The following is from my lecture notes (trying to understand).
Let $(q_n)$ and $(r_n)$ be two sequences of rational numbers. Let $X=\{(q_n)\in \mathbb{Q}: \forall\varepsilon\in \mathbb{Q}^+, \exists n_\varepsilon\in\mathbb{N} \mbox{ such that } |q_n-q_m|<\varepsilon \mbox{ for } n,m\ge n_\varepsilon\}$, that is $X$ is a set of Cauchy sequences of rationals.
Define an equivalent relation ~ on $X$ as follows:
$(q_n) \sim (r_n)$ if $\forall\varepsilon \in \mathbb{Q}^+, \exists n_\varepsilon\in \mathbb{N} \mbox { such that } |q_n-r_n|<\varepsilon \mbox{ whenever } n\ge n_\varepsilon$.
Now define
$$t_n= \begin{cases} 1 & r_n = 0 \\ 1/r_n & \mbox{otherwise} \\ \end{cases}$$
Then $(r_n)(t_n) \sim (1,1,\dots)$.
The part I don't get is how it can be possible that $(r_n)(t_n) \sim (1,1,\dots)$. For if $r_n=0$ then $(r_n)(t_n)=0$. Take $\varepsilon = 1/2$ and the $\sim$ does not hold.
Would appreciate some clarification.