Binary Operations
The union of two sets A and B is denoted A∪B and is defined as A∪B={x|x∈A or x∈B}.
The intersection of two sets A and B is denoted A∩B and is defined as A∩B={x|x∈A and x∈B}.
Two sets are disjoint if they have no elements in common, that is, A and B are disjoint if A∩B=Ø.
The next operation that we will consider is called the intersection of sets. This is the set that we get when we look at elements that the two sets have in common. The intersection of two sets, A and B is the set containing all elements that are in both A and B; the notation for A intersect B is A∩B. So, if x is an element of A and x is an element of B, then x is an element of A∩B.
⛲ Example 1. For the sets C={bear, camel, horse, dog, cat} and D={lion, elephant, horse, dog}, we would get C∩D={horse, dog}.
[Definition] Intersection
The intersection of sets A and B, symbolized by A∩B, is the set containing all the elements that are common to both set A and set B.
The shaded region, region II. in Figure 1 represents the intersection of sets A and B.
⛲ Ex2. Get Ready
E={1,2,3,4,5,6}
F={2,4,6,8,10}
G={1,3,5,7,9}
① Write down the numbers in both E and F.
② Write down the numbers in both E and G.
③ Write down the numbers in both F and G.
🔑 Answers:
① 2,4,6
② 1,3,5
③ none
[Definition] The intersection of two sets S and T is the collection of all objects that are in both sets. It is written S∩T. Using curly brace notation
S∩T={x🙁x∈S) and (x∈T)}
The symbol and in the above definition is an example of a Boolean or logical operation. It is only true when both the propositions it joins are also true. It has a symbolic equivalent ∧. This lets us Write the formal definition of intersection more compactly:
⛲ Ex3: Intersections of sets
Suppose H={1,2,3,5}, I={1,3,4,5}, and J={2,3,4,5}.
Then:
H∩J={2,3,5}, and
I∩J={3,4,5}
⛲ Ex4. Let K={2,4,6,8} and L={6,8,10,12}.
Find K∩L.
✍ Solution: We see that 6, 8 are the only elements which are common to both K and L. Hence K∩L={6,8}.
⛲ Ex5. Let X={Ram, Geeta, Akbar} be the set of students of Class XI, who are in school hockey team. Let Y={Geeta, David, Ashok} be the set of students from Class XI who are in the school football team. Find X∩Y.
✍ Solution: We see that element ‘Geeta’ is the only element common to both. Hence, X∩Y={Geeta}.
⛲ Ex6. Let M={1,2,3,4,5,6,7,8,9,10} and N={2,3,5,7}. Find M∩N and hence show that M∩N=N.
✍ Solution: We have M∩N={2,3,5,7}=N. We note that N⊂M and that M∩N=N.
lntersections
If A and B are sets, then A∩B, read “A intersection B”, is a new set. Its elements are those objects which are in A and in B i.e. those elements which are in both sets.
Example. If O={1,2,3,4} and P={2,4,6,8}, list the elements of the set O∩P.
If A and B and C are sets, their intersection A∩B∩C is the set whose elements are those objects which appear in A and B and C i.e. those elements appearing in all three sets.
⛲ Ex7. If Q={1,2,3,4}, R={2,4,6,8} and S={3,4,5,6}. List the elements of the set Q∩R∩S.
⛲ Ex8. If T={3,5,7,9,11}, V={7,9,11,13} and W={11,13,15},then find
(i) T∩V (ii) V∩W
✍ Solution:
We have, T={3,5,7,9,11}, V={7,9,11,13}, and W={11,13,15}
(i) T∩V={3,5,7,9,11}∩{7,9,11,13}={7,9,11}
(ii) V∩W={7,9,11,13}∩{11,13,15}={11,13}
⛲ Example 9.
(i) If Y={1,3,5,7,9} and Z={2,3,6,8,9}, then find Y∩Z.
(ii) If Y={e,f,g} and Z=Ø, then find Y∩Z.
(iii) If Y={x:x=3n, n∈ℤ} and Z={x:x=4n, n∈ℤ}, then find Y∩Z.
✍ Solution:
(i) Given, Y={1,3,5,7,9} and Z={2,3,6,8,9}
⇒Y∩Z={3,9} [∵3 and 9 are only elements which are common]
(ii) Given, Y={e, f, g} and Z=Ø
⇒Y∩Z=Ø [since, there is no common element]
(iii) Let x∈Y∩Z
⇒x∈Y and x∈Z
⇒x is a multiple of 3 and x is a multiple of 4.
⇒x is a multiple of 3 and 4 both.
⇒x is a multiple of 12.
⇒x=12n, n∈ℤ
Hence, Y∩Z={x:x=12n, n∈ℤ}
⛲ Ex10. If E={x:x is a natural number}, F={x:x is an even natural number},
G={x:x is an odd natural number} and H={x:x is a prime number}, find
(i) E∩F (ii) E∩G (iii) E∩H (iv) F∩G (v) F∩H (vi) G∩H
Answer 7:
E={x:x is a natural number}={1,2,3,4,5,…}
F={x:x is an even natural number}={2,4,6,8,…}
G={x:x is an odd natural number}={1,3,5,7,9,…}
H={x:x is a prime number}={2,3,5,7,…}
(i) E∩F={x:x is a even natural number}=F
(ii) E∩G={x:x is an odd natural number}=G
(iii) E∩H={x:x is a prime number}=H
(iv) F∩G=Ø
(v) F∩H={2}
(vi) G∩H={x:x is odd prime number}
⛲ Ex11: The Intersection of Sets
Given
I={1,2,3,8} J={1,3,6,7,8} K={} |
find
a) I∩J. b) I∩K. c) Ī∩J. d) (I∩J).
Solution:
a) I∩J={1,2,3,8}∩{1,3,6,7,8}={1,3,8}. The elements common to both set I and set J are 1, 3, and 8.
b) I∩K={1,2,3,8}∩{}={}. There are no elements common to both set I and set K.
c) To determine Ī∩J. we must first determine Ī.
Ī∩J={4,5,6,7,9,10}∩{1,3,6,7,8}={6,7}
d) To find (I∩J), first determine I∩J.
(I∩J)={1,3,8}={2,4,5,6,7,9,10}
If A and B are sets and A∩B=Ø then we say that A and B are disjoint, or disjoint sets.
⛲ Ex12. Find the intersection of each pair of sets:
(i) X={1,3,5}, Y={1,2,3}
(ii) L={a,e,i,o,u}, M={a,b,c}
(iii) N={x:x is a natural number and multiple of 3} O={x:x is a natural number less than 6}
(iv) P={x:x is a natural number and 1<x≤6} Q={x:x is a natural number and 6<x<10}
(v) R={1,2,3}, S=Ø
Answers:
(i) X={1,3,5}, Y={1,2,3}
X∩Y={1,3}
(ii) L={a,e,i,o,u}, M={a,b,c}
L∩M={a}
(iii) N={x:x is a natural number and multiple of 3}=(3,6,9,…}
O={x:x is a natural number less than 6}={1,2,3,4,5}
∴ N∩O={3}
(iv) P={x:x is a natural number and 1<x≤6}={2,3,4,5,6}
Q={x:x is a natural number and 6<x<10}={7,8,9}
P∩Q=Ø
(v) R={1,2,3}, S=Ø, So,R∩S=Ø
⛲ Ex13. If T={3,5,7,9,11}, V={7,9,11,13}, W={11,13,15} and Y={15,17}; find
(i) T∩V
(ii) V∩W
(iii) T∩W∩Y
(iv) T∩W
(v) V∩Y
(vi) T∩(V∪W)
(vii) T∩Y
(viii)T∩(V∪Y)
(ix) (T∩V)∩(V∪W)
(x) (T∪Y)∩(V∪W)
Answers:
(i) T∩V={7,9,11}
(ii) V∩W={11,13}
(iii) T∩W∩Y=(T∩W}∩Y={11}∩{15,17}=Ø
(iv) T∩W={11}
(v) V∩Y=Ø
(vi) T∩(V∪W)=(T∩V)∪(T∩W)={7,9,11}∪{11}={7,9,11}
(vii) T∩Y=Ø
(viii) T∩(V∪Y)=(T∩V)∪(T∩Y)={7,9,11}∪Ø={7,9,11}
(ix) (T∩V)∩(V∪W)={7,9,11}∩{7,9,11,13,15}={7,9,11}
(x) (T∪Y)∩(V∪W)={3,5,7,9,11,15,17}∩{7,9,11,13,15}={7,9,11,15}