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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 $=$ and $=$2012-12-02
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    @Amr is equivalent to but may be not equal.2012-12-02
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    oh i see. I thought you were using = as equivalent2012-12-02
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    Does this not follow immediately from the transitivity property for equivalence relations?2012-12-02

1 Answers 1

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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 $|r_n-\sqrt c_n|<1/n$

Case 2: one of the two sequences does not converge to zero

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 $b'_n=c_n/a'_n$

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    If A=0 , we can't prove that is equivalent to .2012-12-02
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    yes i will edit my answer2012-12-02
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    In case 1 , we can't prove that for all n , $c_n=a_n'b_n'$ .2012-12-02
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    I edited it again !2012-12-02
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    In case 1, how can we prove that $b_n'\rightarrow 0$ ?2012-12-02