I am trying to prove that $\mathbb{Q}$ and $\mathbb{Q} \times \mathbb{Q}$ have the same cardinality so I must construct a bijection between the sets.
I have supposed there exists a function $f: \mathbb{Q} \to \mathbb{Q} \times \mathbb{Q}$ where $f$ is 1-1 and onto but I'm not sure where to begin proving this.
I doubt explicitly defining this function would be of much use, (much like how Cantor's Diagonalisation argument requires no formula as such, as it is tedius) I'm simply interested in seeing if it is possible to map every $q \in \mathbb{Q}$ to a $(r, s) \in \mathbb{Q} \times \mathbb{Q}$ and testing whether it is 1-1 and onto.
I think I could use something similar to Cantor's Diagonalisation argument for a bijection from $\mathbb{N} \to \mathbb{Q}$ but I can't wrap my head around it for my case.
Best Answer
For an explicit construction,
You should know from the literature that there exists a bijection $f~:~\mathbb{Q}\to \mathbb{N}$. Further, you should know that there exists a bijection $g~:~\mathbb{N}\to\mathbb{N}\times\mathbb{N}$. Finally, there is a bijection $h~:~\mathbb{N}\times\mathbb{N}\to\mathbb{Q}\times\mathbb{Q}$
We have then $h\circ g\circ f$ is a bijection from $\mathbb{Q}\to\mathbb{Q}\times\mathbb{Q}$