PROBLEM & SOLUTION OF SUBSETS
1) Find all the Subsets of the set B = { 0, 1 }
Ans.) The subsets of A containing one element are { 0 }, { 1 }, also the null set φ and the set B itself are subsets of B.
So, the subsets of the sets B are φ, { 0 }, { 1 }, { 0, 1 } (Ans.)
2) Write all the subsets of the set X = { 1, 2, 3 }
Ans.) The subsets of X containing one element are { 1 }, { 2 }, { 3 }
The subsets of X containing two elements are { 1, 2 }, { 2, 3 }, { 1, 3 }
The null set φ and the set X itself are also subsets of the set A.
So, the subsets of the set X are { φ }, { 1 }, { 2 }, { 3 }, { 1, 2 }, { 2, 3 }, { 1, 3 }, { 1, 2, 3 } (Ans.)
3) Write all the subsets of the set A = { 1, 2, 3, 4 }
Ans.) The subsets of A containing one element are { 1 }, { 2 }, { 3 }, { 4 }
The subsets of A containing two elements are { 1, 2 }, { 2, 3 }, { 1, 3 }, { 1, 4 }, { 4, 2 }, { 3, 4 }.
The subsets of A containing three elements are { 1, 2, 3 }, { 2, 3, 4 }, { 1, 2, 4 }, { 1, 3, 4 }
The null set φ and the set X itself are also subsets of the set X .
so, the subsets of the set A are { φ }, { 1 }, { 2 }, { 3 }, { 4 }, { 1, 2 }, { 2, 3 }, { 1, 3 }, { 1, 4 }, { 4, 2 }, { 3, 4 }, { 1, 2, 3 }, { 2, 3, 4 }, { 1, 2, 4 }, { 1, 3, 4 }, { 1, 2, 3, 4 }. (Ans.)
If, X = { 1, 2, 3, 4 } and Y = { 1, 2, 3, 4, 5, 6 } then every element of X is an element of Y but the elements 5 & 6 of Y are not an elements of X. So, X ⊂ Y
4) If A = { all the students of Oxford University }
And X = { all the student of UK } then find the relation
Then we can describe that as, A ⊂ X .
If X is a finite set containing n elements then the number of proper subset of X = 2ⁿ - 1. (Ans.)
5) The proper subsets of the set A = { 1, 2 } are φ , { 1 } , { 2 }
{ 1, 2 } is a subset of A but not a proper subset,
The number of proper subsets of A = 2² - 1 = 4 – 1 = 3 (Ans.)
6) If A = { 1, 2, 3, 4 } and Q = { x : x – 1, 1< x < 6 , x ∈ N } , then prove P = Q
Ans.) A = { 1, 2, 3, 4 }
Q = { x : x – 1, 1< x < 6 , x ∈ N }
As per given the condition x = 2, 3, 4, 5
So substituting the value of x = { (2 – 1), ( 3 – 1 ), ( 4 – 1 ), ( 5 – 1 ) }
So, the set Q = { 1, 2, 3, 4 }
So, we can say that P = Q (Ans.)
7) If A = { 5, 7, 11, 13, 17 } and B = { 17, 13, 11, 7, 5 }, then prove A = B
Ans.) Two sets A & B are equal if each element of A is an element of B and each element of B is an element of A, in other words, The Set A & B are equal if A is a subset of B and B is a subset of A, that is A ⊆ B and B ⊆ A .
So, both the sets A & B are equal and denoted by A = B (Ans.)
8) Proof P = Q, where P = { letter of the word ROOF } and Q = { letter of the word FOR }
Ans.) As per the given condition, we can find P = { letter of the word ROOF } and Q = { letter of the word FOR }
P = { R, O, F }
Q = { F, O, R }, thus every element of P is a member of Q, or P ⊆ Q
Also, every element of Q is a member of P, or Q ⊆ P, P = Q (Ans.)
9) If P = { x / x = 2n + 1, n ∈ N and n < 8 } and Q = { days in a week } then prove P ↔ Q
Ans.) As per the given condition P = { x/x = 2n + 1, n ∈ N and n < 8 } and Q = { days in a week }
P = { x / x = 2n + 1, n ∈ N and n < 8 } , here n = 1, 2, 3, 4, 5, 6, 7
So, substituting the value of n in the equation x = 2n + 1
= { ( 2.1 + 1), ( 2.2 + 1 ), ( 2.3 + 1), ( 2.4 + 1 ), ( 2.5 + 1 ), (2.6 + 1 ), ( 2.7 + 1 ) }
= { 3, 5, 7, 9, 11, 13, 15 }
Here n(p) = 7
and in Q = { days in a week }
= { Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday }
So, n(Q) = 7
So, we can observe n(P) = n(Q)
So, P ↔ Q (Ans.)
10) If P = { x / x = 2n + 1, n ∈ N and n < 8 } and Q = { days in a week } then prove P = Q
Ans.) As per the given condition P = { x / x = 2n + 1, n ∈ N and n < 8 } and Q = { days in a week }
P = { x / x = 2n + 1, n ∈ N and n < 8 }, here n = 1, 2, 3, 4, 5, 6, 7
So, substituting the value of n in the equation x = 2n + 1
= { ( 2.1 + 1 ), ( 2.2 + 1 ), ( 2.3 + 1 ), ( 2.4 + 1 ), ( 2.5 + 1 ), ( 2.6 + 1 ), ( 2.7 + 1 ) }
= { 3, 5, 7, 9, 11, 13, 15 }
and in Q = { days in a week }
= { Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday }
Monday to Sunday ∈ Q but Monday to Sunday ∉ P,
Therefore P ≠ Q (Ans.)