1. ## Relation Problem !

Let $\displaystyle R$ and $\displaystyle S$ be the relations on z by defined by

$\displaystyle a R b \iff a \equiv b\mod 9$

$\displaystyle a S b \iff a \equiv b \mod 6$

Find $\displaystyle (RUS)^{\infty}$ and $\displaystyle \frac{A}{(RUS)^{\infty}}$

2. Originally Posted by seamoh
> Suppose R and S be the relations on z by defined :a R b &iff; a&equiv;b(mod 9) aSb &iff;a[YOUTUBE]&equiv;[/YOUTUBE]b mod 6) find (RUS)^(infinity )and A/(((RUS))^(infinity));

>

Do you mean find $\displaystyle (R \cup S)^\infty$ as in everything in all the sets in $\displaystyle (R \cup S)$? There are a total of $\displaystyle 19$ such sets.

If $\displaystyle a, b \in R \cup S$ then $\displaystyle a=9r+b=6s+b \Rightarrow 3r=2s$. Thus, $\displaystyle 2|r$ and $\displaystyle 3|s$ (although the $\displaystyle s$ follows from the formula every time you plug in the $\displaystyle r$). So, the elements related to $\displaystyle i \in \mathbb{Z}$ under $\displaystyle R$ and $\displaystyle S$ are the elements, $\displaystyle \{18r+1 : r \in \mathbb{Z} \}$.

What do you mean by the set $\displaystyle A$?