# intersection and union proof

• Feb 25th 2010, 10:11 AM
Laurali224
intersection and union proof
Prove by contradiction or otherwise that (A intersect B)=(A intersect C) and (A union B)=(A union C) if and only if B=C.

SO... I know we have to show this both ways since it's an iff.
Going the first way- Let (A intersect B)=(A intersect C) and (A union B)=(A union C). then for any x in A intersect B, x is in A intersect C. and for any x in A union b, x is in A union C. For x in A intersect B, x is in A and x is in B. for X in A intersect C, x is in A and x is in C. Since x is in A, B, C, and A intersect B equals A intersect C, then sets B and C must be equal.

For A union B and A union C... is where I get a little confused... and also going the other way, letting B=C and proving these two things are true.
• Feb 25th 2010, 01:59 PM
Drexel28
Quote:

Originally Posted by Laurali224
Prove by contradiction or otherwise that (A intersect B)=(A intersect C) and (A union B)=(A union C) if and only if B=C.

SO... I know we have to show this both ways since it's an iff.
Going the first way- Let (A intersect B)=(A intersect C) and (A union B)=(A union C). then for any x in A intersect B, x is in A intersect C. and for any x in A union b, x is in A union C. For x in A intersect B, x is in A and x is in B. for X in A intersect C, x is in A and x is in C. Since x is in A, B, C, and A intersect B equals A intersect C, then sets B and C must be equal.

For A union B and A union C... is where I get a little confused... and also going the other way, letting B=C and proving these two things are true.

So $A\cap B=A\cap C$ and $A\cup B=A\cup C$.

Let $x\in B$ then $x\in A\cup B\implies x\in A\cup C$. Suppose then that $x\notin C$ then $x\notin A\cap C$ and thus $x\notin A\cap B\implies x\notin B$. Contradiction.

Similarly, let $x\in C$ then $x\in A\cup C\implies x\in A\cup B$. Suppose that $x\notin B$ then $x\notin A\cap B\implies x\notin A\cap C$ and so $x\notin C$. Contradiction