## VERIFYING COMMUTATIVE AND ASSOCIATIVE PROPERTIES WITH GIVEN SETS

(i) Commutative Property :

(a)  A u B  =  B u A

(Set union is commutative)

(b)  A n B  =  B n A

(Set intersection is commutative)

(ii) Associative Property :

(a)  A u (B u C)  =  (A u B) u C

(Set union is associative)

(b)  A n (B n C)  =  (A n B) n C

(Set intersection is associative)

Question 1 :

Given A  =  {a, x, y, r, s}, B  =  {1, 3, 5, 7, -10}, verify the commutative property of set union.

Solution :

Commutative property of set union

AUB  =  BUA

AUB  =  {a, x, y, r, s} U {1, 3, 5, 7, -10}

AUB  =  {a, x, y, r, s, 1, 3, 5, 7, -10}   ------ (1)

BUA  =  {1, 3, 5, 7, -10} U {a, x, y, r, s}

BUA  =  {a, x, y, r, s, 1, 3, 5, 7, -10}  ------ (2)

(1)  =  (2)

Question 2 :

Verify the commutative property of set intersection for A  =  {l, m, n, o, 2, 3, 4, 7} and B  =  {2, 5, 3, -2, m, n, o, p}

Solution :

commutative property of set intersection

A⋂B  =  B⋂A

A⋂B  =  {l, m, n, o, 2, 3, 4, 7} ⋂ {2, 5, 3, -2, m, n, o, p}

=  {m, n, o}  --- (1)

B⋂A = {2, 5, 3, -2, m, n, o, p} ⋂ {l, m, n, o, 2, 3, 4, 7}

=  {m, n, o}  --- (2)

(1) = (2)

Question 3 :

For A  =  {x|x is a prime factor of 42}, B  =  {x|5 < x ≤ 12, x ∈ N} and C  =  {1, 4, 5, 6}

verify AU(BUC)  =  (AUB)UC.

Solution :

A  =  {x|x is a prime factor of 42}

A  =  {2, 3, 7}

B  =  {x|5 < x ≤ 12, x ∈ N}

B  =  {6, 7, 8, 9, 10, 11, 12}

C  =  {1, 4, 5, 6}

L.H.S

AU(BUC)

(BUC)  =  {6, 7, 8, 9, 10, 11, 12}U{1, 4, 5, 6}

=  {1, 4, 5, 6, 7, 8, 9, 10, 11, 12}

AU(BUC)  =  {2, 3, 7}U{1, 4, 5, 6, 7, 8, 9, 10, 11, 12}

=  {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}  --- (1)

R.H.S

(AUB)UC

(AUB)  =  {2, 3, 7}U{6, 7, 8, 9, 10, 11, 12}

(AUB)  = {2, 3, 6, 7, 8, 9, 10, 11, 12}

(AUB)UC  =  {2, 3, 6, 7, 8, 9, 10, 11, 12} U {1, 4, 5, 6}

(AUB)UC  =  {1, 2, 3, 5, 6, 7, 8, 9, 10, 11, 12}  ----(2)

(1)  =  (2)

Question 4 :

Given P = {a, b, c, d, e} Q = {a, e, i, o, u} and R ={a, c, e, g}. Verify the associative property of set intersection.

Solution :

P∩(Q∩R)  =  (P∩Q)∩R

L.H.S

P∩(Q∩R)

(Q∩R)  =  {a, e, i, o, u} ∩ {a, c, e, g}

Q∩R  =  {a, e}

P∩(Q∩R)  =  {a, b, c, d, e}∩{a, e}

P∩(Q∩R)  =  {a, e} ---- (1)

R.H.S

(P∩Q)∩R

(P∩Q)  =  {a, b, c, d, e} ∩ {a, e, i, o, u}

(P∩Q)  =  {a, e}

(P∩Q)∩R  =  {a, e}∩{a, c, e, g}

(P∩Q)∩R  =  {a, e} ---- (2)

(1)  =  (2)

Hence proved

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