Cauchy sequence $(1/n)$ and lack of multiplicative inverse In my textbook, there is an excerpt asking why the all-positive Cauchy sequence $1/n$ (and its equivalence class $[(1/n)]$) does not have a mult. inverse. More specifically it asks why $[(n)]$ isn't the inverse, and I'm not sure why. Term-by-term, multiplying the two sequences $(n \cdot 1/n)$ all give $1$ as their term.
 A: The full claim is that it doesn't have a multiplicative inverse in the equivalence classes of Cauchy sequences. 
The reason why $n$ is not a multiplicative inverse is because it is not Cauchy, since $$|m-n|>1$$
for all $m\neq n$.
But to prove the full claim you need to prove more than that because maybe there is some other sequence that is Cauchy and its class serves as multiplicative inverse.
Assume that $a_n$ is such a sequence. Then $\frac{a_n}{n}\to 1$. Take $\epsilon = 1/2$. Then, there is $N$ such that if $n>N$ then $a_n/n\in[1/2,3/2]$. Since $a_n$ is Cauchy, its tail is bounded: $$|a_m-n_a|<1/2\implies \exists M,\ \forall m>n>M,\ a_m<a_n+1/2$$
Therefore, there are $B,M$ such that if $n>M$, then $a_n <B$. Taking $n>\max(N,M,2B)$ we get that $$B<n/2\leq a_n\leq B$$ which is a contradiction.

A consequence of this is that defining the real numbers as equivalence classes of Cauchy sequences of rational numbers, then the real number $0$, which is the equivalence class of $1/n$, doesn't have an inverse. There is no $1/0$ in this model of the real numbers.
A: When we say that a mathematical object of type $X$ "has a multiplicative inverse", what we mean is that it has a multiplicative inverse that is also of type $X$.
In this specific example, $[(n)]$ is not a Cauchy sequence.
