How to Show that $[2]_6$ and $[3]_9$ are disjoint

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

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

$$[2]_6$$ 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?

• There was an error in my question. It should have been [3]_9 – darylnak Dec 2 '18 at 1:55

You can think of the congruence class modulo $$m$$ to be the set of numbers with remainder $$n$$ when divided by $$m$$ (where $$n).

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

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.

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