Show that all rational numbers Q fit into the Hilbert hotel

Question

Can I prove this with help of simple induction grounded on the basic axioms of number theory and a linearity pattern?

$$\frac{-x}{y} \neq \frac{x}{y}$$ $$\frac{-x}{-y} = \frac{x}{y}$$

The complete set of rational numbers according to me. $$P = \left\{\frac{a}{b}, [-\infty \leq (a,b) \leq \infty], \ a,b\in \mathbb Z \right\}$$

A subset of $P$ is $E$.

$E$ is a finite set with 50 unique elements. These elements fit in Hilbert's hotel. $$E = \left\{\frac{a}{b}, [-5 \leq (a,b) \leq 5], \ a,b\in \mathbb Z \right\}$$

I derive the formula $ r = 10k$, where $r$ is the amount of rooms needed.

$r=10k+10$, when k+1. $$ \left\{\frac{a}{b}, [-k \leq (a,b) \leq k], \ a,b\in \mathbb Z \right\}$$

when $k$ goes to infinty $r$ goes to infinity. I have successfully counted the amount of rooms needed for an infinite amount of fractions. This only shows it's possible to fit them in the hotel. But in order to show how, do I need a function to map all fractions systematically?

Answer 1

You didn't show the rational numbers fit. It's obvious that any finite set fits, but the problem is to extend this to infinitely many clients.

You first reserve the chambers with numbers of the form $p_n^m$, for $m\ge1$, where $p_1=3,p_2=5,\dotsc$ is the list of odd prime numbers in increasing order.

Then you reserve the chambers with numbers of the form $2p_n^m$, again for $m\ge1$.

Note that chamber number $0$ is not reserved; in it you place $0$.

Now place the nonzero rational number $m/n$, with $m,n$ coprime and $n>0$, in room number $p_n^m$ or $2p_n^{-m}$ according whether $m>0$ or $m<0$.

Infinitely many rooms will be empty, even among those that have been reserved, but it's not a problem, is it? But you have a precise way to tell where any rational number will be hosted.

Answer 2

First of all, if you really want to be grounded on the basic axioms of number theory, then your definition of $\mathbb Q$ is wrong because the division operation is not defined on $\mathbb Z$. The actual definition of $\mathbb Q$ is a little different. Also, your definition includes $0$ in the denominator.

Now, coming back to your actual question, the formula you derived wont work because your set $E$ contains $\frac 1 2$ as well as $\frac 2 4$ and both are equal, but you counted them twice. A better argument is to say that the number $r\leq 10k$.

But, the the fact that rational people (no pun intended) would fit into the Hilbert's hotel is a very well established fact as we know that the infinity of $\mathbb Q$ is the same as the infinity of $\mathbb N$. This thread may help.

Answer 3

What you have shown is that any finite number of rationals, however large, will fit into Hilbert's hotel. This is not the same thing as showing that all the infinitely many rationals will fit into Hilbert's hotel at the same time.

Induction allows you prove some statement $P(k)$ is true for any natural number $k$. It essentially proves an infinite number of statements in one go. But if the statement $P(\mathbb{N})$ or $P(\infty)$ is something that has a sensible meaning, it will NOT be proved by the induction because $\mathbb{N}$ and $\infty$ are not natural numbers.