Is there a bijection between $\mathbb N$ and $\mathbb N^2$?

Question

Is there a bijection between $\mathbb N$ and $\mathbb N^2$?

If I can show $\mathbb N^2$ is equipotent to $\mathbb N$, I can show that $\mathbb Q$ is countable. Please help. Thanks,

André Nicolas · Accepted Answer · 2013-03-09T04:23:15.860

20

It is easier to write an answer than to find one of its several occurrences on this site.

Let $n$ be a positive integer. Then there exist uniquely determined positive integers $u$ and $v$ such that $n=2^{u-1}(2v-1)$. This is because every positive integer $n$ can be written in one and only one way as a power of $2$ (possibly $2^0$) and an odd number.

Define $g$ by $g(n)=(u,v)$. Then $g$ is a bijection from $\mathbb{N}$ to $\mathbb{N}^2$.

For a different bijection, search for the Cantor Pairing Function.

edited Mar 09 '13 at 04:23

answered Mar 09 '13 at 04:16

André Nicolas

507,029

2

Indeed the first duplicate I found has essentially the same answer by you! – hardmath Mar 09 '13 at 04:29
3

It might be nice to learn some searching secrets. I did remember answering the question, but finding is another matter. – André Nicolas Mar 09 '13 at 04:35
@AndréNicolas You can search for user:me is:answer cantor pairing to boil it down to (currently) seven candidates – Tobias Kienzler Mar 15 '13 at 14:00
How exactly do you write 8 with this? – JobHunter69 May 30 '23 at 17:01
This seems wrong. let n = 2^m. Then 2^(m-u+1)=2v-1 which is impossible if v is a positive integer – JobHunter69 May 30 '23 at 17:02

Julien · Answer 2 · 2013-03-09T05:31:47.870

Here is a non constructive proof.

Of course, there is an injection of $\mathbb{N}$ into $\mathbb{N}\times\mathbb{N}$.

Now consider the map $$ (x,y)\longmapsto 2^x\cdot 3^y $$ By the fundamental theorem of arithmetic, this is an injection of $\mathbb{N}\times\mathbb{N}$ into $\mathbb{N}$.

By Cantor-Bernstein-Schroeder, there exists a bijection between $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$.

Edit: I added this answer because the only constructive proof I knew was the zigzag along the diagonals, which I find slightly tedious to write down. But thanks to Andre Nicolas, now I know a straightforward and easy-to-write-down bijection. Which is actually a little step away from the injection I use above.

Aang · Answer 3 · 2013-03-09T05:01:36.847

4

Yes there exists a quadratic bijection.

See this

edited Mar 09 '13 at 05:01

answered Mar 09 '13 at 04:46

Aang

14,672

score 3 · Answer 4 · answered Mar 09 '13 at 04:11

Yes. Imagine starting at $(1,1)$ and then zig-zagging diagonally across the quadrant. I'll leave you to formulate it.

Hint: for every natural number $>1$ there's a set of elements of $\mathbb{N}^2$ that add up to that number. For $2$, there's $(1,1)$. For $3$, there's $(2,1)$ and $(1,2)$...

score 2 · Answer 5 · answered Mar 09 '13 at 04:15

You're trying to find a way to "count" the things in $\mathbb N^2$ in a manner that gets you to any pair $(a,b)$ in some finite time. Notice that "counting" something is essentially finding this bijection - there is a "first" pair, and a "second pair" and so on.

There are many ways to do this, but most of them come down to ordering the set $\mathbb N^2$ in some meaningful way.

Hint: There are countably many diagonals, i.e. there is a "first diagonal" a "second diagonal" and so on. Does this work?

Is there a bijection between $\mathbb N$ and $\mathbb N^2$?

5 Answers5

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