ForumsThe TavernMath Thread

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daleks
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daleks
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Chamberlain

This thread is just what the title says it is. It is all about Math. You can post math jokes(the jokes don't have to be good), math questions, what you like about Math, or even why you hate Math.

My math joke: Resistance is not futile. It is voltage divided by amps.

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aknerd
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aknerd
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Peasant

Well, there are clearly many different ways to define subject so this is pointless. Here's something more interesting: Prince Rupert's Cube.

What is the largest cube that you could fit "through" a unit cube (1x1x1)? By this I mean the cube must pass through a "tunnel" in the unit cube.

It turns out that the largest cube that meets this requirement is actually LARGER than a unit cube. In order words, it is possible to pass a cube through a smaller cube!

Here's a picture:
http://upload.wikimedia.org/wikipedia/commons/2/28/Prince_Ruperts_cube.png

The white space is the tunnel carved into the cube. As you can see, the center of a face on the larger cube will initially pass through a corner on the unit cube, in the direction of the opposite corner.

While this solution makes logical sense and is actually kind of obvious (most people I've talked to have immediately come up with this solution)*, its still really cool.

*if you don't get it immediately, try visualizing it with squares.

johnmerz
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johnmerz
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Shepherd

[quote=daleks]Are we just counting the main subjects?[/quote]

Yes, apparently so. And nice "cube thing" you have pictured there, aknerd. It makes me dizzy by just looking at it... I do get how it works though.

daleks
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daleks
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Chamberlain

The cube makes sense. I had never thought about it before though.

How do you know all this stuff aknerd?

aknerd
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Peasant

It makes me dizzy by just looking at it...

I think its a bad drawing, but I couldn't find a better one online. Or a video showing a 3d model. I guess the results are trivial enough that not many people care about it. I still think its cool.

How do you know all this stuff aknerd?

I happened to read about this particular thing while thumbing through a math encyclopedia looking for something else. I'm currently about halfway to getting a math major, which is where my general math knowledge comes from.

Oh, on that note: a few pages back I was talking about set theory. Does anybody know the infinite hotel paradox?

If you had a hotel with infinitely many rooms, and every single room was occupied, could you fit in a new customer? Why or why not?

What if a bus arrived carrying infinitely many people, how many of them could you find rooms for? How?

What infinitely many such buses arrived? Could you fit all of their passengers in? How?
daleks
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daleks
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Chamberlain

Does anybody know the infinite hotel paradox?

I think I have heard of it but I don't know it.
If you had a hotel with infinitely many rooms, and every single room was occupied, could you fit in a new customer? Why or why not?

What if a bus arrived carrying infinitely many people, how many of them could you find rooms for? How?

What infinitely many such buses arrived? Could you fit all of their passengers in? How?

Is sharing rooms allowed?
aknerd
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Peasant

Is sharing rooms allowed?


Well, no. But you are allowed to move people around. I guess that should have been clear.
daleks
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Chamberlain

If you had a hotel with infinitely many rooms, and every single room was occupied, could you fit in a new customer? Why or why not?

Yes you could or the other questions you asked would be pointless. I don't know how seeing how infinity is not a number really but more of an idea. I guess the guy could stay in the lobby.
aknerd
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aknerd
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Peasant

Yes you could or the other questions you asked would be pointless.


I realized that when I wrote it. The answer is "yes" to all of the questions, but the point is to figure out how to fit the new people in.

The first case is easy. You just move the person in room 1 to room 2, room 2 to room 3, room n to room n+1. Then, you put the new person in the first room. This is possible, since you will always have a room to move the evicted person to, because there is no "last" room.

The second case is only slightly less simple. You just move everyone to twice their room number, which is equivalent to moving the person in room n to the nth even number. Then, every odd number room will be empty, and there are infinitely many odd numbers, so you can move everyone on the bus into the odd rooms. This is possible because there are just as many odd or even numbers as there are natural numbers.

But what about the last case? It is possible, but why?

I don't know how seeing how infinity is not a number really but more of an idea.

I think you could make the case that all numbers are just "ideas". I've talked before about different ways to think about infinity. Here, we are just dealing with countably infinite sizes, which is probably the simplest kind of infinity.
daleks
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Chamberlain

So were you just using natural numbers? Or could 2.13 be a room.

It always sounds so simple once you say it but I can never think of it before.

aknerd
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Peasant


So were you just using natural numbers? Or could 2.13 be a room.


You can number the rooms anyway you like, as long as there are countable many of them. You run into problems if you have as many rooms as there are, say, irrational numbers.
daleks
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Chamberlain

Are any of you a fan of xkcd? I really like the comics but some of the math ones I don't understand.

aknerd
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Peasant

Of course! Though I have trouble with the programming jokes...

Lately, I have been more into SMBC; it has a similar humor, but it better executed.

Oh, and to answer my own question, a solution I came up with would be to number each of the buses with a prime number. So the first bus is 2, the second is 3, the third is 5, etc. For ease of reference, we'll call the nth prime number p.n, so 5 would be p.3.

Then, for each bus n, we locate all of the guests in the hotel that are in the rooms that are powers of p.n. We move all the guests already in the hotel to the even power rooms, like p.n^2, p.n^100, etc. That leaves all of the odd power rooms for the people on the bus.

Since there will be no overlap, ie p.n^j = p.m^k only if m=n and j=k, everyone will have exactly one room to go to, and each room will only have one person. This can be easily proven: For each p.m^k, the prime factorization would be exactly (p.m)^k, since it is just a prime number times itself k times. Then, no other prime number would be a factor. So, each bus would have its own unique "family" of powers.

Example: the 10th person an the 10th bus would go to room number (p.10)^19 (since 19 is the tenth odd number). Hopefully there is a very quick escalator!

Wiki has a page about this paradox. They solved the last case in a very similar method to mine, except they simplified things a lot by considering the hotel as the first bus, and moving all the guests into the powers of 2, leaving all the other prime powers empty for the rest of the buses. This is probably the simplest solution out there.

daleks
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Chamberlain


Is the question in this comic even possible? It seems to me that the answer would probably be 0 because the voltage would not be able to make it that far.

johnmerz
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johnmerz
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Shepherd

Is the question in this comic even possible?

I personally think that it isn't possible, but, you never know...
daleks
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daleks
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Chamberlain

It seems to me that the answer would probably be 0 because the voltage would not be able to make it that far.

I read it wrong. Voltage has nothing to do with it. I still have no idea what the answer is.
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