I want to prove that if $\{a_n\} \sim \{u_n\}$ and $\{b_n\} \sim \{v_n\}$, then $\{a_nb_n\} \sim \{u_nv_n\}$, where each sequence is a Cauchy sequence of rationals.
Here is my attempt.
Proof:
Since all the given sequences are Cauchy, they are bounded by some number. Choose M to be the max of these bounds, that way it bounds all the sequences.
Both $\{a_n-u_n\}$ and $\{b_n-v_n\}$ converge to zero. In other words, for all positive rational epsilon, there exists positive natural numbers $N_1$ and $N_2$, for all $n\in \mathbb{N}^+$, such that if $n \geq max\{N_1, N_2\}$, then $\{a_n\}$ and $\{b_n\}$ are $\frac{\epsilon}{2M}$ close to $\{u_n\}$ and $\{v_n\}$, respectively. So we get,
$|a_nb_n - u_nv_n| = |a_nb_n -a_nv_n +a_nv_n - u_nv_n|$
$|a_nb_n - u_nv_n| = |a_n(b_n-v_n) + v_n(a_n - u_n)|$
$|a_nb_n - u_nv_n| \leq |a_n||b_n-v_n| + |v_n||a_n-v_n|$
Now, since $\{a_n\}$ and $\{v_n\}$ are bounded by $M$ it follows that:
$|a_nb_n - u_nv_n| \leq M|b_n-v_n| + M|a_n-v_n|$, and we know that the sequences are eventually $\frac{\epsilon}{2M}$ close so...
$|a_nb_n - u_nv_n| \leq M|b_n-v_n| + M|a_n-v_n| \leq M\frac{\epsilon}{2M} + M\frac{\epsilon}{2M} = \epsilon$.
Therefore $\{a_nb_n-u_nv_n\}$ converges to zero and $\{a_nb_n\} \sim \{u_nv_n\}$.