Question

P={x:x is a real number between 2 and 7 } q={x:x is a irrational number between 2 and 7}

Answer 1

Compete question:

Test for the commutative property of union and intersection of the sets

P={x:x is a real number between 2 and 7}

Q={x:x is an irrational number between 2 and 7}

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