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I was reading a proof on the counting of rational numbers and this formula came out at the beginning but I'm not even sure what it does.

$$f(m,n)=1+...+(m+n-2)+m$$

where $m$ and $n$ are naturals.

Supposedly, $f(1,1)=1$ associates one and only one element of the naturals to the set $$\mathbb{N} \times \mathbb{N} = \{ (m,n) : m, n \in\mathbb{N}\}$$

Any help ? While we're at it, what does "$\times$" mean between the two sets ?

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From what you say about the purpose of defining the function, I think it must mean that $f(m,n)$ is the sum of all numbers from $1$ up to $m+n-2$, with an extra $m$ added, which you could write as $f(m,n)=(m+n-2)(m+n-1)/2+m$.

The values of $f(m,1)$ are all the triangular numbers. The values of $f(m,2)$ are all the numbers which are $1$ more than a triangular number greater than $1$, and so on, so the values of $f(m,n)$ for fixed $n$ are all the numbers $k$ for which $k$ is $n-1$ more than the largest triangular number less than or equal to $k$. Thus every positive integer appears once and once only as a value of the function.

The purpose of this is to show that, although $\mathbb N\times\mathbb N$ appears larger than $\mathbb N$, they are essentially the same size (they are both countably infinite), because there is a one-to-one correspondence between their elements.

For sets $S,T$, the product $S\times T$ is the set of all pairs of the form (element of $S$, element of $T$).

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Here X means cartesian product of natural numbers with natural numbers.

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    This is probably better as a comment, since it doesn't speak to what I assume is the main question, the function $f(m,n)$.2017-01-22