Let R and S are "Congruence modulo 4" and "Congruence modul0 6" relations respectively on set of integers, R={(a,b)| a=b(mod 4), S={(a,b)| a=b(mod 6)

Question

Let R and S are "Congruence modulo 4" and "Congruence modul0 6" relations respectively on set of integers, R={(a,b)| a=b(mod 4), S={(a,b)| a=b(mod 6). Calculate:- 1. R union S 2. R intersection S 3. R-S 4. S-R 5. R XOR S "=" above means congruence modulo. I didn't understand this question at all. I'm new to relations. Also, explain why do we have to take LCM of 4 and 6. Saw the solution

Answer 1

For example:

$$R\cup S=\left\{\;(a,b)\in\Bbb Z\times\Bbb Z\;/\;a=b\pmod 4\;\;\color{red}{\text{or}}\;\;a=b\pmod 6\;\right\}$$

So for example, we have that

$$(3,7)\,,\,(-5,1),\,(14,18)\in R\cup S\;,\;\;\text{but}\;\;(1,3),\,(-3,2)\notin R\cup S$$

Or also

$$R\cap S=\left\{\;(a,b)\in\Bbb Z\times\Bbb Z\;/\;a=b\pmod 4\;\;\color{red}{\text{and also}}\;\;a=b\pmod 6\;\right\}$$

For example,

$$(0,12),\,(-35,-23), (-11,1),\,(5,17)\in R\cap S\;,\;\;\text{but}\;\;(0,4),\,(0,6),\,(-4,5),\,(2,15)\notin R\cap S$$

You now must try the rest by yourself. There are thousands of sites in the web which deal with this basic and very important stuff. Just google "relations on sets" or something like that, and of course scores of books, too.