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 [quote="zeek"]I've always seen the (a,b) notation too. I've never seen the ]a,b[ notation. If you really want to stick with using brackets, this might make more sense: [a,b] when a & b are in the interval. a],[b when they are outside the interval. [a,[b when a is inside and b is outside.[/quote]
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Antrax
Posted: Wed Feb 11, 2004 10:28 pm    Post subject: 1

We were taught [] for closed interval and () for open interval ~shrug~
Antrax

------------------
"Look, that's why there's rules, understand? So that you think before you break 'em" - Lu-Tze, Thief of Time
Mercuria
Posted: Wed Feb 11, 2004 9:39 pm    Post subject: 0

... i've never seen ]a,b[ either... always (a, b)

there's always been discrepency about using a bracket or a parenthesis next to infinity, though...
dethwing
Posted: Wed Feb 11, 2004 9:20 pm    Post subject: -1

*shrug* Makes more sense to me.
zeek
Posted: Wed Feb 11, 2004 7:48 pm    Post subject: -2

I've always seen the (a,b) notation too. I've never seen the ]a,b[ notation.

If you really want to stick with using brackets, this might make more sense:
[a,b] when a & b are in the interval.
a],[b when they are outside the interval.
[a,[b when a is inside and b is outside.
Quailman
Posted: Wed Feb 11, 2004 7:42 pm    Post subject: -3

You think that just because you saw it on the internet, it must be true?
mith
Posted: Wed Feb 11, 2004 7:29 pm    Post subject: -4

Most people?

http://mathworld.wolfram.com/OpenInterval.html
dethwing
Posted: Wed Feb 11, 2004 7:10 pm    Post subject: -5

Most people use ]a,b[ to mean open intevals. (a,b) looks like an ordered pair to me.
Posted: Wed Feb 11, 2004 5:31 pm    Post subject: -6

You use letters for math. Funny men!
Beartalon
Posted: Wed Feb 11, 2004 5:19 pm    Post subject: -7

It's the standard notation, which I forgot after several years. Silly me too.

[This message has been edited by Beartalon (edited 02-11-2004 12:19 PM).]
TANSTAAFL
Posted: Wed Feb 11, 2004 5:18 pm    Post subject: -8

yeah, (a,b) is the range between but not including a and b. [a,b] is the notation my class uses for when the range includes the points a and b.
Quailman
Posted: Wed Feb 11, 2004 3:40 pm    Post subject: -9

Actually, I didn't understand why the range was described that way when I first looked at the problem, and never bothered to consider why after puzllling with it for a bit. It wasn't until you pointed it out that I understood. I thought (-infinity,-1) would include -1. Silly me.

[This message has been edited by Quailman (edited 02-11-2004 10:40 AM).]
Beartalon
Posted: Wed Feb 11, 2004 3:34 pm    Post subject: -10

(apparently, one should wonder why he didn't read that the first time, wouldn't one? Some one didn't, and now every one knows)
Quailman
Posted: Wed Feb 11, 2004 3:32 pm    Post subject: -11

(One would assume that that is why the range was described as "for all t in (-infinity,-1)U(-1,1)U(1,infinity)", wouldn't one.)
Beartalon
Posted: Wed Feb 11, 2004 3:26 pm    Post subject: -12

(but when t=1 or t=-1, this equation will not equal 1)
TANSTAAFL
Posted: Mon Feb 09, 2004 10:34 pm    Post subject: -13

figures, i was trying to hard... I just had to look at the fill in the blank equation she gave us a little harder.

x2-y2=1

I kept thinking I had to add the two terms, not subtract.

[This message has been edited by TANSTAAFL (edited 02-09-2004 05:34 PM).]
Lilifreid
Posted: Mon Feb 09, 2004 10:27 pm    Post subject: -14

Try looking at the new terms to see what you can do to/with them to get rid of the t's. If that doesn't make sense, post what you got when squaring them (or IM me).
TANSTAAFL
Posted: Mon Feb 09, 2004 10:15 pm    Post subject: -15

I know, I've been around here long enough that I shouldn't be asking for help with my homework, but I can't make it to the Professor's office hours due to my work schedule, and the book doesn't go over any problems this complex.

We are working on parametric equations, here is what the problem says

quote:

Consider the curve C parametrized by:

x = (t2+1)/(t2-1)
y = (2t)/(t2-1)
for all t in (-infinity,-1)U(-1,1)U(1,infinity)

By squaring both x and y, find an equation of C in terms of just x and y (no t):

_______ - ______ = 1

So I squared each side, but I still don't see how I can solve for t in terms of x or y in order to find the equation.

I'm not asking for someone to give me the answer, but maybe point me in the right direction?