Defintion: Two sequences $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ of rationals are said to be equivalent if for every $\varepsilon >0$ there exists an $N\ge 1$ such that for all $n>N$ we have that $|a_n-b_n|<\varepsilon$.
Exercise: Show that if $(a_n)_{n=1}^\infty = (b_n)_{n=1}^\infty$ are equivalent sequences of rationals, then $(a_n)_{n=1}^\infty$ is a Cauchy sequence if and only if $(b_n)_{n=1}^\infty$ is a Cauchy sequence.
Proof: Given that $(a_n)_{n=1}^\infty = (b_n)_{n=1}^\infty$. Suppose that $(a_n)_{n=1}^\infty$ is a Cauchy sequence. Then we have the following three inequalities.
For every $\varepsilon >0$ there exists an $N_1$ such that for all $j,k>N_1$ we have that $|a_k-a_j|<\frac \varepsilon 3$.
For every $\varepsilon >0$ there exists an $N_2$ such that for all $j,k>N_2$ we have that $|a_j-b_j|<\frac \varepsilon 3$ and $|b_k-a_k|<\frac \varepsilon 3$.
Let $N=max\{N_1,N_2\}$. Then for all $j,k>N$ the above inequalities hold. Adding the three inequalities and using the triangle inequality twice we get $|a_j-b_j+b_k-a_k+a_k-a_j|<\varepsilon$. From which it follows $|b_k-b_j|<\varepsilon$. Thus, for every $\varepsilon>0$ there exists an $N\ge 1$ such that for all $j,k>N$ we have that $|b_k-b_j|<\varepsilon$. Hence, $(b_n)_{n=1}^\infty$ is a Cauchy sequence.
By a similar argument, we can prove the implication in the other direction.
I am new to Analysis. Is this proof correct?