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I’m not sure how to prove this. Specifically, I don’t fully understand congruence class modulo m to prove these sets are disjoint.

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Hint $\ 3 + 9x = 2+6y \iff \color{#c00}1 = 6y-9x = \color{#c00}3(2y-3x)$

Or $\ \ \ n\equiv 2\pmod{\!6}\,\Rightarrow\, n\equiv \color{#0a0}2\pmod{\!3}\ \ $ by $\ \ 2+6j = 2+3(2j)$

but $\,\ \ n\equiv 3\pmod{\!9}\,\Rightarrow\, n\equiv \color{#0a0}3\pmod{\!3}$

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$[2]_6$ is the set of all integers which have $2$ as remainder when divided by $6$, and $[3]_9$ is the set of all integers which have $3$ as remainder when divided by $9$. So an integer $x$ in both congruence classes could be written as $$x=2+6k=3+9\ell$$ which implies $$3-2=1=6k-9\ell.$$ Can you why there is a problem?

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  • $\begingroup$ There was an error in my question. It should have been [3]_9 $\endgroup$ – darylnak Dec 2 '18 at 1:55
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You can think of the congruence class modulo $m$ to be the set of numbers with remainder $n$ when divided by $m$ (where $n<m$).

Once you have that, no integer can have two different remainders, then it must be in at most one of these sets, maybe neither.

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If I understand your notation, $[2]_6$ consists of all the numbers congruent to $2$ modulo $6$. Those are the numbers that leave a remainder of $2$ when you divide by $6$, so $$ [2]_6 = \{ \ldots , -10, -4, 2, 8, 14, \ldots\}. $$ Can you finish now? Write down $[3]_6$ and check whether it and ${2}_6$ have any numbers in common.

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  • $\begingroup$ There was an error in my question. It should have been [3]_9 $\endgroup$ – darylnak Dec 2 '18 at 1:56
  • $\begingroup$ Well write that one down and see if it overlaps $[2]_6$. $\endgroup$ – Ethan Bolker Dec 2 '18 at 2:19

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