If A and B are sets,then A cap (B - A) is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) By the definition of set difference,$B - A$ is the set of elements that are in $B$ but not in $A$.Therefore,any element $x \in (B - A)$ satisfies

Vedclass · Accepted Answer

$\phi$

Vedclass · Answer

(A) By the definition of set difference,$B - A$ is the set of elements that are in $B$ but not in $A$.Therefore,any element $x \in (B - A)$ satisfies $x 
otin A$.Since $A \cap (B - A)$ consists of elements that are in both $A$ and $(B - A)$,and no element can be in both $A$ and $(B - A)$ simultaneously,the intersection is the empty set.Thus,$A \cap (B - A) = \phi$.

If $A$ and $B$ are sets,then $A \cap (B - A)$ is

Similar Questions

Show that for any sets $A$ and $B$,$A = (A \cap B) \cup (A - B)$ and $A \cup (B - A) = (A \cup B).$

$A - (A - B)$ is

Let $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ and $B = \{2, 3, 5, 7\}$. Find $A \cap B$ and hence show that $A \cap B = B$.

If $X = \{4^n - 3n - 1 : n \in N\}$ and $Y = \{9(n - 1) : n \in N\}$,then $X \cap Y =$

For any two independent events $E_1$ and $E_2$,$P\{(E_1 \cup E_2) \cap (\bar{E}_1 \cap \bar{E}_2)\}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module