matt grime

Right, scanned through it now. Odd that two of the big names in it are Karoubi and Deligne, who might recognize being called physicists. Topologists, and homomlogical algebraists, if they're different, have had such things as Z_2 graded objects for ages. We just never call them super. Damn physicists stealing our ideas.... (note there is _humour_ in this post)

The thing is, mathematicians have very little interest in the Z_2 grading, at least with that phrasing, in and of themselves anymore than any other grading. Whereas mathematical physicists find these specifically coming up all the time, so I find that quote disingenuous, but then I'm reading out of context, and Jon often has his tongue in his cheek when writing such things.

Johnny, what the hell are you talking about? Now isn't the time to introduce differentials or 1 forms into this topic. Stop it, please. Dave was doing a good job of discussing derivatives without your off topic intrusion into unnecessary parts of mathematics, not to say inconsistent. In post 13 you declare dt=1, at no point does Dave declare dt is a number, he sticks with the proper idea that d/dt is an operator. then in post 17 you state that Dave wanted it to be a number. Well, he didn't, you set it to be one, despite the fact that, when treated properly it is a differential, not a number. You are not treating it properly by the way. In fact I get the impression that this is all kind of new to you and you're not well versed in these. THe ponit is that derivatives are about limits, not difference equations. There are formal simliarites but that is all. Indeed the differences are so large that they ought not to be discussed in the same arena at this stage. Forget the "classical" definition of derivative, it is THE definitoin of derivative, and fortunately limits commute with finite sums (when everything exists).

They don't actually (have to) exist: it is just a way of thinking abuot doing algorithms. THink of it as a black box, if you will. The questions are: is there a black box that can take in information and give out information. INformation is naturally discrete bits and we we care when or when not an answer can be returned IN THEORY.

So, have you worked out P(A|B)? As the above post indicates, it's straightforward.

What on earth is a negative relation?

That depends on what you claim 0 to the 0 is. A lot of people would say that it is undefined, though it makes most sense to declare it to be 1.

Where's the delete option?

Yes, That is correct, so, how many eigenvalues (you can read that off) how many eigne vectors, hence you can work out the algebraic and geometric multiplicities. The better base is {2,2x,x^2}. This is the basis wrt which T is in Jordan Normal Forn.

T is a 3 by three matrix, so write it out with respect to the natural basis 1,x,x^2, say (actually, there is a better basis one could choose making the question more obvious, can you see what it is?)

Hint for the second one: If a*b=0 then one of a or be is zero.

many commands in LaTeX are "what you'd guess". you want an alpha? it's slash alpha. you want to underline? \underline{ text to underline}. there are also underbrace, overbar and other things too.

And if you take two non-zero vectors, u and v, then uxv may be zero. Thus you cannot formally invert multiplication for all vectors. (And this only works in R^3 anyway). The point is that there is almost always no useful way to define a multplication operation on vectors that is invertible (ie there will always be two non-zero vectors that mulitply to give the zero vector) . The notable exception is R^2 where we can define (a,b)*(c,d)=(ac-bd,bc+ad), and we get a space that is isomorphic to the complex numbers.

so what is the algebraic method then?

What is division? it's the inverse operation to multiplication. so, what ways do you have of multiplying two vectors to give a vector?

inside the latex math environment \left( \begin{array}{ccc} entry & entry &entry \\ entry & entry &entry \\ \end{array} \right) would make a 3 column matrix with entries centred on in the column, Vary the 3 c's to make more columns. separate entries with & and \\ is a newline. ie the above would make a 2 row matrix. the command \left<symbol> makes a left bracket as big as the matrix. you can try varying <symbol>, to say, | to get a determinant.

Yeah, and some of us think that "u" isn't perfect for precision for "you"....

Do not confuse mathematics with the things it models. In the "real world" all caclulations are done with rational multiples with respect to some base measurement, but that says nothing about the acutal mathematical properties of numbers, it merely states something about the "real world". Numbers are not things that exsit in any physical sense, but there are things that exist in the physical world that numbers are useful for desribing properties of. God that was an ugly sentence. But, basically, pi is a real number, it is irrational, even transcendental, and it is the the ratio of a geometric NOT real world quantities. It can theoretically be evaluated in base ten to arbitrary precision, given enough time by one of the many series formulae for it. pi squared of six is for example the sum of the reciprocals of the squares of the natural numbers.

It's inverse isn't defined on R, arctan isn't an invertible function from R to R so this is a non-example. You don't seem to understand that the domain and range are part of the definition are functions - it makes little sense to say tan is invertible with continuous and inverse and isn't.

I have written many explanations of the proof of this simple fact, some even in this thread I beleive, certainly in another one linked to from here. Understand the terms you're using. 0.99.. is the limit of the partial sums of the series 0.9 + 0.09+0.009+... that is it is the limit of the cauchy sequence of rationals x_n = 0.9...9 with n nines. Let y_n be the constant cauchy sequence y_n=1. Then |x_n-y_n| = 10^{-n} hence in the reals, which is the completion of the rationals wrt euclidean distance, the limits are equivalent. OK? If you think the proof is "too hard" sorry, but that's what the proof is. All the other arguments provided are illustrative at best, but at least they don't introduce spurious nonsensical things such as your "proof", such as "supposing infinity is a number", and when we multiply by ten we get something in that place holder "infinity". Plus you also state that 1 and 0.99... are different (presumably as real numbers, though you never mention anything about the set in which the argument takes place). There is no need to introduce infnity at all. (Even analysis can be done entirely without using the word infinity. Limits are things defined in terms of finite objects that get arbitrarily "large")

"How can one number equal another different number"? It can't if they are *different*. But, again, like so many people, you're confusing a *real* number with its representation. Why can people accept that 1/2 and 2/4 are the same number but cannot do that for 0.99... and 1?

You're not getting this. You're doing arithemetic with finitely long strings of 9's and drawing a conclusion about an infinite string that doesn't necessarily hold (and indeed doesn't) since they are *different*.

Please, for the love of mathematics, can I ask you never to use that explanation again?

But in the first post, the author raised the very good question about limits. Yes this is because of limits, and limits are how the real numbers are defined (or at least, that is one of the equivalent definitions - being able to take limits in a well behaved sense is the defining characteristic of the reals), and in the decimal representation of real numbers, .9 recurring and one are different representations of the same real number, jsut like 1/2 and 2/4 are different rational representations of the same rational number.

Ok, the conditionals are easy if we assume the host knows where the goat is. There are 3 possibilities when you pick. Goat1, Goat2, Car. If you pick either goat1 or goat2 and after seeing the host open a door you swap you win. In the third option if you swap after being shown a goat you lose. So 2/3 of the time you win. The conditional is in the fact that of you pick goat1 he must show goat2, and vice versa. In the case where the host guesses a door (and we assume if he opens the car's door you can't swap), then if you pick goat1 he shows goat2 with prob. 1/2, so the probability of a win after swapping is 1/6, simililarly with picking goat1. in the third option you lose always with a swap. so swapping means you win 1/6+1/6=1/3 of the time.