Did you know?

[RazielDemon]

That a circle is defined as a conic section that passes through the two imaginary circular points at infinity in algebraic geometry?


anyways, can someone explain Hilberts Hotel to me? I reccon it's never full to start with.

[Justice]

I thought a circle was defined as "an infinite set of points equidistant from the fixed, constant center, C."

cheers, and
-=</|Awesome Party!|\>=-

[Juggalo]

O <<<<<< Circle is something round and 2D.

[MVB]

Juggalo that is not a circle. If you zoom in and look closely you'll notice it's actually an object composed of straight faces and angles, and is not a circle at all.

gg loser, learn2life

[Juggalo]

It's a frickin' cricle! :furious:

God geometry pisses me off.

*googles boobies*

ahhh, that's better..... AND THEY'RE CIRCULAR TOO!

[Justice]

No, they're actually 3-dimensional ovals.

You want flat circle boobs .... OK then

( o Y o )

JUGGS LIKES FLAT WOMEN

cheers, and
-=</|Awesome Party!|\>=-

[Weavern]

They are conical due to the effects of gravity :p The definition of a circle is a set of points an equidistance from the focii. The definition of an oval is a series of points which form a triangle of set parimeter using the two focii as a corners of the circle. Thus the informal string rule for ovals holds true.

However the cone explanation works but you first need to define what a cone is as explained by euclidian algebraic proofs. Which are evil.

Wiki explains hibert's hotel if not google.

[RazielDemon]

Yah, I know the idea, but If it's possible to shove the people allong 1 room, then the room at the end was never full to begin with, and so the hotel was not fully booked.

[Justice]

What's Hibert's Hotel?

Some weird Math problem?

cheers, and
-=</|Awesome Party!|\>=-

[RazielDemon]

In a hotel with a finite number of rooms, once it is full, no more guests can be accommodated. Now imagine a hotel with an infinite number of rooms. You might assume that the same problem will arise when all the rooms are taken. However, there is a way to solve this: if you move the guest occupying room 1 to room 2, the guest occupying room 2 to room 3, etc., you can fit the newcomer into room 1. Note that such a movement of guests would constitute a supertask.

It is even possible to make place for an infinite (countable) number of new clients: just move the person occupying room 1 to room 2, occupying room 2 to room 4, occupying room 3 to room 6, etc., and all the odd-numbered new rooms will be free for the new guests.

If an infinite (countable) number of coaches arrive, each with an infinite (countable) number of passengers, you can even deal with that: first empty the odd numbered rooms as above, then put the first coach's load in rooms 3n for n = 1, 2, 3, ..., the second coach's load in rooms 5n for n = 1, 2, ... and so on; for coach number i we use the rooms pn where p is the i+1-st prime number. You can also solve the problem by looking at the license plate numbers on the coaches and the seat numbers for the passengers (if the seats are not numbered, number them). Regard the hotel as coach #0. Interleave the digits of the coach numbers and the seat numbers to get the room numbers for the guests. The guest in room number 1729 moves to room 1070209. The passenger on seat 8234 of coach 56719 goes to room 5068721394 of the hotel.

An even stranger story regarding this hotel shows that mathematical induction only works in one direction. No cigars may be brought into the hotel. Yet each of the guests (all rooms had guests at the time) got a cigar while in the hotel. How is this? The guest in Room 1 got a cigar from the guest in Room 2. The guest in Room 2 had previously received two cigars from the guest in Room 3. The guest in Room 3 had previously received three cigars from the guest in Room 4, etc. Each guest kept one cigar and passed the remainder to the guest in the next-lower-numbered room.

[Tank0]

pfft i am moving to the holiday inn , way to many people and cigars to get some sleep in this hotel