Question

1. . Let A = {1,2,3,4,5}, B = {1,3,5,7,9}, and C = {2,6,10,14}.

a. Compute the following sets: A∪B, A∩B, B∪C, B∩C, A\B, B\A.

b. Compute the following sets: A∩(B∪C), (A∩B)∪(A∩C), A∪(B∩C), (A∪B)∩(A∪C).

c. Prove that A∪B = (A\B)∪(A∩B)∪(B\A).

2. Let C0 = {3n : n ∈ Z} = {...,−9,−6,−3,0,3,6,9,...} C1 = {3n+1 : n ∈ Z} = {...,−8,−5,−2,1,4,7,10,...} C2 = {3n+2 : n ∈ Z} = {...,−7,−4,−1,2,5,8,11,...}.

a. Prove that the sets C0, C1, and C2 are pairwise disjoint.

b. Compute C0 ∪C1 ∪C2.

3. Let R>0 be the set of positive real numbers, that is, R>0 = {x ∈ R : x > 0}.

Prove that {e x : x ∈ R} = R>0 and {logx : x ∈ R>0} = R.

Answer 1

1. Given sets are -

A = {1,2,3,4,5} , B = {1,3,5,7,9} , C = {2,6,10,14}

(a) We will find -

We know that '' A union B '' is a set containing all the elements present in set A or set B . So ,

: ''A intersection B '' is a set containing all the elements present in both the sets A and B , so ,

Similarly ,

: is the set of all the elements presents in set B or set C .

AND ,

is the set of all the elements present in both the sets B and C .

But , here we see that there is no any element common in set B and C i.e., present in both the sets B and C .

Thus , the intersection set of sets B and C will be a null set ( empty set ) represented by ''phi''

Then ,

A/B : represents the set containing all those elements which are present in set A but not in set B . So ,

A/B = { 2,4 }

Similarly ,

B/A : represents the set containing all those elements which are presents in set B but not in set A . So ,

B/A = { 7,9 }

(b) Let us consider -

• 

We have , A = {1,2,3,4,5} and ,

, therefore ,

• 

We have ,

and ,

, THEREFORE ,

• 

We have , A = {1,2,3,4,5} and ,

, therefore ,

• ​​​​​​​

We have ,

, and

, therefore ,

(c) We have to prove that -

First of all , let us consider -

...............(i)

Then for RHS , we have -

Thus , we get -

i.e.,

...........(ii)

Thus , from (i) and (ii) , we get -

LHS = RHS

Hence , we have -

PROVED !!