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}.

Answer 1

Answer:

P=(x: x is a real numbers between 2 and 7

Q = (x: x is an irrational number between 2 and 7 )

We know that P contains all the rational numbers lying between 2 and 7 .

And PUQ and QUP each of them contains all the real numbers between 2 and 7 .

Hence Q is the proper set of P

PnQ and QnP =Q .

Hope it helps u

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Answer 2

Method (1)

• See attachment

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Method (2)

Commutative Property of union of sets

(A U B)' = (B U A)

Here P = {3, 4, 5, 6}, Q = {√3, √5, √6}

P U Q = {3, 4, 5, 6} U { √3, √5, √6 }

P U Q ={ 3, 4, 5, 6 , √3, √5, √6 } ——(1)

Q U P = { √3, √5, √6 } U {3, 4, 5, 6}

Q U P = {√3, √5, √6 , 3, 4, 5, 6 } ——(2)

From (1) and (2)

• PUQ = QUP

It is verified that union of sets is commutative.

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Commutative Property of intersection of sets (P U Q) = (Q U P)

$\sf{P \cap Q = \{3, 4, 5, 6 \} \: \: \bigcap \: \: \{ √3, √5, √6 \} = \{ \} \: \: \: \: \: - - - - (1) }$

$\sf{Q \cap P = \{ √3, √5, √6 \} \: \: \bigcap \: \: \{3, 4, 5, 6 \}= \{ \} \: \: \: \: \: - - - - (2) }$

From (1) and (2)

• $\sf{Q \: \cap \: P = P \: \: \cap \: \: Q }$

It is verified that intersection of sets is commutative.

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Method (3)

• The set Q is proper subset of set P as set of real numbers contain irrational numbers also.
• Hence, P U Q =P and P n Q = Q