Exercise 2.2 | Class 9 | Advanced Mathematics | Sets

1. Express the following sets both in roster and rule form:
a) Set of even prime numbers
Ans.
A = {2}
A = {\(x: x\) is the even prime number}

b) Set of odd numbers lying between 4 and 20
Ans.
B = {5, 7, 11, 13, 15, 17, 19}
B = {\(x: x = 2n -1, 4 < x < 20; x\in N\)}

c) Set of the positive real roots of the equation \(x^3-2x^2-x +2=0\)
Ans.
C = {2, 1}
C = {\(x: x^3-2x^2-x + 2 = 0; x\in R^+\)}

d) Set of all multiples of 5, which are natural numbers.
Ans.
D = {5, 10, 15, 20, 25, …..}
D = {\(x: x = 5n; x\in N\)}

e) Set of all integers whose square is less than 64
Ans.
E = {\(0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 7\)}
E = {\(x: x^2 < 64; x\in Z\)}



2. Write True or false.
If \(A\) = {\(a, b, c, d, e\)} then,
(i) {\(a\)}\(\notin A\)
Ans. True.
[A set doesn’t belong to another set ]

(ii) {\(a, b\)}\(\in A\)
Ans. False.
[A set doesn’t belong to another set ]

(iii) {\(c, d, e\)}\(\subset A\)
Ans. True.
[A subset itself again a set]

(iv) \(\phi\)\(\in A\)
Ans. False.
[\(\phi\) itself is a set of the power set]

(v) \(\phi\)\(\subset A\)
Ans. True.
[\(\phi\) is a subset of each and every set]

(vi) \(A\)\(\in A\)
Ans. False.
[A set doesn’t belong to another set ]

(vii) {\(e\)}\(\subset A\)
Ans. True.

(viii) \(A\)={\(A\)}
Ans. False.

(ix) {\(0, a\)}\(\subset A\)
Ans. False.
[\(0\notin A\)]

(x) {\(0, a\)}=\(a\)
Ans. False.

(xi) {\(b, c, d\)}={\(c, d, b\)}
Ans. True.
[Order of elements do not matter]

(xii) {\(\phi\)}\(\subset\){\(a\)}
Ans. True.

(xiii) {\(a, d, c\)}={{\(a\)}, {\(d\)}, {\(c\)}}
Ans. False.

(xiv) {\(\phi\)}\(\subset A\)
Ans. False.
[A set doesn’t belong to another set ]

(xv) {\(\phi\)}\(\subset \phi\)
Ans. True.

(xvi) If \(A\subset B\) then \(A^c\subset B^c\)
Ans. False.
For example, if \(A\) = {\(1, 2, 3\)}, \(B\)= {\(1, 2, 3, 4, 5\)}
and \(U\) = {\(1, 2, 3, 4, 5, 6, 7\)}
Then, \(A^c=U-A\)= {\(4, 5, 6, 7\)}
and \(B^c=U-B\)= {\(6, 7\)};
⸫ \(A^c\)is not a subset of \(B^c\).



3. If \(A\subset B\), \(B\subset C\), show that \(A\subset C\).
Solution:
Let us consider an element \(x\) such that \(x\in A\)
then, by definition of subset, \(x\in B\) for all \(x\in A\) ...(i)
Similarly, for the 2nd case, if \(x\in B\)
then, by definition of subset, \(x\in C\) for all \(x\in B\) ...(ii)
Now, combining eqn (i) and (ii), we get,
    \(x\in C\) for all \(x\in A\)
Here, we see that all the elements of the set \(A\) exist in the set \(C\), so we conclude \(A\subset C\).

Alternative Solution
Using Venn diagram, we can show \(A\subset C\) when it is given that \(A\subset B\) and \(B\subset C\).
\(A\subset B\) can be shown as

\(B\subset C\) can be shown as

The given conditions make sure that \(A\subset B\subset C\), which is shown as

\(A\subset B\subset C\) gives us\(A\subset C\).



4. If \(A\) = {\(a, b, c, d\)}, what is \(P(A)\) and \(n[P(A)]\)?
Solution:
Here, \(A\) = {\(a, b, c, d\)}
so, \(n(A) = 4\)
thus, \(n[P(A)] = 2^4 = 16 \)
⸫ \(P(A)\) ={\(\phi\), {\(a\)}, {\(b\)}, {\(c\)}, {\(d\)}, {\(a, b\)}, {\(a, c\)}, {\(a, d\)}, {\(b, c\)}, {\(b, d\)}, {\(c, d\)}, {\(a, b, c\)}, {\(a, b, d\)}, {\(a, c, d\)}, {\(b, c, d\)}, {\(a, b, c, d\)}}



5. Write all the subsets of the sets {\(a\)} and \(\phi\).
Solution:
All the subsets of {\(a\)} are \(\phi\) and {\(a\)};
and all the subsets of \(\phi\) is \(\phi\).



6. Represent the sets in the same Venn diagram:
\(U\) = {\(1, 2, 3, 4, 5, 6, 7, 8\)}, \(A\)= {\(1\)}, \(B\) = {\(1, 4, 7\)}, \(C\) = {\(2, 4, 5, 8\)}
Solution:
Here, \(U\) = {\(1, 2, 3, 4, 5, 6, 7, 8\)}
\(A\)= {\(1\)}, \(B\) = {\(1, 4, 7\)}, \(C\) = {\(2, 4, 5, 8\)}
\(\dagger\) \(A\cap B\) = {\(1\)}
\(\dagger\) \(B\cap C\) = {\(4\)}
\(\dagger\) \(A\cap C=\phi\).
\(\dagger\) \(A\cap C\cap C=\phi\).

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\(\dagger\) It is not supposed to use the concept intersection here. But the way it is done here is quite convenient.
Click Here for Algebraic Operation on Sets.

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