Xerxes

So, anyway. Returning to the OP. First as has been said, more-or-less, if a function is bijective, you may assume it has an inverse. So what's a bijection? Here. Let f: A --> B. If, for all z in B there exists x in A such that f(x) = z, the function f is said to be surjective. If for some x and y in A, f(x) = f(y) in B implies x = y in A, f is said to be injective. A function that is both injective and surjective is said to be a bijection. How to "determine" bijectivity? One way would be to graph your function, i.e. x vs f(x). If any "line" passing through the +/- f(x) axis and parallel to the +/- x axis intersects the curve f(x) vs x at least once and no more than once, then you have your bijection, an invertible function. Micky Mouse? Of course, hope it helps.

Good. In which case, the statement that "it necessary that [math] f(g(x)) = g(f(x) = x [/math]" is ...ahem.. misleading.

I'm confused. In your set-up you had [math]f: A\rightarrow B, g: B \rightarrow A[/math]. This implies [math] x \in A, f(x) \in B[/math]. If [math] g\circ f = id_A[/math] then [math]g(f(x)) = x[/math] as you say. But if [math]f\circ g = id_B[/math] how can [math]f(g(x)) = x[/math]? This implies [math]x \in B[/math] also, and that A = B. Is this what you meant? Wouldn't it be clearer if you said for all [math] x \in A, y \in B, g(f(x)) = x, f(g(y)) = y[/math]? Certainly I see no requirement that in general [math]g(f(x)) = f(g(y))[/math] Or am I missing the point?

Dammit, man! I've admitted my errors, what more do you want? My wallet? My wife? (Dream on) Who is Hatten, by the way?

Woops! yes, you are right, silly me. But this doesn't alter the general truth of the rest of my post, I believe (your PNG doesn't show it): for arbitrary x, the values of sin(x) and cos(x) cannot be used to recover x unless one restricts the codomain of arcsin and arccos in the way I suggested. Which is all totally irrelevant to the present thread. Ignore me.

Are you sure? cos(0) = 1, what's cos(180)? Working in radians I find that [math]\cos(0) = \cos(\pi) = \cos(2\pi) =1[/math]. Likewise [math]\cos(\tfrac{\pi}{2}) = \cos(\tfrac{3}{2}\pi) = 0[/math], surely? More generally, for all x, sin(x) and cos(x) are in [0,1]. Setting arcsin as the inverse sine function, likewise arccos, as I recall it is convention to set [math]\arcsin(\sin(x)) \in [-\tfrac{\pi}{2},\tfrac{\pi}{2}][/math] and [math]\arccos(\cos(x)) \in [0,\pi][/math]. These are called the "principle values" of arcsin and arccos, respectively. I'm way, way rusty on trig, but this as I rememeber it.

D.H.: Maybe you missed this bit I never meant to imply that the reals are merely a set, monoid, group, etc, but they most certainly can be viewed as all these things. I respectfully invite you to open any text on Abstract Algebra, and you will see what I mean. And in your text you will see that the field over which a vector space is defined is itself the prototypical vector space. Or, if you have no such text, maybe you'd like to look here: http://en.wikipedia.org/wiki/Real_number or here http://en.wikipedia.org/wiki/Examples_of_vector_spaces just for starters I am not wrong, don't be so rude.

You do realize I was horsing around, don't you?

Why does everyone get this Pauli quote wrong? It should be "....not even eligible to be wrong" Apparently he (P.) was a great wit; my favourite Pauli-ism was in reference to Einstein's (and other's) attempt to find a Unified Field Theory: "What God has torn asunder, let no man join" (a reference to the Anglican Church marriage ceremony, I believe). Right. I'm starting now. I'll call it BEN EXPLAINED+++++. OK with you?

You fear his daughter might marry your son (or v.v)? Phhht. Unlikely. And if he votes the way that you would prefer that he did not? So? It's democracy, right, live with it.. (Sorry mods, my wife's birthday, I'm pissed)

Well, then, what would they be calling "it"? And, I'm sorry, I really don't understand what you mean by "tweaking the distributive property". Do we have a choice here? Surely, distributivity is what it is (i.e. no tweaking allowed)?

Oh dear-- let me tell you all a sad story A couple of years ago some very brave folk started up a science WiKi called WiSci. Even though we had few editors, we were building, and it could have been really good. But, sadly, all editors (except me) chose to spend the next 3 months composing a letter of rebuttal to the Intelligent Design geeks. Of course they never responded. It killed the WiKi, which I was sad about, but which I did warn about. It seems to me that this sub-forum might be in danger of going down the same road with all this "rebuttal of Farsight" crap. Who cares what he thinks? I mean who really cares? (Hands up, please!)

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 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.

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) 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)

Hmm... It obviously doesn't come across, but I am a professional scientist (not in physics, though, self-evidently) , so I do understand this procedural stuff

Hmm. There's a slight discrepancy here. YT thinks that students might use this site as a research tool (which I seriously doubt). For example; Ben and Severian, among others, assume we all know what a W, Z and Higgs boson is (are?), without offering the tutorial. What is a trawling student to do without the sort of information such a tutorial might provde? In my opinion, the time would be much better spent teaching morons like me about this stuff, rather than just preening each other's feathers with esoteric terms. Anyway,it looks like this sub-forum is yet another scrapping school-yard. One of too many, in my opinion. I'll stick with math; I think I may have more chances there.

I must have made my point very poorly. YES, refute ideas, equations, whatever, in the thread in which they were offered. I am merely objecting to starting up two whole new threads, bearing a member's name with (it appears to me, at least) the sole purpose of humiliating him Again, I don't follow this argument. Am I dim? (don't answer!) Sorry to break it to you, chum, this site is pretty low, low down on my student's crib list Well, I see I am about as popular as burnt toast here. Ah well.

Ben, I'm addressing this to you, as you started the two threads I objected to, but I believe it applies generally No, they don't, why should they? If it ain't science, it ain't on their patch. Period. No of course it isn't. The "bullying" comes in when you, Ben, open threads specifically targeted in a highly derogatory way, against another forum member. Do I agree with Farsight? Probably not (I say probably as I don't know enough physics to be completely sure). Does he have a right to post his so-called theories? I hope so. But more to the point.... Have you, or anybody else, shown some non-expert like me why he is wrong? I see a lot of vague comments about Lorentz invariance, WW coupling at 1 KeV (what the eff is that, not that I care; please don't tell me). I think you unfairly underestimate the ability of casual visitors here to detect fringe when they see it. I also think you overestimate the influence that fora such as this have in the real world. Like, you think the Southern states re-introduced Intelligent Design in schools by reading stuff on a ChatRoom board? I don't Whichever way, who the eff cares? I don't get it. YT, and others, perhaps, said that "ignoring isn't an option". I fail to see why not. Ignored people generally lose heart in my experience. As for "cancerous memes" (what's a meme on a Friday night?), sorry I find it grotesque.

Ya, sure, that's right, I agree. But my point is this: Trash Farsight in his own threads (or in his responses to other people's) if you disagree, but why waste bandwidth starting multiple new threads specifically for that purpose? It looks suspiciously like bullying to me, which I think the forum should oppose. Certainly I find it personally distasteful (so there...)

I wish I were Farsight - so much loving attention! Two (at least) whole threads to trashing him!. Seriously, I don't get it. Who cares? Like who really cares what Farsight thinks?

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]

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

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

Yeah, OK, I apologize too for my foul-mouthed outburst. I guess I'm a bit touchy on the subject, that's all. Sorry all round for spoiling the atmosphere here.

Correct me if I'm wrong, but I had sort of understood you to be a physicist? Is this the way physicists interact? Perhaps? It was in such bad taste that you are unlikely to be taken seriously (at least by me) on any forum of your choice. So f.u.c.k you, you are an ignorant and insensitive moron.