if one is bounded the other one is also bounded... If $(a_n)$ is a bounded sequence of positive rational numbers and equivalent with sequence $ (b_n)$, show that $ (b_n)$ is also bounded. 
I was thinking that since $(a_n)$ and $ (b_n)$ are equivalent then $|(a_n)-(b_n)|$< $\epsilon$
 A: By definition, $a_n\underset{n\to+\infty}\sim b_n$ means that there exists a sequence $(\varphi_n)_{n\in\mathbb{N}}$ such that $\lim\limits_{n\to+\infty}\varphi_n=1$ and:
$$\forall n\in\mathbb{N},\quad b_n=\varphi_n a_n.$$
Since $\lim\limits_{n\to+\infty}\varphi_n=1$, the sequence $(\varphi_n)_{n\in\mathbb{N}}$ is bounded, i.e., there exists $M\in\mathbb{R}_+$ such that
$$\forall n\in\mathbb{N},\quad\lvert\varphi_n\rvert\leq M.$$
By assumption, the sequence $(a_n)_{n\in\mathbb{N}}$ is bounded, hence there exists $R\in\mathbb{R}_+$ such that
$$\forall n\in\mathbb{N},\quad\lvert a_n\rvert\leq R.$$
Now, for all $n\in\mathbb{N}$ one has:
$$\lvert b_n\rvert=\lvert\varphi_na_n\rvert\leq RM.$$
Hence the sequence $(b_n)_{n\in\mathbb{N}}$ is bounded.
Note. The hypothesis that the sequence $(a_n)$ is a sequence of positive rationals is useless. It works for any sequence of complex numbers (or even in any (semi-)normed vector space with appropriate definition of equivalent sequences).
A: If you're defining two sequences as equivalent if $\lim_{n\rightarrow{\infty}}\frac{a_n}{b_n}=1$, then for every $\epsilon>0$, $|a_n-b_m|<\epsilon$ for some $n,m\in\mathbb{N}$. If $\{a_n\}$ is bounded, then for some $M$, $a_n\le{M}$ for all $n\in\mathbb{N}$. If $\{b_n\}$ was unbounded, then you could easily find $m\in\mathbb{N}$ sufficiently large that $|a_n-b_m|\ge|M-b_m|>\epsilon$ for any fixed $\epsilon>0$. Therefore, $\{b_n\}$ must also be bounded.
