Q:

How can a bijection be made from $\mathbb{N}$ to $\mathbb{Q}$ using diagonalization? 

I'm studying Cantor's diagonalization, but something seems unclear to me. 
There is this table for diagonalization:
╔═══════════════════════╗
║ X  1   2   3   4   5  ║
║ 1 1/1 1/2 1/3 1/4 1/5 ║
║ 2 2/1 2/2 2/3 2/4 2/5 ║
║ 3 3/1 3/2 3/3 3/4 3/5 ║
║ 4 4/1 4/2 4/3 4/4 4/5 ║
║ 5 5/1 5/2 5/3 5/4 5/5 ║
╚═══════════════════════╝

And there can be a bijection between the diagonal
From 2/1 -> 1/2
From 3/1 -> 1/3 (diagonals)
From 4/1 -> 1/4
I don't understand where the bijection really is?  I guess 2/1=2 ($\mathbb{N}$) is mapped to 1/2 ($\mathbb{Q}$)
But what about 2/5?  I can only see that 2/5 will make |$\mathbb{N}$| < |$\mathbb{Q}$|
However, I can only think that |$\mathbb{N}$x$\mathbb{N}$| = |$\mathbb{Q}$|  
What am I missing or misunderstanding and where's the bijection?  Can you give me 2 elements that are mapping to each other? 

A:

Consider this sequence and see if it helps:
1 maps to 1/1
2 maps to 2/1
3 maps to 1/2 
4 maps to 1/3
5 maps to 3/1 (As the 2/2 is already included in the list we are building)
6 maps to 4/1
7 maps to 3/2
8 maps to 2/3
9 maps to 1/4
10 maps to 1/5
11 maps to 5/1 (As the others are already done on this diagonal.) 
12 maps to 6/1
13 maps to 5/2
14 maps to 4/3
15 maps to 3/4
16 maps to 2/5
Thus there is keeping track of what is already handled if there are common factors.