matt grime

post away with it.

No, you won't, since you are going against the notation used by a not unreasonable number of mathematicians - every one educated at Cambridge, not to mention every analyst I've ever met and so on. The only time logs are ever taken in *maths* (ie not physics or engineering) where base e is not the chosen base is log base 2 for use in coding theory. Of course that is guaranteed to have rebuttals, but I carefully qualify this as any reasonable mature treatment of analysis. I must confess I have exactly one analysis textbook to hand, Goursat's classic text, and it uses log to mean base e, as would almost any other proper analysis book I@d be prepared to wager, and I've never seen ln used in complex analysis proper, writing as someone who's taken (several) graduate courses in (complex) analysis. log is just the usual standard, johnny in *pure mathematics*, it is the only one that makes sense, just as radians are the only unit of angle that make sense really. ln just signifies another dmbing down *sigh*, what next? Calling it the Argand plane?

What kind of texts Johnny? Speaking as a professional mathematician who's worked in both the US and the UK I'd say that log is the one I'd expect to read in any paper or decent book. Of course if you're basing this on reading some engineering text...

given a matrix A you need to solve the equations Ax=0 and see what x can be - often it will give you a "system of paramters" (not standard notation) and anything satisfying that "system" is avector in the null space.

Both parts are right, and if it helps don't see the word "vector" before the word space, just see sub, since they are in this case the same thing

Yes. so you've got a vector space, and all is well - it;s the span of the vectors (1,2,1) and (3,3,6)

Obviously - i only ever said I wqs finding the critical points on the curve. As I really don't own a calculator I cannot find them and really can't then work out the distance. I'm ignoring the matty thing.

You need to have more confidence in yourself. This, after all, is maths, and it is more important you understand how you got the answer than what the actual (numerical) answer is.

I've no idea, I didn't do a) - who cares what the actual number is, all we know si that we need to minimize the distance between (u,v) and 8,1 where u,v is in the curve. We know the distance, and we know minimizing it is the same as minimizing the square fothe distance, namely minimize (u-8)^2+(v-1)^2 and we know that v-1=u^{3/2} thus w minimize (u-8)^2 + u^3 differentiate it to find the critical points and we get the same thing as in b. I don't even own a calculator so i can't find the critical points, as they won't be at integers by the look of it.

Well, if you put an integral sign at the front, of course the integral of 1/x wrtx id log(x), give or take a constant. Note, it is common practice to use log for natural log, rather than ln. Just because a question doesn't mention sometihng doesn't mean the answer won't. I presume you're ok with integrating x to get x^2/2, but the "question" deosn't mention quadratics does it?

Given a point (u,v) on the curve what is the equation of the tangent line at that point? In particular what is its gradient. Now consider the line from (8,1) to (u,v) what is its gradient? When are those two lines perpendicular, ie when is that line feom (8,1) to (u,v) normal to the curve? You have done Normal lines, right? The tangent is [math]3u^{1/2}/2[/math] and the line from 8,1 to u,v has slope [math]\frac{v-1}{u-8}[/math] but of course v=1+u^{3/2}, or v-1=u^{3/2} Recall that if the tangent's gradient is D that the normal has gradient -1/D, so we must need the u.v such that [math]\frac{u^{3/2}}{u-8}= \frac{-2}{3u^{1/2}}[/math] rearrange and solve 3u^2=-2(u-8) Edit: what's wrong with my tex? here it is in ascii: The tangent is 3u^{1/2}/2 and the line from 8,1 to u,v has slope (v-1)/(u-8) but of course v=1+u^{3/2}, or v-1=u^{3/2} Recall that if the tangent's gradient is D that the normal has gradient -1/D, so we must need the u.v such that (u^{3/2})/(u-8)= (-2)/)3u^{1/2} rearrange and solve 3u^2=-2(u-8)

Well, in c) you ought to say that the function only has a one sided second derivative at x=0, and some people, like me, would point out that therefore f'' doesn't exist at x=0.

If I were to ask ou to solve the equation assuming a,b,c,and d are non-zero real numbers of: a(x+1/b)+d=c You could do it without thinking. This question is no harder (though we just must remember that AB doesn't equal BA for all matrices).

Just as long as no one claims that 1 and 0.999... are distinct real numbers. Makes me want to scream when they do that.

The answer to the last question is "yes", and I've proved at least twice in this thread that K is nondegenerate on sl_2, by showing that for any v there is a w such that K(v,w) is not zero, and hence K is nondegenerate.

The real numbers: 1. They are a field 2. They are ordered 3. They are complete. these have subaxioms, however, you don't need to think about these, but understand that the real numbers are the set of numbers which are defined to allow us to do analysis - or if you're doing A-levels, they let us do differentiation by assumption. Roughly they contain the rational numbers, they contain all the possible limits of sequences of rational numbers, these limits are unique (this means , in particular, that 1=0.99.. is a very basic consequence of the axioms, and must be true), that we can add subtract, multiply and divide (except for zero) and that + and * are commutative etc, and that we can compare elements with < and that either x<y, y<x or y=x and exactly one of these holds.

You must show that the bilinear form it defines is non-degenerate, which is straight forward. If v is any vector in the space v=(a,b,c) define w=(c,b,a) then K(v,w) = |a|^2+|b|^2+|c|^2 You cannot apply this matrix to a general Y since it is a *bilinear form* thus needing two inputs. You must show that if v is in sl_2 that K(v,w)=0 for all w implies v=0, or as above that for any v there is some w with K(v,w)=/=0

So, you're asking "is there an ordering on the complex numbers that respects the multiplicative nature of C, a la the ordering of R, eg given two numbers either x<y, y<x or x=y, and if z>0, then x<y implies xz<yz?" The answer is "of course not" since R is the unique complete ordered field and R and C are not isomorphic as fields, never mind as totally ordered ones, and they are both complete. If this is what you thought the issue was then it is a strange thing to say "the square root of -1 leads to the contradiction". Certainly if one assumes i>0, the i^2=-1>0, contradiction, and if i<0, then i*i>0*i, -1>0 again, a contradiction. However, you see this only is a contradiction to the assertion "there is a an ordering of C compatible with some conditions as above", which is a false assertion. You surely can't have expected people to come to that conclusion that that is what you were after based upon the title of the thread which you have said is a) self explanatory and b) deliberately vague. For if that were so then every true proposition would lead to a contradiction - it would contradict the assertion that some false proposition were true. Putting it another way, you're saying "are there any statements about C that are false owing to the fact that i^2=-1", and obviously there are: for instance the statement, "there is no element that squares to give -1" is contradicted by the existence in C of i. This is hardly earth shattering.

I avoided the existential point about numbers since it is completely immaterial in the context you stated it. When I said that the square root of minus one may or may not exist it was in the mathematical sense of: the square root of minus 1 does not exist (ie is not an element of) in the set of real numbers, similarly, in Z_4, the integers mod 4 there is no element that squares to give 3, but there is one in Z_2.

The only comment I have is if this notional reader knew the proper definitions of a set then they'd not need to have that extra assumption explicitly italicized. There ought to be no confusion, and indeed there isn't one. You can't just pick up a maths book and expect to understand the material immediately if you do not know what the notation is.

What issue, Johnny? You haven't raised an issue. What on are you getting at? sqrt(-1) is just some object that may or may not exist within some field. If you were to say, does sqrt(-1)=1 lead to any contradiction (in any field except one of characteristic two) then yes, because you're saying 1=-1. But just (sqrt(-1)? it is not a well formed question. an object can not lead to contradictions until you try to relate it to other things.

Does the question even have any meaning? I'd say not. How about: does Table Mountain lead to any contradiction? The square root of minus one is simply, when it exists, and element that squares to give minus one. Implicit in that is that we are working in some field or such. It is just an object it leads to nothing; reasoning about it may do. For instance the statement "F is a finite Field with p elements" and the statement "there is necessarily an element that squares to give -1" are contradictory, but "the square root of minus one" on its own?

E E is famous on Usenet for being a crank. I think we can safely say that he's nothing to worry about. He allegedly has also said that the real numbers are "wrong" as they are an infinite set.

It's a reductio ad absurdum idea. Do you not understand how that works in general or in this particular example? Here is a link explaining this particular example: http://sweb.uky.edu/~jrbail01/fermat.htm

Of course K(e,x)=K(f,X)=K(g,X)=0 implies X=0, but that isn't what you want to show, since it may be that all three things are nonzero but that K is still degenerate: eg suppose K(e,X)=1, K(f,X)=-1 K(g,X)=1, then K(e+f,X)=0. You need to actually calculate ad(X)ad(Y) where X and Y some general elements of sl_2. I will give you for free (and because it's effing tedious) that K(X,Y) = 4tr(XY) where tr is the trace. In particular if the basis is ordered as f,e,g ie the diagonal one in the middle that the matrix of K is given by: [math]\left( \begin{array}{ccc} 0 & 0 & 4 \\ 0 & 8 & 0 \\ 4 & 0 & 0 \end{array} \right)[/math] the latex isnt' appearing so it's: 004 080 400 in plain ascii