We will define the relation ~ on $\mathbb N \times \mathbb N$ by $(a,b)\sim (c,d)$ iff $a + d = b + c$.
Prove that the operation given by: $[(a,b)][(c,d)] \stackrel{\text{def}}= [(ac + bd, ad + bc)]$ is well-defined.
My attempt at this proof:
Let $(a,b)$ and $(a',b')$ be elements of $[(a,b)]$, and similarly, $(c,d)$ and $(c',d')$ be elements of $[(c,d)]$.
My guess is that we need to show that:
$(a'c' + b'd', a'd' + b'c')$ is an element of $[(ac + bd, ad + bc)]$.
So we know the following:
$aa' + bb' = ab' + ba' \\ cc' + dd' = cd' + bc'$
But now I'm wondering what to work with. It would be great if someone could give some assistance. Thank you!
Edit: I am also seeking assistance on the following problems:
Q: Prove that $\mathbb N \times \mathbb N$/~ contains an additive identity, i.e. find an element [(i, j)] ∈ $\mathbb N \times \mathbb N$/~ with the property that
$[(i,j)] + [(c,d)] = [(c,d)] $
Q: Prove that every element of $\mathbb N \times \mathbb N$/~ has an additive inverse, i.e. for any $[(a,b)] \in \mathbb N \times \mathbb N/~$, show that there exists $[(c,d)] \in \mathbb N \times \mathbb N/~$ such that
$[(a,b)] + [(c,d)] = [(i,j)]$ where [(i,j)] is the additive identity.