This claim is proven by building a bijection [tex] \phi: \mathbb{N}\to\mathbb{Q} [/tex] by following a diagonal procedure.
Imagine you have an infinite matrix, whose coefficient are all possible fractions:
[tex]\left[\begin{array}{cccc}\frac{1}{1}&\frac{2}{1}&\frac{3}{1}&\ldots\\&&\\\frac{1}{2}&\frac{2}{2}&\frac{3}{2}&\ldots\\&&\\\frac{1}{3}&\frac{2}{3}&\frac{3}{3}&\ldots\end{array}\right][/tex]
Now, count them "diagonally": this is how we map every natural number, using the same positions:
[tex]\left[\begin{array}{cccc}\phi(1)&\phi(2)&\phi(6)&\phi(7)\\\phi(3)&\phi(5)&\phi(8)&\phi(13)\\\phi(4)&\phi(9)&\phi(12)&\ldots\\\phi(10)&\phi(11)&\ldots&\ldots\end{array}\right][/tex]
The attached image, taken from aminsaied.wordpress.com/2012/05/21/diagonal-arguments/, should make it clearer
