Name: Set Theory & Interval Notation
Uploaded: 2019-07-19
Duration: 10 min 50 s
Description: Let's learn the concept of sets, relations, and the types of sets, such as empty or null set, finite set, infinite set, set union and set intersection, etc.

Question

Test for the commutative property of union and intersection of the sets $P = {x : x$ is a real number between $2$ and $7$ } and
$Q = {x : x$ is an irrational number between $2$ and $7$ }.

(Consider a proper single integer value for the rational number or irrational number; Note: Don't take any decimal values)

Accepted Answer

Given sets can be written in the roaster form as follows:
$P = {3, 4, 5, 6}, Q = {\sqrt{3}, \sqrt{5}, \sqrt{6}}$

Then, $P \cup Q = {3, 4, 5, 6} \cup {\sqrt{3}, \sqrt{5}, \sqrt{6}} = {3, 4, 5, 6, \sqrt{3}, \sqrt{5}, \sqrt{6}} . . . . . . . (i)$

$Q \cup P = {\sqrt{3}, \sqrt{5}, \sqrt{6}} \cup {3, 4, 5, 6} = {\sqrt{3}, \sqrt{5}, \sqrt{6}, 3, 4, 5, 6} . . . . . (i i)$

From equations $(i)$ and $(i i)$ , we get

$∴ P \cup Q = Q \cup P$

Hence, union of sets is commutative.

Similarly, $P \cap Q = {3, 4, 5, 6} \cap {\sqrt{3}, \sqrt{5}, \sqrt{6}} = \{\} . . . . . (i i i)$

$Q \cap P = {\sqrt{3}, \sqrt{5}, \sqrt{6}} \cap {3, 4, 5, 6} = \{\} . . . . . . (i v)$

From $(i i i)$ and $(i v)$ , we get

$∴ P \cap Q = Q \cap P$

Thus, intersection of sets is commutative.

Important Questions on Set Language

Learn and Prepare for any exam you want !

Test for the commutative property of union and intersection of the sets P={x:x is a real number between 2 and 7} and Q={x:x is an irrational number between 2 and 7}. (Consider a proper single integer value for the rational number or irrational number; Note: Don't take any decimal values)

Important Questions on Set Language

Learn and Prepare for any exam you want !