# Thread: Hilbert's Hotel is Troubling Me

1. ## Hilbert's Hotel is Troubling Me

I am about to do an inspiring lesson on Hilbert's hotel to my 15 year old pupils (they are bright, don't worry) but I have an issue that needs resolving for my own benifit.

The problem is when infinitely many coaches turn up with infinitely many people on them. The solution, as I'm led to believe, is this:

1) Move all the current guests from their room m to 2m, freeing up all the odd numbered rooms.
2) Label each coach q = 1, 2, 3, 4, ...
3) Label each seat on each coach n = 1, 2, 3, 4, ...
4) Place people from the 1st coach into rooms numbered 3^n, 2nd coach into 5^n, 3rd coach into 7^n and qth coach into the (p+1)^n where p = (q + 1)th prime number.

Thus everyone is accomodated. My problem is that the hotel is now not full. In fact, every odd numbered room which has 2 or more distinct factors is empty.

Please can someone tell me where I have got things wrong or whether the hotel is indeed now not full.

Thanks,

James (this is my first post so please be kind).

2. Originally Posted by jac1999
The problem is when infinitely many coaches turn up with infinitely many people on them. The solution, as I'm led to believe, is this:
1) Move all the current guests from their room m to 2m, freeing up all the odd numbered rooms.
2) Label each coach q = 1, 2, 3, 4, ...
3) Label each seat on each coach n = 1, 2, 3, 4, ...
4) Place people from the 1st coach into rooms numbered 3^n, 2nd coach into 5^n, 3rd coach into 7^n and qth coach into the (p+1)^n where p = (q + 1)th prime number.
Thus everyone is accomodated. My problem is that the hotel is now not full. In fact, every odd numbered room which has 2 or more distinct factors is empty.
Well of course the hotel is not full.
If you want it full each odd numbered room must be occupied.

The mapping $\Phi (j,k) = 2^{j - 1} \left( {2k - 1} \right)$ is a bijection $\Phi :\mathbb{Z}^ + \times \mathbb{Z}^ + \leftrightarrow \mathbb{Z}^ +$.
That is a fun fact to prove.

Assign the $k^{th}$ person on the $j^{th}$ coach to room # $2\Phi (j,k)-1$.
So the first person on the first coach is assigned to room #1.

That will fill the hotel.

3. Thanks Plato. I appreciate the answer.

I guess then that we have to be happy that the hotel can absorb an infinite number of people from an infinite number of coaches and, at the end of it, have an infinite number of spare rooms if we use the solution I laid out.

My class will love this tomorrow (I think I'll leave the bijection out)

4. There is a nice graphical representation of a bijection between $\mathbb{N}\times\mathbb{N}$ and $\mathbb{N}$: see Wikipedia.

5. That is useful, thanks.
Looks very similar to the method of showing the rationals are countably infinite. Why is the 'path' different in this one?

6. I agree, though I don't know what picture exactly you have in mind for rationals. I think that showing that $\mathbb{N}\times\mathbb{N}$ is equinumerous with $\mathbb{N}$ is simpler than the proof for rationals because you don't need to identify certain pairs, exclude division by zero, etc.

I am used to the picture that differs from the one in Wikipedia: from (0, 1) one goes to (0, 2) instead of (2, 0) and then descends, rather than ascends, from (0, 2) to (2, 0). I believe this bijection has a simple polynomial formula in one direction, different from the one above. However, this is not important at all.

7. The one I'm talking about looking like the one at this link
Proof that rational numbers are countable - from Homeschool Math
However I have issues with this as it doesn't have negative numbers or zero.
Although I agree that NxN = N is easier to show than Q=N.

By the way, how do people get maths symbols on here?
[IMG]file:///C:/DOCUME%7E1/JAC/LOCALS%7E1/Temp/moz-screenshot-1.png[/IMG][IMG]file:///C:/DOCUME%7E1/JAC/LOCALS%7E1/Temp/moz-screenshot-2.png[/IMG][IMG]file:///C:/DOCUME%7E1/JAC/LOCALS%7E1/Temp/moz-screenshot-3.png[/IMG][IMG]file:///C:/DOCUME%7E1/JAC/LOCALS%7E1/Temp/moz-screenshot.png[/IMG]

8. Yes, this is the picture I was talking about.

You can write LaTeX code inside the [tex] tag. Also, when you are writing a message, there is a Sigma on the toolbar that surrounds the selected text with [tex]. I also had to search for this. I think, how to write math should be more prominently shown in the forum help section.

9. Originally Posted by jac1999
The one I'm talking about looking like the one at this link
Proof that rational numbers are countable - from Homeschool Math
However I have issues with this as it doesn't have negative numbers or zero.
Although I agree that NxN = N is easier to show than Q=N.

By the way, how do people get maths symbols on here?
Why would you have any issues with that argument?
The countable union of countiable sets is a countable set.

You know that $f(x)=-x$ is a bijection.
The finite set $\{0\}$ is countable.

See this forum.

10. I don't have any issues with it mathematically for precisely the reasons you make. But I do have issues with it as a 'job done' pictoral representation.

Only a small issue, though.