matt grime

Do not worry every one, for if the person who didn't know about sets was reading the definition for the first time and had taken the time to read it all properly they'd Know that sets do not have repeated elements and that they are unordered by definition. This whole nonsense about trying to infer something from notation alone is silly - why should they even infer that {a,b,c} is a set?

In order to show that the killing form is degenerate it suffices to find two elements X and Y such that Tr(Ad(X)Ad(Y)) =0, and in the above it so happens you can pick X and Y any two basis elements - you may not be so lucky in general, but yes you've got the gist. I posted a better reply at physicsforums.com

Let the basis be e,f,g in the same order as you wrote them. then what is the linear map defined by ad(x) fpr some x in the algebra? It sends e to [xe]=0, t to [xf] =xf-fx, and g to [xg]=xg-gx=0 So, just write out X as a b 0 c and mutliply out the above expressions and write [xe] as a combination of basis vectors. [xe] is simply -b.f. Repeat for where it sends f and g, and you've got how it acts on the basis as a linear map thus you can write out its matrix wrt that basis just like any other map.

Strange, cos every one else is agreed that the answer is 6.25, half the longest side given.

The triangle was established to be a scaled 3-4-5 right angled triangle, and thus the hypotheneuse is a diameter of the circle required.

A guinea is still one pound one shilling, or 1 and 1 twentieth of a pound, or £1.05 in new money, I imagine.

It is purely sepculative, obviously, but it does seem a more than reasonable explanation. Ok, perhaps no one sat down and said: right let's pick a base with lots of divisors, but it evolved in such a way. Another example, and one perhaps more readily agreeable, is the old predecimal currency in the UK. There used to be 240 pennies in the pound, making it very useful for trading in commoditities since it is readily divisible by 2,3,4,5,6,8,10, etc. In fact the most annoying thing is that it isn't divisible by seven, and there are seven days of the week. A fact that is enshrined in the old unit of a Guinea, which was 245 pence (1 pound and one shilling), which is divisible by 7, and which is still the unit in which horses are priced.

Why isn't 360 easy to work with? It is divisible by 2,3,4,5,6,8,9,10,12,15,20,30,45, etc. In short it is a very useful base, that probably comes from older civilizations use of base 60. Radians are also used, and preferred in all but baby uses of geometry (ie high school) since they have much nice analytic properties - it is the natural unit for physics eg in measuring angular momentum, and is the natural unit for sin and cosine, for then differentiating sin gives cos and differentiating cos gives -sin: using degrees yields multiplicative constants to deal with. This is analogous to the fact that natural logs and base e are indeed natural. If you dislike 360, then you onw't like radians since there are 2pi radians in a circle. Some places also use gradians: there are 400 gradians in a circle.

then there was no need to do "for the transformation T to be one to one, the columns oof matrix A have to be lineraly independent (by a particular therom). That is T is one to one iff Ax=0 has only the trvial solution. However, because there will always be at least one free variable in the reduced forms of A, therefore the columns of A are not linearly independent. Thus the transformation T can not be one-to-one."

I can't say I bothered to solve f(f(x))=x. This is maths: understanding why the method works is more important than actually getting the correct answer. though why in case 2 have you specified a=b=0 AND ab!=1: ab=0 automatically. Nor do I see why you're looking at those two cases in particular. Just solve f(f(x))=x

Erm, no, that's not right. f could be a function from R to C_{infty}, it could be a function on any number of domains, to any number of codomains, who says the domain is even a subset of the reals? I know you probably automatically think the domain is implicit, but it isn't. Indeed the most sensible interpretation of your function is as a function from the extended complex numbers (or reals) to itself. I cannot emphasize enough that the domain and codomain of a function are part of the definition of the function. f as written is not a function merely an expression. If you wish to check if the expression f(f(x)=x then do so, but do not state it is "inverse". Sadly this kind of abuse of notation is common, and perpetuated by people who ought to know better, ie even the much vaunted academic undergrad calc texts.

And again, you've not stated the domain or codomain rendering the question meaningless.

We can't say when f is or isn't inverse since you've failed to state the domain and codomain of the function. f isn't even *defined* on the real numbers unless b=0, or unless a/b=c (when the function is identically 1/b).

why on earth would i need l'hopital? it's obvious from archimedes principle. why? it's obivous, and frequently the *definition* It's the *definition*, look it up.

Fine, so don't use 0!=1, then you'll have to ignore all the mathematical formulae that do use that definition. The limit is just -1 in your above post as surely absolutely anyone can see? The abs value of your f(k) is the radius convergence, being the abs value of the ratio of consecutive terms in the series as the index tends to infinity. Pointless and completely off topic, but what's new? Newton probably used Newton's method to work out sqrt(2), and all the other square roots, but, once, more, this is neither here nor there. When are you going to actually learn about the things that are explained to you, instead of wandering off on tangents?

Look at the power series, it's obvious how to work out the radius of convergence from that. and even it it weren't then surely the fact that if I put any x in such that x<-1 and have r, say, 1/2, then the resulting power series cannot converge otherise the square root of a negative number would be real, hence its radius of convergence is 1. But you're entirely missing the point. It is just a matter of convenience and consistency to adopt 0!=1, and x^0 extends to 0^0=1 for the uses you have come up with, and there is no other useful definition for 0!, so it is unabiguous, but there are other ways to define 0^0 that contradict the extension of x^0 to x=0, thus we cannot declare 0^0 to be unambiguous. There is no absolute truth in mathematics, there are just things we construct, and if the construction is universal then there is no harm in adopting it universally, hence 0!=1 is both useful and consistent. Set theory often defines "emtpy products" and "empty" functions, as well as empty sets, and in this situation the combinatorics mean that 0!=1 is sensible. But why are you so opposed to the definition of 0! as 1? Why not? If you don't like the convention, do not adopt it, but that is your choice.

Computing 2^1/2 using the formula for (1+x)^r does not prove anything, what about all the other possible cases? do they all converge? I see you're about to go off on a pointless tangent and ignore what has been explained to you. So that's this thread off limits then.

So, we see you only have defined ! for strictly positive integers, and thus *you* have no way to define 0!. What makes you tihnk that this means 0! is defined at all? The rest of the world is happy to extend the definition of ! to include 0 by declaring 0!=1, which greatly simplifies many expressions, though it is not *necessary* to do so - it is simply an extension that *makes sense*. If you are prepared to accept certain axioms of set theory then it also makes sense since there is exactly 1 permutation of an empty set: the empty permutation, but that is merely a convention too. And as for the second reply: you claim that you can raise numbers to non integral powers via applying the "binomial theorem" to (1+x)^r this is patentyly a false statement. It is only true for non-natural r if |x|<1, and does not let you define 3^{sqrt(2)} for instance. Thus, when I asked you to define what it means to raise numbers to non-rational, or integral powers you gave an answer that would, if it were correct, allow us to define the square root of -1 simply by putting x=-2 and r=1/2 into the expansion of (1+x)^r and thuis yield i as the sum of lots of real numbers. Of course -2 is outside the circle of convergence for that power series.

5 column vectors in R^4? Can they be independent?

That is one way to define x^r *for some x* but not the correct way for all x, for instance according to that theory the square root of minus 1 can be gotten by adding up real numbers. But of course you're sidestepping convergence issues when you write (1+x)^r, since the only converges if |x|<1. So, no, that isn't the correct definition for exponentiation. It is in fact exp{rlog(x)} as I repeatedly said, with some choice of branch of log, and a very restricted domain. Why a priori do you think 0^0 is defined? Is it the value of f(x) = x^0 when x=0? Is it the value of g®=0^r when r=0, or is it the value of h(t)=t^t when t is 0? Not that 0 must necessarily be in the domain of any of those functions, and indeed it is specious to define functions without domains, so I will declare the domain of f to be R with the *definition* that f(0)=1, thus we have a nice continuous function, the function f(x)=1, which is the convention taken in defining Taylor series. We note g can only reasonably have domain the strictly positive integers to begin with, but we extend it to strictly rational positive r, and thence by continuity to the closure of the striclty positive rational r, which is the positive reals (including 0) which means that g(x)=0 for all x. This is at least the third time that I've explained to you why it is that we say 0^0 is undetermined - there are two equally sound ways to define it giving different answers. But in Taylor series we are clearly in the first case, that of f needing to be extended to allow f(0). I leave it to you to decide what h(x)'s domain can be. You're sidestepping the point, also, that it is only you that thinks that symbols 0! and 0^0 must have some god-given meaning. They do not, and even the choice that 3!=3.2.1 is just a definition. It so happens that 3! is the number of distinct orderings of 3 objects, and undoubtedly such things led to the study of !, but the definition of n! is not "the number of ways of ordering n objects", it is the unique solution to x_n = n(x_n-1) with x_0=1 and n=>0. We set n!:=x_n it can then be used to describe things. How do *you* define n!? Why are you not asking what (-1)! is? Why must it be anything?

what makes yo think that exponentiation is a function that is allowed to raise a number to the power m if m isn't an natrual number? What is (-1)^{1/2}? Not real is it? and this only holds if m isn't zero. so we have something about a case when a=/=0, let us see why you think this means anything for the case a=0 (acutally, you never explain what the non-zero case has to do with the zero case, at least not in a way that is mathematical) which is a good reason not to define 0^0 using this method, however we all know sin(x)/x is 1 when x is zero don't we? that is one formula for it based upon the usual standard that 0^0:=1 did we say it was indeterminate? we said that by considering some expressions you couldn't determine what 0^0 was, that doesn't mean that it can't be determined in general. have you ever heard of removable singularities? evidently not, which is a shame since i already explained to you why there is ambiguity in defining 0^0. so let's do it again: If we consider the function f(x)=x^0 for all x not equal to zero it is, by continuity defined to be 1 (if we're thinking of x a real number). note that x^r is not defined necessarily as a function from R to R, and this lack of understanding of what a function is is something that is hindering you by the way, but we can define it for all strictly positive x as exp(rlog(x)), anyway I digress, the point was, that we can define a function from R\{0} to R via x^0 which is identically 1, thus we can extend to state 0^0=1 if we *fix* the exponent. now, conversely consider the function g®=0^r, for r strictly positive, which it is not unreasonable to say is identically 0 by contiinuity again. Then we can extend, by continuity to a function such that 0^0=0. Now, consider the function h(x)=x^x for x strictly positive. This hasn't a hope in hell of being extended to a function when x=0. see, there are different functions in play here. I mean you're blindly assuming that it makes any sense of even talking about raising a real number to a real power, which isn't true at all. Now, in the case of taylor series we are in the first situation where we have x^0 which we extend by continuity to state 0^0=1. What makes you think that the same symbol must have the same meaning in all contexts? Not really, at least nothing that we can't explain: you just won't accept that these are just conventions that we adopt as need be to suit our own needs. It is common in mathematics: without some axioms of set theory we cannot define a basis of all vector spaces, with it we can show there is a game that both players win with probablity 1. Tough. That is jsut a consequence of an axiomatic system.

or you could use the fact that |x|^3 is just x^3 if x is positive, and -x^3 if x is negative, and thus clearly differentiable at all points but 0. so it suffices to check the derivative at 0, from the right you get the derivative being as for x^3, and the left as -x^3, both of which give 0 at the origin, so it is differentiable.

If you wished to explain concisely why did you post that verbose ramble? So you just want the number of different ways of selecting r objects from n objects with order important, and no replacement. Why were you unable to say that? That being the common explanation. The number being [math]^nP_r[/math], abused in ascii to nPr, or n pick r. nCr is the binomial coefficient n choose r, where order is unimportant. They are are, respectively, the well known numbers [math] \frac{n!}{(n-r)!} =^nP_r[/math] [math]\binom{n}{r}=\frac{n!}{r!(n-r)!}[/math] and obviously 0!=1. Compared to what you wrote how is that considered to be overly long? It even gives a formula for the numbers. Now, what on earth is all that nonsense about random?

Johnny, I am more than familiar with all of the standard, and lots of the non-standard parts of mathematics, including [math] ^nC_r[/math]. Would you possibly consider that a maths phd may know more than you? Just explain what you mean by an r-permutation, and if you really feel like it, why not state why it matters that the set is "given to you at random" whatever that may mean. I suspect you're just asking "how many ways are there of selecting r objects from n, order unimportant (or perhaps you want order important)". Is that all you're asking, and why didn't you say so? I suspect, from the examples, that an r-permutation is an ordered r-tuple selected from the set of n elements without repetition. As it is, you've asked, and this is a quote, "How many elements in the generated set, with P(n,r) elements in it?" which makes no sense, and stated, "Suppose we have a set given to us by someone else, and they chose it at random. So you recieve the set from them. Since this can randomly vary, let n denote the number of elements in the set you recieve." What does that even mean? To receive a set randomly? What if, heaven forfend, it weren't finite?

No, "random" isn't clear from context. Random requires probability, which requires a sample space and a measure on it. And, no, i''ve no idea what an r-permutation is - i can think of several things it might be, but seeing as you've not even explained what random means, and apparently can't, then i think you should state all of your assumptions and meanings. So, please, try and explain what it means to "give" a random set to some one; why is that set finite; what is the sample space, and what is the measure on the sample space. However, it isn't even clear if that is important, as with many of the things you include in a post. Are you simply asking how many r perms of a set of cardinality n ( a natural number) there are? what ever an r perm may be. (ie the random thing is completely unnecessary - does the fact that the set is given "randomly" affect the number of r-perms??)