Question

show that for any set A and B
A U (B-A)=(A U B)

Answer 1

Assume \(x\in A\cup (B-A)\). Then \(x\in A\) or \(x\in B-A\).

If \(x\in A\), then \(x\in A\cup B\). If \(x\in B-A\), then you still have that \(x\in A\cup B\).

Thus for any element in \(A\cup(B-A)\) you have that it also belongs to \(A\cup B\), so the LHS is a subset of the RHS, i.e. \(A\cup(B-A)\subseteq A\cup B\).

Now assume \(x\in A\cup B\). Then \(x\in A\) or \(x\in B\).

If \(x\in A\), then \(x\) is in any union containing \(A\), namely \(A\cup (B-A)\). If \(x\in B\), then \(x\in B-A\), so \(x\in A\cup (B-A)\).

Therefore, \(A\cup B\subseteq A\cup(B-A)\).

Since \(LHS\subseteq RHS\) and \(RHS\subseteq LHS\), you have that \(LHS= RHS\).

Answer 2

THIS is gd...but can you show it in a easier way.......like using the properties

Answer 3

You can always rely on this sort of procedure to prove set equality :P

Answer 4

we can understand it by venn diagram too

Answer 5

how?pls