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December 11

Functions

I have a question from an Algebra II course. It was assigned as a challenge problem last month, but i would like to know how to do it now in retrospect.

ƒ(x)= 1/x

g(x)= 2x/(x²+16)

Find f[g(x)] and simplify

Simplify the function and describe its domain.

Thanks for the help —Preceding unsigned comment added by 72.73.109.166 (talk) 02:33, 11 December 2008 (UTC)[reply]

If you want our help, you must show us what you've done on the problem so far. StuRat (talk) 03:51, 11 December 2008 (UTC)[reply]

In this case finding ''g(ƒ(x)) seems more like a "challenge problem" than finding ƒ(g(x)), since it has a more interesting bottom line answer that can't be seen without going through the simplification: it ends up being the same as g(x) except that the coefficients "1" and "16" in the denominator get interchanged.

But do be more specific if you want help, since the way you phrased the question makes it hard to do anything except hand you answer without being sure whether your difficulty has been addressed. Michael Hardy (talk) 04:22, 11 December 2008 (UTC)[reply]

f(x)= 1/x means that anything can be written instead of x, f(0)=1/0 which is undefined, so 0 is outside the domain of f. f(1)=1/1 which can be simplified to f(x)=1, f(2)=1/2, f(y)=1/y, f(x+1)=1/(x+1), f(1/x)=1/(1/x) which can be simplified to f(1/x)=x, and so on. So now you can compute f(g(x)), and perhaps the result can be simplified. Bo Jacoby (talk) 07:16, 11 December 2008 (UTC).[reply]

The domain of the function is the same as the domain of f.

Topology Expert (talk) 09:12, 11 December 2008 (UTC)[reply]

I really can't understand how this is not trivial. If you really are stuck, I suggest you search Wikibooks to find a good book on function compositions but to tell you the truth, this is basic algebra.

By the way, are you sure that the instructor called this a 'challenging problem'?

Topology Expert (talk) 11:29, 11 December 2008 (UTC)[reply]

Isn't this a vaguely skeptical question? This problem was assigned one month ago; he may not remember exactly what the instructor said... besides, being one month ago, it is just retrospective, certainly not homework .--PMajer (talk) 15:48, 11 December 2008 (UTC)[reply]

I'm afraid I am only a freshman in high school, just starting functions in an Algebra course. When it was assigned as a challenge problem, we hadn't even seen functions yet. I have gotten as far as 1/(2x/x²+16)

But i am now truly stuck, obviously because my algebra isnt up to par. This is not a homework problem, so I am not gaining any advantage over classmates by getting help here. I just wondered how the problem was simplified. Thanks though —Preceding unsigned comment added by 72.73.109.166 (talk) 00:05, 12 December 2008 (UTC)[reply]

I'll just ask my teacher because Wikipedia wasnt very much help. Thanks —Preceding unsigned comment added by 72.73.109.166 (talk) 00:22, 12 December 2008 (UTC)[reply]

Do you know what

{\frac {\,1\,}{\frac {x}{y}}}

equals? It's a basic rule for reciprocals, if you don't know it, you won't get anywhere. You can probably work it out for yourself, though - what's

{\frac {\,1\,}{\frac {1}{2}}}

(ask your calculator if you don't know)? --Tango (talk) 00:28, 12 December 2008 (UTC)[reply]

Thank you Tango. That did help. I am now past my stuck point. It is basic algebra, with a touch of help from one of you —Preceding unsigned comment added by 72.73.109.166 (talk) 00:48, 12 December 2008 (UTC)[reply]

This is another instance where messing around is getting us nowhere. The OP wanted an answer. Tango’s response was probably the most helpful but still sort of ‘messed around’. I mean you have to know what ${\frac {\,1\,}{\frac {x}{y}}}$ is; you can’t just derive it (you can in fact (in a way using basic group theory) but that is besides the point and people who learn basic algebra generally don’t appreciate ‘derivations’). It got to the point, where the OP will probably not come back and ask a question here ever again. Yes, we helped him in the end but it literally took one whole day to get an answer that is trivial to anyone who knows algebra. Tango’s answer should have been first up without pointless discussion.

So I suggest that before we start losing our jobs, in future, ask the OP whether he/she wants the answer and if so, give it to him/her (along with the method (in the order of method and then answer)). I can just imagine how long it would take for someone to know how many grams constitutes 5kg if he/she were to ask here (if the person who asks this question has already completed school, there is no need to teach him the method because he probably just wants the answer to he can move on with life).

Topology Expert (talk) 10:24, 12 December 2008 (UTC)[reply]

But that person would benefit so much more by being given a link to SI prefix than by being told "5000g". "Give somebody a fish, and they'll eat for a day. Teach someone to fish, and they'll eat for life." Just because you aren't in school any more doesn't mean it's not worth learning stuff. The fact that they've needed to do such a conversion this time suggests it's very likely they will need to again at some point. They can't come to us every time. (This example actually holds some significance for me - a year or two ago my housemate, a computer science student [who really should have known how to do basic arithmetic!], was following a recipe and needed my help to convert 200g into kg because that's what the scales were marked it - I took the time to explain to her about how "kilo" means 1000 and how to divide by 1000 you need to shift the decimal point 3 places and now, I hope, she knows how to do it herself in future.) --Tango (talk) 10:49, 12 December 2008 (UTC)[reply]

I certainly agree with you. My point and advice to all volunteers is: give the answer as well as the method instead of messing around (which I admit, gets me easily carried away). However, first give them the method.

P.S Of course it is worth learning after finishing school (this certainly applies to every mathematician, for example). But there are ignorant people out there who would rather not learn and in that case, there is no point in forcing them to.

Topology Expert (talk) 12:30, 12 December 2008 (UTC)[reply]

The point is that it is for their own good. If you just give them the answer, such people will ignore the method. If you only give them the method they can't ignore it. --Tango (talk) 12:33, 12 December 2008 (UTC)[reply]

Yes, but are we their parents? What about this: Ask them whether they want to know the method. If no, give them the answer. If yes, give them the method and ask them to answer a similar problem (someone can make that problem up). If they answer correctly, give them the answer.

This is probably the best option but I still don't believe that you should force someone to do something that they don't want to; especially when they have specifically requested otherwise.

P.S Most problems on the reference desk are so easy, I would not worry about re-answering them a few times.

Topology Expert (talk) 12:40, 12 December 2008 (UTC)[reply]

Integration problem

I want to integrate (x^2+2x)^2. If I use the reverse chain rule to find (x^2 + 2x)^3/3, differentiating again gives (x^2 + 2x)(2x + 2). So what's the actual antiderivative? I appreciate all help. —Anonymous Dissident^Talk 03:51, 11 December 2008 (UTC)[reply]

The easiest way I see is to expand and integrate each term individually. Never try to reverse the chain rule in the way you did; you were assuming that the derivative of (f(x))^n is n(f(x))^(n-1), but it's actually n(f(x))^(n-1) times the derivative of f(x) with respect to x. --Bowlhover (talk) 04:06, 11 December 2008 (UTC)[reply]

"Reversing the chain rule" like that works fine if the derivative of the "inside" function is constant. But in this case it's not constant, and what "Bowlhover" told you will serve you far better than that. Michael Hardy (talk) 04:23, 11 December 2008 (UTC)[reply]

"Reversing the chain rule" also works when you can actually reduce the integrand into a product of two terms, one of which resembles the derivative of the other. In that situation, you can use integration by substitution. However, here you are, indeed, forced to expand and integrate one term at a time. Confusing Manifestation(Say hi!) 05:17, 11 December 2008 (UTC)[reply]

isn't that the reverse power rule? not the reverse chain rule. 96.242.34.226 (talk) 22:50, 12 December 2008 (UTC)[reply]

Finding the derivative of

(f(x))^{n}

with respect to x requires both the power rule and the chain rule, so whether you call the reverse process the "reverse power rule" or the "reverse chain rule" depends on which rule you think is more important here. Personally, I think both names are prone to confusion when taken out of context. Eric. 131.215.158.213 (talk) 01:45, 13 December 2008 (UTC)[reply]

Reuleaux triangle

After reading BKell's question on Reuleaux triangle above, I looked at the animation on that page and have two questions. (1) If the triangle is a solid paint brush, what area of the unit square doesn't get painted - and more to the point how do you go about working this out?. And possibly harder (2) If the edges of the 'triangle' are a paintbrush, how much of the unit square gets painted? -- SGBailey (talk) 07:30, 11 December 2008 (UTC)[reply]

If you follow the external links there you can find a page on Applications of shapes of constant width one of the applications is a machine to drill an almost square hole. --Salix (talk): 09:11, 11 December 2008 (UTC)[reply]

Nice questions; (1) in fact is easier (it will take more time describing it than computing) and it is possibly somewhere in internet, if not in wikipedia, for it regards the shape and the area of the hole mentioned by Salix alba, and has therefore an application. The other point (2), I guess, just requires some work; I will give you some hints at the end. Please have a pencil. Say that the square is in a Cartesian coordinate system and has vertices (0,0) (0,1) (1,0) (1,1). The first thing to observe is that there is essentially one position up to rotations where all the three vertices $(A,B,C)$ of the Reuleaux triangle lie on the edges of the square simultaneously. This is: triangle and square co-axial, one vertex is at the midpoint of one edge of the square, and the other two are at some points P, Q on opposite edges, each at a distance $\scriptstyle 1-{\sqrt {3}}/2$ from a vertex of the square. If you consider all the corresponding rotated positions, you then have 12 points disposed symmetrically on the boundary of the square. These points define 2 interior subintervals on each edge. There are 4 further small arcs of curves to be considered, in the interior of the square, and with endpoints the 8 P's above, each arc being close to a vertex of the square; I will describe them in a moment. But first let's number the 12 arcs (8 intervals plus 4 small curves) in cyclic order: $I_{0},I_{1},I_{2}...$ , say for any k mod 12 (clockwise, of course); it turns out that these, all together, bound a convex octagonal shape spanned by the Reuleaux triangle through a whole rotation (i.e., the hole). During the rotation A is in some $I_{k},$ , B is in $I_{k+4}$ , C is in $I_{k+8}$ , where in any case two of these are intervals, and one is a curve arc. To describe one of the curves, say $I_{1}$ , close to (1,1), I have found the following nice parametrization for you. Let A, B, C pass through respectively $\scriptstyle I_{5},\ I_{9},\ I_{1}$ and let $\scriptstyle \pi /6\leq t\leq \pi /3$ be the (acute) angle in A between the base of the square and the straight line from A to B. Then $\scriptstyle A=(\cos(t),\ 0)$ , $\scriptstyle B=(0,\ \sin(t))$ and $\scriptstyle C=(\sin(t+\pi /6),\ \sin(t+\pi /3))$ . You can easily check that this way A, B, C have distance 1 from each other. But this $C=C(t)$ is a parametrization of an arc of the whole ellipse contained in the square $\scriptstyle [-1,1]\times [-1,1]$ with center in (0,0), axes on the diagonals of the square, and tangent at the edges of the square. It is indeed a particular case of a Lissajous curve (if you go there you'll find exactly the ellipse). Once you know it's an ellipse you can easily compute the area; you do not need integrating, nor even writing the equation of the ellipse: just use elementary geometry, and a proportionality argument to squeeze linearly the ellipse along one of its axes onto a circle (the ratio of suitable areas does not change after stretching). I quickly computed with Maple the area of the hole as 98.7% of the unit square (it's 0.9877004070.. would somebody like to check; I'd be glad). As a last remark, note that, proving that the Reuleaux triangle is actually always inside the curve described by A in a whole rotation still requires a small metric comparison. The point is that, while the Reuleaux triangle points his little head $A$ , say in $I_{7}$ , and sniffs the origin (0,0), his (/her?) large bottom crawls around (1,1), however whithout getting it very close to that point, because it's quite large: just check that the distance from A in either $I_{6}$ or $I_{7}$ or $I_{8}$ to any point in $I_{1}$ is never less than 1 (and it is, even if you rectify the curve arcs $I_{1}$ and $I_{7}$ , that make it easier the comparison). Notice that for A in $I_{7}$ the envelope of the corresponding family of bottom-circle arcs is relevant to your point (2): the four rotated copies of this curve should bound the annulus-shaped painted area from inside, while the exterior boundary is clearly the same of your point (1). The delicate point is, that there are here other two curves to look at: the envelopes of the bottom positions of the Reuleaux around (1,1) while A is shifting through $I_{6}$ , and symmetrically, through $I_{8}$ . These two curves, and their rotated copies, apparently, do not touch the inner boundary of the painted region of (2), even though they get very close to it, at some point, what needs some computation. Is that clear enough?--PMajer (talk) 11:06, 12 December 2008 (UTC)[reply]

Thanks -- SGBailey (talk) 09:58, 14 December 2008 (UTC)[reply]

Functions

Hi. I'm trying to solve this problem that one of my friends gave me.

If the function f: N→N satifies the inequality below, prove that f(n)=n.

f(n+1)>f(f(n))

I've tried solving this on my own, but I couldn't get anywhere and it's been troubling me for nearly a week. A full solution or even a small hint would be a lot of help. Please help me!! Thanks.

(note: I am a Korean middle school student and it's been quite a while since i've used English, and thus my expressions may sound a little awkward at best. Please excuse my rusty English. But I have no trouble understanding mathematical terms so please help me^^) Johnnyboi7 (talk) 08:40, 11 December 2008 (UTC) —Preceding unsigned comment added by Johnnyboi7 (talk • contribs) 08:32, 11 December 2008 (UTC)[reply]

The function satisfies the inequality for some n and you have to prove that for that particular n, f(n) = n?

Topology Expert (talk) 09:19, 11 December 2008 (UTC)[reply]

no, i meant that there exists a function such that for every natural number n, the inequality holds true, and should such function exist, that function would only be f(n)=n.Johnnyboi7 (talk) 11:36, 11 December 2008 (UTC)[reply]

First prove that f(0) = 0 and f(n) > 0 whenever n > 0. Do this by considering the smallest value taken by the function. Then use induction. If you get stuck, just ask again.

By the way, I know that in Chinese at least, the word for high school (中学) sounds like middle school, which means something a bit different in English. I imagine it might be the same in Korean as in Chinese. Are you sure you're talking about middle school and not high school? 67.150.245.141 (talk) 09:55, 11 December 2008 (UTC)[reply]

Thanks a lot!! I'll try what you just said

By the way, middle schools (中學校/중학교grade 7~9) in korea (south for those who were wondering) are roughly equivalent to junior high schools in the US, and currently I am in 9th grade.

And wow, I never knew that the chinese call high schools middle schools!! Thanx for that extra information lol Johnnyboi7 (talk) 11:36, 11 December 2008 (UTC)[reply]

All right. Middle school in most states of the U.S. goes up to grade 8, and this seemed like too difficult a problem for that level, or even grade 9, really. 67.150.247.111 (talk) 11:58, 11 December 2008 (UTC)[reply]

ohh I've come across a minor bump along the road: the problem states that f:N→N and 0 is not a natural number so the value f(0) does not exist.Johnnyboi7 (talk) 13:11, 11 December 2008 (UTC)[reply]

Then start with 1 instead (and consider f(1)). Remember, thinking is important; don't just try out values. First see whether this holds for a few functions you invent and understand why. Then have a go at proving the general statement (and no, this isn't too difficult for ninth grade; in fact I think it should be easy enough for seventh graders). Topology Expert (talk) 13:47, 11 December 2008 (UTC)[reply]

If your definition of natural numbers has 1 as the smallest natural number then you can start by showing f(1) = 1 and f(n) > 1 whenever n > 1, and continue from there. Gandalf61 (talk) 13:45, 11 December 2008 (UTC)[reply]

A helpful start is to show that

f(n)\geq n

. The way I would suggest is defining m as the least natural number such that

f(m)<m

and then showing that contradicts

f(m)>f(f(m-1))

, and hence it is impossible for such an m to exist. Dragons flight (talk) 00:04, 12 December 2008 (UTC)[reply]

did you guys know that this problem is actually from IMO 1977, problem no. 6?????? lol I'm not sure if this easy enough for even 12th graders, let alone 7th or 9th graders220.122.189.57 (talk) 08:26, 12 December 2008 (UTC)[reply]

I guessed that this was an IMO as soon as I saw it! But really, IMO problems only test number theory and Euclidean geometry so I don't really 'appreciate' IMO (if you know what I mean).

Topology Expert (talk) 09:52, 12 December 2008 (UTC)[reply]

What is a "cz" a unit of?

While editing HMAS Sirius (O 266), I came across the statement that this replenishment ship has a capacity of "34,806 cz fuel". This statement is confirmed by the source,[1], but I do not know, and cannot find through Google, what a "cz" is in relation to the given figure. Does anybody know what a cz is a unit of, and how much this is in a more widely known measurement? -- saberwyn 09:14, 11 December 2008 (UTC)[reply]

Another page on the same website refers to "the transfer of 2,500 CZ (2.5 million litres) of fuel"[2], which suggests a CZ is a cubic metre (1000 litres). However a suitable meaning for CZ does not show up on acronymfinder.com, and it doesn't seem to be a usual abbreviation. Maybe it's a mistake or a jargon term used in a very narrow environment. --Maltelauridsbrigge (talk) 12:48, 11 December 2008 (UTC)[reply]

"Maybe it's a mistake or a jargon term used in a very narrow environment". I'm thinking so. The only other place I've seen this unit is in reference to Royal Navy replenishment ships

Try http://www.bta.mil/fedebiz/PRIVATE/EDIT/document/lfwg/flissue.pdf. Presumably we should have an article on some of these Units of measure? -- SGBailey (talk) 14:27, 11 December 2008 (UTC)[reply]

The supplied source confirms the above assertion that a cz is military-speak for a cubic metre. I'll work out where best to insert this information soon.

Thanks to both of you for your help! -- saberwyn 20:51, 11 December 2008 (UTC)[reply]

Reciprocal of a distance time graph

A sort of follow up to my previous question: If you draw the reciprocal of a distance/time graph. So you draw the graph distance against 1/time that should show you whether the acceleration is constant right? But which part of it shows you that? I would assume the gradient but it doesn't seem to work. Any advice? Harland1 (^t/_c) 23:22, 11 December 2008 (UTC)[reply]

No, this is not a useful way of determining if acceleration is constant. What makes you think it would be? Algebraist 23:27, 11 December 2008 (UTC)[reply]

What is a useful way then? Harland1 (^t/_c) 23:30, 11 December 2008 (UTC)[reply]

The most obvious (not necessarily best; I haven't thought about this much yet) way is to first draw a graph of distance against time, and from this graph read off the value at 0 (c say) and the derivative at 0 (a say). Then draw a graph of sqrt|distance-c-a*time| against time. Acceleration is constant iff this graph is a straight line. Algebraist 23:34, 11 December 2008 (UTC)[reply]

You could just measure the gradient at a few points and plot those against time, giving you a velocity/time graph, and then see if that is a straight line. It actually only tells you if the average acceleration between the points you measure is constant, but that's all your method will do as well, since to plot your graph you are going to select a few points, do the calculating for them, plot them and then connect them by eye (at least, I can't see a better way). --Tango (talk) 23:44, 11 December 2008 (UTC)[reply]

The better way probably involves actual statistical analysis, which I've never been comfortable with. Algebraist 23:46, 11 December 2008 (UTC)[reply]

For appropriate values of "better", maybe. I'd say any method which avoids statistics is automatically better! --Tango (talk) 23:48, 11 December 2008 (UTC)[reply]

Thanks, but when you've got your graph how do you find the acceleration off it? The gradient? Harland1 (^t/_c) 05:37, 12 December 2008 (UTC)[reply]

Acceleration is the gradient of a velocity/time graph. --Tango (talk) 10:43, 12 December 2008 (UTC)[reply]

If you graph distance versus time on special graph paper that has both axes marked off logarithmically, and the graph you get is a straight line with a slope of 2 (or 1, or 0), then you have constant acceleration. Unfortunately this doesn't go both ways: it's possible to have constant acceleration and yet have a distance–time graph that doesn't come out to be a straight line. However, if you have constant acceleration, then as time goes to infinity your graph will have an asymptote which is a line with slope 2 (or 1, or 0). —Bkell (talk) 07:31, 12 December 2008 (UTC)[reply]

(See Log-log graph.) —Bkell (talk) 07:34, 12 December 2008 (UTC)[reply]

I would prefer not to use the word 'gradient' because it has a few different meanings in mathematics. Rather, 'slope' would be the more correct term and would be less prone to confusion. Topology Expert (talk) 13:18, 12 December 2008 (UTC)[reply]

I only know of one meaning of "gradient". The gradient in question is the 1-dimensional case, but it's the same basic idea as in any dimension. --Tango (talk) 16:17, 12 December 2008 (UTC)[reply]

The gradient of a function is always a vector (or a column matrix; whichever you prefer). On the other hand, the slope is a scalar. In this case, we are dealing with scalars; not vectors so 'slope' would be the most appropriate term (in fact, gradient would not be correct).Topology Expert (talk) 16:27, 12 December 2008 (UTC)[reply]

And in any case, since we are dealing with the one-dimensional case here, slope would be more preferred than gradient (but as I have already mentioned, gradient would in fact, be incorrect). Topology Expert (talk) 16:29, 12 December 2008 (UTC)[reply]

In fact, the gradient of a differentiable function would be a linear transformation of R in the one-dimensional case whereas the slope would not. Topology Expert (talk) 16:41, 12 December 2008 (UTC)[reply]

No, the derivative of the function at a point would be a linear function (a covector). The gradient at a point is a vector, dual to the derivative. Of course, the difference is even less important than normal in this 1d case. Algebraist 17:30, 12 December 2008 (UTC)[reply]

The gradient of a function of n variables is an n-dimensional vector. Thus, the gradient of a function of one variable (which is what we have here) is a 1-dimensional vector, also known as a scalar (well, the space of 1-vectors is isomorphic to the field of scalars (considered as a vector space over itself), you can argue that they aren't strictly equal). --Tango (talk) 17:07, 12 December 2008 (UTC)[reply]

The field of scalars is the space of 1-vectors (literally!). However, the 'gradient' of a function is a linear transformation of R whereas the 'slope' of a function is not. Topology Expert (talk) 18:11, 12 December 2008 (UTC)[reply]

You could argue that (x) and x are different things, although they are clearly isomorphic. That's not really important, though. When I learnt about gradients they were vectors of partial derivatives, so the 1D case is exactly the same. --Tango (talk) 18:47, 12 December 2008 (UTC)[reply]

Nope. I am still not convinced. In my opinion, isomorphic objects are the same (literally). So saying that they are different things (or even arguing about it) would be mathematically incorrect. But as I said, the gradient of a function at a point is a linear transformation from R to R (and in fact a linear isomorphism if the function's derivative at that point is non-zero and therefore any differentiable function from R to R is an immersion iff its slope is nowhere 0) whereas its slope can never have such a property. Topology Expert (talk) 20:22, 12 December 2008 (UTC)[reply]

My comment about isomorphisms was just to head off anyone disagreeing with me on those grounds, I didn't intend it to become a major debate. However, since you insist on making a big deal out of it: the multiplicative group of powers of t and the additive group of integers are clearly not equal - a group is a pair, a set and a binary operation. Two pairs are equal if their first components are equal and their second components are equal. Two sets are equal if they are contained in eachother. t is not an integer, so clearly the sets aren't equal, so the groups aren't equal. They are, however, isomorphic and, generally, when talking about groups we actually talk about isomorphism types of groups (ie. groups modulo isomorphism). For that reason we generally talk about isomorphic groups as being "the same", but it is wrong to call them equal. Anyway, a gradient is a vector field, that is, a function from R^m to Rⁿ. The case we're talking about is just the case where m=n=1. There is nothing special about 1 compared to any other number of dimensions, it's still a gradient. --Tango (talk) 23:11, 12 December 2008 (UTC)[reply]

The gradient of a function at a point is not a linear map, as I mentioned just above. Algebraist 23:15, 12 December 2008 (UTC)[reply]

The gradient of a function from R to R is a 1x1 Jacobian and therefore is a linear map. Topology Expert (talk) 09:28, 13 December 2008 (UTC)[reply]

1x1 matrices are isomorphic to the underlying field (ie. the real numbers), and the slope of a function is a real number, so if gradient and slope aren't equal, they are at least isomorphic (and in this case it is standard to identify them). --Tango (talk) 14:22, 13 December 2008 (UTC)[reply]

Always identifying isomorphic objects is a really bad idea, essentially because there may be many different isomorphisms by which you could identify them, and it may matter which you use. Algebraist 23:18, 12 December 2008 (UTC)[reply]

OK. Consider [0,1] and [-1,1]; they are mathematical objects. Inside, R, they are obviously not equal (i.e when R is the universal set). Suppose we consider them 'isolated', i.e, don't assume an imbedding (in any category) of them inside some ambient space. We still can't speak of set equality. But if we consider the natural isomorphism from [0,1] to [-1,1] (i.e an isomorphism that preserves structure in every category that applies) and view both sets as a subset of [-1,1], they will be the same. Whether sets are equal, depends largely on how they are imbedded inside some ambient space. The obvious (inclusion) imbedding of both sets inside R makes them 'not equal' but it is possible to imbed both sets in R in such a way that makes them equal (when I speak of an 'imbedding', I speak of an imbedding in all categories that apply). I know that a lot of mathematics studies isomorphism classes and their properties (such as K-theory) but in most cases, isomorphism classes are assumed to consist of one object (like homotopy theory where in fact, two loops imbedded in some ambient space may be homotopic, but still not 'equivalent' (isotopic)). If I am wrong, I am still not convinced of this. Topology Expert (talk) 19:22, 13 December 2008 (UTC)[reply]

Two sets are equal if they contain the same elements. Embeddings (note the spelling) don't come into it. --Tango (talk) 22:22, 13 December 2008 (UTC)[reply]

Many authors spell 'embedding' as 'imbedding' (for example Munkres) so that is not incorrect. I still don't agree with you: two sets are equal (as subsets a universal set) iff the corresponding inclusion imbeddings are equal. Therefore, imbeddings do come into it. Topology Expert (talk) 12:35, 14 December 2008 (UTC)[reply]

That's a very strange definition of set equality and not one I have ever seen before. You also seem to be using a strange definition of subset, since by the definition I know, the inclusion of a subset into the set is unique. --Tango (talk) 14:51, 14 December 2008 (UTC)[reply]

Are you saying that I don't know what a subset is? If no, then I am sorry for the misinterpretation. If yes, I am disappointed because this was an unecessary comment. But if we have two sets A and B that are subsets of a set C, then A = B if and only if, the corresponding inclusion maps i_A and i_B into C are equal. So imbeddings to come into set equality. Topology Expert (talk) 15:10, 14 December 2008 (UTC)[reply]

There is no need in introduce a superset to both sets, two sets are equal simply if and only if they contain the elements, ie.

(A=B)\iff (\forall a\,a\in A\,\iff \,a\in B)

. I don't see how you can talk about the inclusions being equal - the concept of equality of functions is only defined if they are between the same sets, so you are assuming A=B in your definition. --Tango (talk) 17:23, 14 December 2008 (UTC)[reply]

Yes I am implying that A = B by saying that the inclusions are equal (they can't be equal otherwise). And two sets A and B are equal iff their inclusion maps to A U B are equal. Topology Expert (talk) 17:29, 15 December 2008 (UTC)

Ok, what is your definition of equality of maps, then? --Tango (talk) 18:32, 15 December 2008 (UTC)[reply]