# Class 11 RD Sharma Solutions – Chapter 1 Sets – Exercise 1.4 | Set 2

**Question 9. Write down all subsets of the following subsets: **

**(i) {a}**

**(ii) {0,1}**

**(iii){a,b,c}**

**(iv){1,{1}}**

**(v) {ϕ}**

**Solution:**

We know, if A is a set and B is a subset of A, then B is a subset of A, then B is called a proper subset of A if B ⊆ A and B≠A, ϕ, illustrated by B ⊆ A or B ⊂ A.

For any set S with n elements, the power set has 2^{n }elements.(i) Since n=1, the power set has 2

^{1 }= 2 elements.The subsets are {a} and ϕ, but the set has no proper subsets.(ii) Since n=2, the power set has 2

^{2}= 4 elements. The elements of the power subset are ϕ, {0}, {1}, {0,1}.(iii)Since n=3, the power set has 2

^{3 }= 8 elements.The elements of the power subset are ϕ, {a}, {b}, {c}, {a,b}, {b,c}, {a,c}, {a,b,c}.(iv) Since n=2, the power set has 2

^{2}= 4 elements. The elements of the power subset are ϕ, {1}, {{1}}, {1,{1}}.(v)Since n=1, the power set has 2

^{1}= 2 elements, which are ϕ and {ϕ}.

**Question 10. Write down all proper subsets of the following subsets: **

**(i) {1,2}**

**(ii) {1,2,3}**

**(iii){1}**

**Solution: **

We know, if A is a set and B is a subset of A, then B is a subset of A, then B is called a proper subset of A if B ⊆ A and B≠A, ϕ, illustrated by B ⊆ A or B ⊂ A.

(i) The proper subsets are {1}, {2}

(ii) The proper subsets are {1}, {2}, {3}, {1,3}, {2,3}, {1,2}

(iii) The subsets are {1} and ϕ, but the set has no proper subsets.

**Question 11. What is the total number of proper subsets of a set consisting of n elements?**

**Solution**

We know that, for any finite set A, having n elements, the total number of subsets of A has 2

^{n}elements. However, the number of proper subsets is 2^{n}– 1, where A is not included.

**Question 12. If A is any set, prove that A ⊆ ∅ <=> A = ∅**

**Solution:**

In order to prove that two sets A and B are equal, we need to show the following:

A ⊆ B and B ⊆ A.

Since, ∅ is a subset of every set, therefore, A ⊆ ∅.

Therefore, ∅ ⊆ A.

Hence, A = ∅

Let us assume now, that A =∅

Now, every set is a subset of itself,

∅ = A ⊆ ∅

Hence, proved.

**Question 13. Prove that A ⊆ B, B ⊆ C**,** and C ⊆ A => A = C**

**Solution:**

We have, A ⊆ B, B ⊆ C and C ⊆ A, therefore, A ⊆ B ⊆ C ⊆ A.

Now, A is a subset of B and B is a subset of C,

So A is a subset of C, that is A ⊆ C.

Also, it is given, C ⊆ A.

We know, if A ⊆ C and C ⊆ A => A = C.

Hence, proved.

**Question 14. How many elements has P(A)=?, if A = ∅.**

**Solution:**

An empty set has zero elements.

Therefore, size of A = n = 0.

Power set of A P(A) = 2

^{n}= 2^{0}= 1 element.

**Question 15. What universal set(s) would you propose for the following? **

**(i) The set of right triangles.**

**(ii) The set of isosceles triangles. **

**Solution:**

(i)The right triangle is a subset of the triangles. Therefore, the set of right triangles is a subset of set of triangles.=> Set of right triangles ⊆ Set of triangles, which becomes Universal set (U) in this case.

(ii)The isosceles triangle is a special case of the triangles where two sides are equal. Therefore, the set of isosceles triangles is a subset of set of triangles.=> Set of isosceles triangles ⊆ Set of triangles, which becomes Universal set (U) in this case.

**Question 16. If X= {8**^{n} – 7n -1 : n ∈ N} and Y = {49(n-1): n ∈ N}. Prove that X ⊆ **Y.**

^{n}– 7n -1 : n ∈ N} and Y = {49(n-1): n ∈ N}. Prove that X

**Solution:**

To prove

X ⊆ Y, we need to show that each element of X belongs to Y.We have,

X= {8

^{n}– 7n -1 : n ∈ N}Y = {49(n-1): n ∈ N}

So, let x ∈ X => x = 8

^{m}– 7m – 1 for some m ∈ N=> x = (1 + 7 )

^{m}– 7m – 1= {n \choose x}

=

=({m \choose 0}1^m + {m \choose 1}1^{m-1} 7 +….+ {m \choose m}7^{m}) – 7m -1

=49({m \choose 2}7^2 + {m \choose 3}7^{3} +…+ {m \choose m}7^{m-2}), m >=2

= 49t

_{m}, m >=2 where t_{m }= {m \choose 2}7^2 + {m \choose 3}7^{3} +…+ {m \choose m}7^{m-2}For m = 1, we have,

X = 8 – 7 x 1 – 1

= 0

Hence, X contains all positive integral multiples of 49.

Hence, Y contains all positive integral multiples of 49 and 0, for n =1.

Therefore,

X ⊆ Y.

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