Find explicit bijection from N to N x N

[complexnumber]

Homework Statement

Find a linear ordering of \mathbb{N} \times \mathbb{N} and
use it to construct an explicit bijection f : \mathbb{N} \to \mathbb{N} \times \mathbb{N}.

Homework Equations

The Attempt at a Solution

I know how to find a bijection by graphically by drawing the \mathbb{N} \times \mathbb{N} in a matrix and then traverse the elements diagonally. However, I cannot find a formula that maps an n to a (x,y) pair.

 

[hgfalling]

Your idea about traversing the elements of the N X N matrix diagonally is good.

Now, we want to write down that idea in a formula form.

Let's start with the first few n:
n=1:(1,1)
n=2:(1,2)
n=3:(2,1)
n=4:(1,3)
n=5:(2,2)
n=6:(3,1)

Now let's consider the pairs in the left column of the matrix, where the pair is (x,1). What pattern is there to these x-values: {1,3,6,10,15...}?
Next consider the pairs that make up a diagonal, for example {(1,3),(2,2),(3,1)}. What pattern is there in these?

Can you use those two patterns to write down an expression for the ordered pair that corresponds to a particular n value?

 

[complexnumber]

Thanks for your reply. I found a bijection from \mathbb{N} \times \mathbb{N} to \mathbb{N} by
<br /> \begin{align*}<br /> n = \left(\sum^{x+y-2}_{k=1} k \right) + x<br /> \end{align*}<br />

However, is this a bijection from \mathbb{N} to \mathbb{N} \times \mathbb{N} asked by the question? Given a n I cannot use this equation to find a (x,y) pair.