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I'm stuck with this problem: prove that for all $n_1,n_2 \in \Bbb N^*$ and for all $r_1,r_2 \ \in \Bbb Z$,

$|[r_1] \mod n_1| = |[r_2] \mod n_2|$.

My thoughts: I must find a bijective function $f: \{a \in \Bbb Z : a \cong r_1\mod n_1 \} \to \{b \in \Bbb Z : b \cong r_2\mod n_2 \} $.

I don't know where to begin.

Thanks.

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1 Answer 1

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What is a typical element of $[r_1]$ modulo $n_1$ ? You have a bijection with $\mathbb{Z}$. The same holds for $[r_2]$. Can you conclude ?

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  • $\begingroup$ A typical element is $r_1 + kn_1$, with $k \in \Bbb Z$. Thus, there is a function $g: [r_1] \mod n_1 \to \Bbb Z$ given by $g(r_1)=r_1 + kn_1$...But I don't think this is a bijection. $\endgroup$
    – Lstoi
    May 1, 2014 at 21:31

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