# A question about Cauchy sequences

Let $\langle a_n\rangle$ , $\langle b_n\rangle$ , $\langle c_n\rangle$ be Cauchy sequences of rational numbers, and $\langle c_n\rangle$ is equivalent to $\langle a_nb_n\rangle$. Prove or disprove that there are two Cauchy sequences $\langle a_n'\rangle$ , $\langle b_n'\rangle$ of rational numbers such that

(1) $\langle a_n\rangle$ is equivalent to $\langle a_n'\rangle$ ;

(2) $\langle b_n\rangle$ is equivalent to $\langle b_n'\rangle$ ;

(3) $\langle c_n\rangle=\langle a_n'b_n'\rangle$ .

If it is true, can we prove it intuitionistically?

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Why don't you let $<a_n>=<a'_n>$ and $<b_n>=<b'_n>$ –  Amr Dec 2 '12 at 11:32
@Amr <c_n> is equivalent to <a_nb_n> but may be not equal. –  Sencodian Dec 2 '12 at 11:37
oh i see. I thought you were using = as equivalent –  Amr Dec 2 '12 at 11:38
Does this not follow immediately from the transitivity property for equivalence relations? –  Simon Hayward Dec 2 '12 at 12:11

Case 1: both $a_n,b_n$ converge to zero. Let $a'_n=r_n,b'_n=c_n/r_n$. Where $r_n$ is a sequence of rationals such that for all n $|r_n-\sqrt c_n|<1/n$
Let $lim_{n\rightarrow \infty} a_n=A$ . Assume WLOG that A is different from $0$. Let $a'_n$ be a sequence of nonzero rationals converging to $A$, and let $b'_n=c_n/a'_n$
In case 1 , we can't prove that for all n , $c_n=a_n'b_n'$ . –  Sencodian Dec 2 '12 at 12:25
In case 1, how can we prove that $b_n'\rightarrow 0$ ? –  Sencodian Dec 2 '12 at 18:10