# Thread: prove A∩B = B∩A

1. ## prove A∩B = B∩A

Let the sets R,S be defined as following finite sets
R= {xÎZ+ / x is divisible by 2 but less than 20}
S= {yÎ Z+ / y is divisible by 3 but less than 30}
For the sets R and S prove the following set identities
i)A∩B = B∩A
ii)A—B ≠ B—A

2. Hi

Is $R=\{ y\in\mathbb{N} \,|\, y\text{ divisible by }2\text{ and }y\leq20\}$ ?

Originally Posted by naseerhaider
i)A∩B = B∩A
This is a general result : $A\cap B$ is the set which contains all the elements which lie in both $A$ and $B$. What does it mean for $B\cap A$ ?

ii)A—B ≠ B—A
To show that two sets are not equal, you only need to find an element which is in only one of the two sets. Can you find an element of $R-S$ which is not in $S-R$ ? (to help you, you can write down the elements of $R$, of $S$, of $R-S$ and of $S-R$)

Hope that helps

3. Originally Posted by naseerhaider
Let the sets R,S be defined as following finite sets
R= {xÎZ+ / x is divisible by 2 but less than 20}
S= {yÎ Z+ / y is divisible by 3 but less than 30}
For the sets R and S prove the following set identities
i)A∩B = B∩A
ii)A—B ≠ B—A
A = {2.4.6.8.10.12.14.16.18}
B={3.6.9.12.15.18}
intersection symbol is not there so just i say...
left/right hand side -LHS/RHS in (i)
LHS={6.12.18}RHS={6.12.18}
that is the first one is equal.
A-B= {2,3,4,8,9,10,14,15,16}
B-A= {3,9,15}
therefore the second one also proved.