2

I'm reading Roger Penrose's "The road to reality". In chapter 16, discussing set theory and Cardinality , he mentions it is easy to construct a bijection (or "1-1 correspondence" - in his phrasing) between $\mathbb{N}$ and $\mathbb{Q}$. and offers $(a,b)\mapsto \frac{1}{2}((a+b)^{2}+3a+b)$ as an exercise to the reader. How would one go about proving something like this?

H.Rappeport
  • 1,536
  • Have a look at page $9$ here (http://goo.gl/WjGRmH) for inspiration. – Jack D'Aurizio Oct 15 '17 at 16:59
  • @JackD'Aurizio $\mathbb{N}\times\mathbb{N}$ is fundamentally different because you're not dealing with reducible fractions and sign. – orlp Oct 15 '17 at 17:01
  • The point is we do not usually build a bijection between $\mathbb{N}$ and $\mathbb{Q}$ in a single step, neither the mentioned map provides it. By exploiting the Cantor-Schroeder-Bernstein theorem we have $|\mathbb{Q}|=|\mathbb{N}\times\mathbb{N}|$, then by exploiting the given map and the Cantor-Schroeder-Bernstein theorem again we have $|\mathbb{N}\times\mathbb{N}|=|\mathbb{N}|$. – Jack D'Aurizio Oct 15 '17 at 17:04
  • @JackD'Aurizio I am sure the average reader of "The road to reality" doesn't know these theorems... – amsmath Oct 15 '17 at 17:05
  • @JackD'Aurizio I just realized that I misread the question. I thought that the provided map actually was a claimed bijection, and not a building block. I'm aware of everything you've mentioned :) – orlp Oct 15 '17 at 17:07
  • @amsmath: then the claim it is easy to construct a bijection between $\mathbb{N}$ and $\mathbb{Q}$ through the following map is hiding something, because the shown map only gives a bijection between $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$. Or between $\mathbb{Z}$ and $\mathbb{Z}\times\mathbb{Z}$ with a minor tweak, but still, $\mathbb{N}\times\mathbb{N}\neq\mathbb{Q}$ and $\mathbb{Z}\times\mathbb{Z}\neq\mathbb{Q}$. – Jack D'Aurizio Oct 15 '17 at 17:08
  • @JackD'Aurizio Well, Roger Penrose is a physicist. ;-) – amsmath Oct 15 '17 at 17:10
  • @JackD'Aurizio Which number is not covered by this map, acting only on coprime $a,b$? – amsmath Oct 15 '17 at 17:13
  • @amsmath: $48$, for instance, which is the image of $(3,6)$. – Jack D'Aurizio Oct 15 '17 at 18:04
  • @JackD'Aurizio And why is $48$ not covered by the map? Is it injective? – amsmath Oct 15 '17 at 18:33
  • @JackD'Aurizio Ok, I got it now (by looking at the grid below). But have you also taken into account negative values for $a$. We are talking $\mathbb Q$, not $\mathbb Q_+$. – amsmath Oct 15 '17 at 18:39
  • @JackD'Aurizio For example, $f(-5,12) = f(2,4) = 23$. – amsmath Oct 15 '17 at 18:54

1 Answers1

1

As JackD'Aurizio's resource mentions, these kind of bijections often use diagonalization. In your example, write a small grid:

$$\begin{matrix} (0, 0)&(0, 1)&(0,2)&(0,3)&(0,4)&(0, 5)\\ (1, 0)&(1, 1)&(1,2)&(1,3)&(1,4)&(1, 5)\\ (2, 0)&(2, 1)&(2,2)&(2,3)&(2,4)&(2, 5)\\ (3, 0)&(3, 1)&(3,2)&(3,3)&(3,4)&(3, 5)\\ (4, 0)&(4, 1)&(4,2)&(4,3)&(4,4)&(4, 5)\\ (5, 0)&(5, 1)&(5,2)&(5,3)&(5,4)&(5, 5) \end{matrix}$$

Now apply the given bijection. What do you notice?

orlp
  • 10,508
  • Thank you for your answer, I see the diagonal "threading". But how does one formulate this into a formal proof? induction? (To be clear, a proof for this being a bijection between $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$ would be enough, I see it takes some more work to show the correspondence from pairs to rationals). – H.Rappeport Oct 15 '17 at 17:38
  • @H.Rappeport Hint: what is the size of the diagonal that starts at $(0, b)$? What is the cumulative sum of all diagonals before the one that starts at $(0, b)$? On what diagonal does $(a, b)$ lie? – orlp Oct 15 '17 at 17:41