jx = -x, and j(-x) = x.

Eg. (-x)^2 = (jx)^2 = j^2(x^2) = x^2.

[Bignose]

Technically the neg number was the first use of a complex number.

 

??????

 

[math]-1 \neq \sqrt{-1} [/math]

 

why would you think that?

[phyti]

Complex analysis is a more recent concept.

As stated, subtraction is not presented as a complex operation.

Think of j as i^2.

[BenTheMan]

[math]e^{0} = 1[/math]

[math]e^{\pi i} =-1[/math]

[math]e^{2\pi i} = 1[/math]

[geoguy]

I never had a hard time understanding why the result was a positive. But I had difficulty understanding why a negative times a positive couldn't be either a negative or a positive answer. This difficulty isn't undertanding the proof so much as in visualizing the concept.

[Xerxes]

Multiplying by a negative is a reflection

In that context, j is a 180 deg rotation,

Well which is it chaps, make up your minds - rotation or reflection? For the record I contend it's neither, but that's by the way.

so jx = -x, and j(-x) = x.

Eg. (-x)^2 = (jx)^2 = j^2(x^2) = x^2......Think of j as i^2.

Wow, j = -1? Why not just say so, then? You don't need imaginary units for that.

[math]e^0 = 1[/math]

[math]e^{\pi i} =-1[/math]

[math]e^{2\pi i} = 1[/math]

Very pretty' date=' I agree, part of my favourite identity, if you must know. But how is it relevant? Surely you're not suggesting that, since [math']e^{\pi i} = 1[/math] and [math]e^{2\pi i} = -1[/math] this somehow "proves" that 2(-1) = 1?

[timo]

Surely you're not suggesting that, since [math]e^{\pi i} = 1[/math] and [math]e^{2\pi i} = -1[/math] this somehow "proves" that 2(-1) = 1?

He surely didn't mean that since you copied his premises incorrectly. Other than that, if you agree that [math] -x = x e^{i\pi} [/math] you can indeed make a similar statement (I strongly assume saying 2*(-1)=2 was just another tyop of yours):

[math] 2 \cdot (-1) = 2e^{0 i \pi} \cdot 1 e^{1 i \pi} = 2 e^{(0+1)i \pi} = -2. [/math]

[Xerxes]

He surely didn't mean that since you copied his premises incorrectly.

Ouch! And me saying it's part of my favourite identity? So sorry folks.

]Other than that, if you agree that [math] -x = x e^{i\pi} [/math]

Trivially, yes, of course. It's been a while since I looked at the internal structure of the Euler identity, but I suspect it makes use of the property which is (unaccountably) causing difficulty to some here. Like, assuming the result you want to prove??

(I strongly assume saying 2*(-1)=2 was just another tyop of yours):

Aaargh, Even worse - I said 2(-1) = 1!! Will I shoot myself now, or later?

 

*Blush, blush* Sorry for any confusion

[BenTheMan]

hypercube---

 

what else would a negative times a negative equal?

[the tree]

Well which is it chaps, make up your minds - rotation or reflection?

Two transformations have that have the same result may as well be the same transformation. I said reflection because I was thinking of a 1d number line as I said it, but neither description is less accurate than the other.

what else would a negative times a negative equal?

Another negative. If we decide to do away with the associative property of multiplication and change the distributive law a little so that x(y.z)=xy.xz then it'd be easy to prove, even though you'd be working with some extremely esoteric maths, it'd still be valid.

[ADAM341]

Besides Tom's proof (which was very nice), I really, really liked the tree's visualization of the concept. Multiplying by a negative is a reflection, so multiplying a negative by a negative is in effect the reflection of a reflection which would be the original back again.

 

Yeah thats a great reply indeed with the suggestions

 

to the above resource.Well I know that even negatives do make a positive

 

and the theory goes on and on.....

[Xerxes]

Two transformations have that have the same result may as well be the same transformation.

Um, maybe you mean "two consecutive transformations that deliver the identity transformation"? Then I would agree

I said reflection because I was thinking of a 1d number line as I said it, but neither description is less accurate than the other.

 

yes it is. Rotation is an operation only available to you in a space of dimension greater or equal to two. Obviously the real line R disqualifies here.

 

Look. I admit that your description of "multiplying a negative by a negative" as a double reflection has some visual merit, but I'm not convinced that this can be true in general.

 

How, for example, under this model, would you describe multiplying a positive by a positive (or a negative by a positive)? How can these be reflections, or any other sort of transformation, for that matter? I think you are going down the wrong road here. Sorry

[BenTheMan]

Another negative. If we decide to do away with the associative property of multiplication and change the distributive law a little so that x(y.z)=xy.xz then it'd be easy to prove, even though you'd be working with some extremely esoteric maths, it'd still be valid.

 

Well if you get rid of the associative property, doesn't it mean that the negative integers would no longer be a group?

[Xerxes]

Well if you get rid of the associative property, doesn't it mean that the negative integers would no longer be a group?

I assume this question is rhetorical? otherwise I can give chapter and verse....what gives here? Do we really need a group theory tutorial?

[BenTheMan]

ummmm I thought that the first property of a group was associativity of the group operation, in this case multiplication? a.(b.c)=(a.b).c? Am I completely wrong?

 

You'll forgive my maths, I hope. And perhaps the post was poorly thought out/worded, because negative integers weren't a group to begin with.

[Xerxes]

Well, associativity of the operation is certainly a prerequisite for a set to qualify as a group. I cannot convince myself that [math]\mathbb{Z}^-[/math] isn't associative under arithmetic multiplication: [math]-1 \times(-2 \times -3) = -6 = (-1\times-2)\times-3[/math]. But it's not closed under this operaration, for the reason given in the OP; [math] -x\times-y \notin \mathbb{Z}^-[/math]. Maybe this is what you meant?

 

It is closed under arithmetic addition, however, which is also associative, and there is an inverse -(-x) = x. The issue of the identity might prove somewhat controversial, however; is [math]0\in\mathbb{Z}^-[/math]? If you are willing to concede it may be (I am agnostic on this), then [math]\mathbb{Z}^-[/math] will be a group.

 

But the OP was asking about multiplication, so I'm talking tosh!

 

I think Ben had the most useful comment - what else could it be; consider [math]-1\times-1 = (-1)^2[/math] Suppose [math] (-1)^2 = -1[/math]. Then [math]\sqrt-1 = -1 ==> -1 = i[/math] which is false by definition, therefore [math] -1\times -1 \neq -1.[/math]

[the tree]

Um, maybe you mean "two consecutive transformations that deliver the identity transformation"? Then I would agree

What? No. I meant what I said.

 

Well if you get rid of the associative property, doesn't it mean that the negative integers would no longer be a group?

If it does then what'd be the problem? Does all logic and all proofs fall apart the moment you move away from groups?

[Dave]

If it does then what'd be the problem? Does all logic and all proofs fall apart the moment you move away from groups?

 

No. But group theory provides us with some very useful existing results (isomorphism theorems, Sylow's theorems, to name just a few). So if you want to do something useful with your new objects, you're going to have to start from the ground upwards.

 

[This rest of the post is not necessarily related to the tree in particular, or the quote above].

 

The amount of sarcasm and use of rhetorical questions in this thread is way over the top and completely unnecessary. If you must banter over such issues then try not to offend everybody when you press the post button. Acting high and mighty simply makes you look unnecessarily arrogant.

[Xerxes]

What? No. I meant what I said.

I which case, reiterate my earlier point; [math]\mathbb{R}[/math] is a 1-space, rotation makes no sense here, reflection does. This was your point, and I am supporting it! (well, less than half-heartedly, I confess; I said why in an earlier post)

 

If it does then what'd be the problem? Does all logic and all proofs fall apart the moment you move away from groups?

The point, as I understood it, being somewhat related to this.

 

[math]\mathbb{R}[/math] can be viewed as:

 

a Set

a Monoid

a Group

an Abelian Group

a Field

a Ring

a Vector Space

a Topological Space

a Manifold

 

and so on. When trying to make proofs in [math]\mathbb{R}[/math] we have to be clear under which of the above we are working, all have different axioms, and therefore different allowable proofs.

 

Here's a simple example: take the case of distributivity [math]a(b+c) = ab+ac, [/math] where [math]a,b,c \in\mathbb{R}[/math]. Here [math]a \in\mathbb{R}[/math] is an element in a field, and [math]b,c\in\mathbb{R}[/math] are elements in an abelian group (to be complete let's just say that the pair R as a field and R as an abelian group are referred to as a Vector Space; well, there are additional axioms, of couse. In order to avoid this sort of annoying ambiguity, in math, the fields "over which" vector spaces are defined are usually referred to as F or K)

[the tree]

OK I never said that playing with the axioms that make R the nice little ordered field that it is would be a good idea, just that it could be done.

 

You would no longer be working with the Reals and like I said it'd be extremely esoteric and it'd probably be completely pointless. However, so long as you take the care to define axioms, you can construct a logical proof from them.

[Xerxes]

OK I never said that playing with the axioms that make R the nice little ordered field that it is would be a good idea, just

that it could be done.

I'm sorry, you lost me there. R is a field, yes. It has a total order, yes. But as I tried to explain in my last post, this is by no means the only way of thinking of R

You would no longer be working with the Reals and like I said it'd be extremely esoteric and it'd probably be completely pointless.

Again, I don't follow you. R is R, with the properties I listed earlier. All elements in R are real, by definition. So how is one "no longer working with the Reals"?

 

What's "esoteric" and "pointless" about considering R as a vector space, a topological space etc. I don't get it.

[the tree]

I'm fairly sure if you were to tweak the distributive property in the fashion I suggested, there'd really be no-one calling it the Reals. You'll note I said that was pointless, and I didn't mention other ways of thinking about the Reals.