Test form 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}
Answers
Answer:
In both cases commutative law hold.
Step-by-step explanation:
We are given that
P = {x : x is a real number between 2 and 7}
Q = {x:x is an irrational number between 2 and 7}
We have to show
1) P∪Q = Q∪P 2) P∩Q = Q∩P
1) P∪Q = Q∪P
L.H.S = P∪Q
Putting values we get
P∪Q = {x : x is a real number between 2 and 7} ∪ {x:x is an irrational number between 2 and 7}
∵ Irrational number ∈ Real numbers
PUQ = {x : x is a real number between 2 and 7}
PUQ = P ......(1)
R.H.S = Q∪P
Q∪P = {x:x is an irrational number between 2 and 7} ∪ {x : x is a real number between 2 and 7}
∵ Irrational number ∈ Real numbers
Q∪P = {x : x is a real number between 2 and 7}
Q∪P = P .......(2)
From (1) and (2)
Hence prove
P∪Q = Q∪P
2) P∩Q = Q∩P
L.H.S = P∩Q
Putting values we get
P∩Q = {x : x is a real number between 2 and 7} ∩ {x:x is an irrational number between 2 and 7}
∵ Irrational number ∈ Real numbers
P∩Q = {x:x is an irrational number between 2 and 7} = Q
P∩Q = Q ......(1)
R.H.S = Q∩P
Q∩P = {x:x is an irrational number between 2 and 7} ∩ {x : x is a real number between 2 and 7}
∵ Irrational number ∈ Real numbers
Q∩P = {x:x is an irrational number between 2 and 7} = Q
Q∩P = Q .......(2)
From (1) and (2)
Hence prove
P∩Q = Q∩P