Question

Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(i)\[\left\{ {x:x} \right.is\;an\,{\rm{ }}even\,{\rm{ }}natural\,{\rm{ }}\left. {number\,} \right\}\]
(ii)\[\left\{ {x:x} \right.is\;an\,{\rm{ }}odd\,{\rm{ }}natural\,{\rm{ }}\left. {number\,} \right\}\]
(iii)\[\left\{ {x:x} \right.is\;a\,{\rm{ }}positive\,{\rm{ }}multiple\,{\rm{ }}of\,{\rm{ }}\left. 3 \right\}\]
(iv)\[\left\{ {x:x} \right.is\;a\,{\rm{ }}prime\,{\rm{ }}\left. {number\,} \right\}\]
(v)\[\left\{ {x:x} \right.is\;a\,{\rm{ }}natural\,{\rm{ }}number\,{\rm{ }}divisible\,{\rm{ }}by\,{\rm{ }}3{\rm{ }}\,and\,{\rm{ }}\left. 5 \right\}\]
(vi)\[\left\{ {x:x} \right.is\;a\,{\rm{ }}perfect\,{\rm{ }}\left. {square} \right\}\]
(vii)\[\left\{ {x:x} \right.is\;a\,{\rm{ }}perfect\,{\rm{ }}\left. {cube} \right\}\]
(viii) \[\left\{ {x:x + 5 = 8} \right\}\]
(ix) \[\left\{ {x:2x + 5 = 9} \right\}\]
(x) \[\left\{ {x:x \ge 7} \right\}\]
(xi) \[\left\{ {x:x \in N{\rm{ and\, }}2x + 1 > 10} \right\}\]

Answer 1

Hint: Let us take the set of natural numbers as the universal set consisting of both even and odd numbers. Assume the universal set to be \[U = \left\{ {1,2,3,4,5,6,.....} \right\}\] and then accordingly give the complements for each part.

Complete step-by-step answer:
Let us assume universal set to be \[U = \left\{ {1,2,3,4,5,6,.....} \right\}\] , even natural numbers \[E = \left\{ {2,4,6,8,10.....} \right\}\] and odd natural numbers \[O = \left\{ {1,3,5,7,9.....} \right\}\]
 (i) As we know, \[U = \left\{ {1,2,3,4,5,6,.....} \right\}\] and \[E = \left\{ {2,4,6,8,10.....} \right\}\]
So, complement of even natural number is \[\left\{ {x:x} \right.is\;an\,{\rm{ }}even\,{\rm{ }}natural\,{\rm{ }}{\left. {number} \right\}^\prime }\] = \[{\left\{ {2,4,6,8,10.....} \right\}^\prime }\]
\[ \Rightarrow U - E = \left\{ {1,2,3,4,5,...} \right\} - \left\{ {2,4,6,8,10,...} \right\}\]
\[ \Rightarrow \left\{ {1,3,5,7,9,....} \right\}\]
\[ \Rightarrow \left\{ {x:x} \right.is\;an\,{\rm{ }}odd\,{\rm{ }}natural\,{\rm{ }}\left. {number\,} \right\}\]
(ii) As we know, \[U = \left\{ {1,2,3,4,5,6,.....} \right\}\] and \[O = \left\{ {1,3,5,7,9.....} \right\}\]
So, the complement of odd natural numbers is \[\left\{ {x:x} \right.is\;an\,{\rm{ odd\, }}natural\,{\rm{ }}{\left. {number\,} \right\}^\prime }\] = \[{\left\{ {1,3,5,7,9.....} \right\}^\prime }\]
\[ \Rightarrow U - O = \left\{ {1,2,3,4,5,...} \right\} - \left\{ {1,3,5,7,9,...} \right\}\]
\[ \Rightarrow \left\{ {2,4,6,8,10,....} \right\}\]
\[ \Rightarrow \left\{ {x:x} \right.is\;an\,{\rm{ even\, }}natural\,{\rm{ }}\left. {number\,} \right\}\]
(iii) As we know, \[U = \left\{ {1,2,3,4,5,6,.....} \right\}\] and positive multiple of 3= \[\left\{ {3,6,9,12,15....} \right\}\]
So, complement of positive multiple of 3 is \[\left\{ {x:x} \right.is\;a\,{\rm{ }}positive\,{\rm{ }}multiple\,{\rm{ }}of\,{\rm{ }}{\left. 3 \right\}^\prime }\] = \[{\left\{ {3,6,9,12,15....} \right\}^\prime }\]
\[ \Rightarrow U - \left\{ {3,6,9,12,15,....} \right\}\]
\[ \Rightarrow \left\{ {1,2,3,4,5,....} \right\} - \left\{ {3,6,9,12,15,....} \right\}\]
\[ \Rightarrow \left\{ {1,2,4,5,7,....} \right\}\]
\[ \Rightarrow \left\{ {x:x} \right.is\;{\rm{ }}not\,{\rm{ }}a\,{\rm{ }}positive\,{\rm{ }}multiple\,{\rm{ }}of{\rm{ }}\left. 3 \right\}\]
(iv) As we know, \[U = \left\{ {1,2,3,4,5,6,.....} \right\}\] and prime number= \[\left\{ {2,3,5,7,11,....} \right\}\]
So, the complement of the prime number is \[\left\{ {x:x} \right.is\;a\,{\rm{ }}\,prime\,{\rm{ }}{\left. {number} \right\}^\prime }\]= \[{\left\{ {2,3,5,7,11,....} \right\}^\prime }\]
\[ \Rightarrow U - \left\{ {2,3,5,7,11,....} \right\}\]
\[ \Rightarrow \left\{ {1,2,3,4,5,....} \right\} - \left\{ {2,3,5,7,11,....} \right\}\]
\[ \Rightarrow \left\{ {1,4,6,8,9,....} \right\}\]
\[ \Rightarrow \left\{ {x:x} \right.is\;{\rm{ }}not\,{\rm{ }}a\,{\rm{ }}prime\,\left. {{\rm{ }}number} \right\}\] or \[ \Rightarrow \left\{ {x:x} \right.is\;{\rm{ }}a\,{\rm{ composite}}\left. {{\rm{ }}number} \right\}\]
(v) As we know, \[U = \left\{ {1,2,3,4,5,6,.....} \right\}\] and natural numbers divisible by 3 and 5= \[\left\{ {15,30,45,60,75....} \right\}\]
So, complement of natural number divisible by 3 and 5 is:\[\left\{ {x:x} \right.is\;a\,{\rm{ }}natural\,{\rm{ }}number\,{\rm{ }}divisible\,{\rm{ }}by\,{\rm{ }}\,3\,{\rm{ }}\,and\,{\rm{ }}{\left. 5 \right\}^\prime }\]= \[{\left\{ {15,30,45,60,75....} \right\}^\prime }\]
\[ \Rightarrow U - \left\{ {15,30,45,60,75....} \right\}\]
\[ \Rightarrow \left\{ {1,2,3,4,5,....} \right\} - \left\{ {15,30,45,60,75....} \right\}\]
\[ \Rightarrow \left\{ {1,2,3,4,5,6,....} \right\}\]
\[ \Rightarrow \left\{ {x:x} \right.is\;{\rm{ }}\,not\,{\rm{ }}a\,{\rm{ }}natural\,{\rm{ }}number\,{\rm{ }}divisible\,{\rm{ }}by{\rm{ }}3{\rm{ }}\,and\,{\rm{ }}\left. 5 \right\}\]
(vi) As we know, \[U = \left\{ {1,2,3,4,5,6,.....} \right\}\] and perfect square= \[\left\{ {1,4,9,16,25,....} \right\}\]
So, the complement of a perfect square is \[\left\{ {x:x} \right.is\;a\,{\rm{ }}perfect\,{\rm{ }}{\left. {square} \right\}^\prime }\]= \[{\left\{ {1,4,9,16,25,....} \right\}^\prime }\]

\[ \Rightarrow U - \left\{ {1,4,9,16,25....} \right\}\]
\[ \Rightarrow \left\{ {1,2,3,4,5,....} \right\} - \left\{ {1,4,9,16,25,....} \right\}\]
\[ \Rightarrow \left\{ {2,3,5,6,7,8....} \right\}\]
\[ \Rightarrow \left\{ {x:x} \right.is\;{\rm{ not\, }}a\,{\rm{ }}perfect\,{\rm{ }}\left. {square} \right\}\]
(vii) As we know, \[U = \left\{ {1,2,3,4,5,6,.....} \right\}\] and perfect cube= \[\left\{ {1,8,27,64,125,....} \right\}\]
So, the complement of the perfect cube is \[\left\{ {x:x} \right.is\;a\,{\rm{ }}perfect\,{\rm{ }}{\left. {cube} \right\}^\prime }\]= \[{\left\{ {1,8,27,64,125,....} \right\}^\prime }\]
\[ \Rightarrow U - \left\{ {1,8,27,64,125,....} \right\}\]
\[ \Rightarrow \left\{ {1,2,3,4,5,....} \right\} - \left\{ {1,8,27,64,125,....} \right\}\]
\[ \Rightarrow \left\{ {2,3,4,5,6,7,9....} \right\}\]
\[ \Rightarrow \left\{ {x:x} \right.is\;{\rm{ \,not\, }}a\,{\rm{ }}perfect\,{\rm{ }}\left. {cube} \right\}\]
(viii) As we know, \[U = \left\{ {1,2,3,4,5,6,.....} \right\}\] and \[\left\{ {x:x + 5 = 8} \right\}\] = \[x + 5 = 8\therefore x = 3\]
Hence, \[\left\{ {x:x = 3} \right\}\]
So, complement of \[\left\{ {x:x = 3} \right\}\] is \[{\left\{ {x:x = 3} \right\}^\prime }\]
\[ \Rightarrow U - \left\{ 3 \right\}\]
\[ \Rightarrow \left\{ {1,2,3,4,5,....} \right\} - \left\{ 3 \right\}\]
\[ \Rightarrow \left\{ {1,2,4,5,6,....} \right\}\]
\[ \Rightarrow \left\{ {x:x \ne 3} \right\}\]
(ix) As we know, \[U = \left\{ {1,2,3,4,5,6,.....} \right\}\] and \[\left\{ {x:2x + 5 = 9} \right\}\] = \[2x = 4\therefore x = 2\]
Hence, \[\left\{ {x:x = 2} \right\}\]
So, complement of \[\left\{ {x:x = 2} \right\}\] is \[{\left\{ {x:x = 2} \right\}^\prime }\]
\[ \Rightarrow U - \left\{ 2 \right\}\]
\[ \Rightarrow \left\{ {1,2,3,4,5,....} \right\} - \left\{ 2 \right\}\]
\[ \Rightarrow \left\{ {1,3,4,5,6,....} \right\}\]
\[ \Rightarrow \left\{ {x:x \ne 2} \right\}\]
(x) As we know, \[U = \left\{ {1,2,3,4,5,6,.....} \right\}\] and \[\left\{ {x:x \ge 7} \right\}\] = \[\left\{ {7,8,9,10,11,....} \right\}\]
So, complement of \[\left\{ {x:x \ge 7} \right\}\] is \[{\left\{ {x:x \ge 7} \right\}^\prime }\] = \[{\left\{ {7,8,9,10,11,....} \right\}^\prime }\]
\[ \Rightarrow U - \left\{ {7,8,9,10,11....} \right\}\]
\[ \Rightarrow \left\{ {1,2,3,4,5,6,7,8,9....} \right\} - \left\{ {7,8,9,10,11....} \right\}\]
\[ \Rightarrow \left\{ {1,2,3,4,5,6,.....} \right\}\]
\[ \Rightarrow \left\{ {x:x < 7} \right\}\]
(xi) As we know, \[U = \left\{ {1,2,3,4,5,6,.....} \right\}\] and \[\left\{ {x:x \in N{\rm{ \,and\, }}2x + 1 > 10} \right\}\] = \[2x > 9 = x > 4.5\]
So, complement of \[\left\{ {x:x \in N{\rm{ \,and\, }} 2x + 1 > 10} \right\}\] is \[{\left\{ {x:x \in N{\rm{\, and\, }}2x + 1 > 10} \right\}^\prime }\]
\[ \Rightarrow U - \left\{ {5,6,7,8,9....} \right\}\]
\[ \Rightarrow \left\{ {1,2,3,4,5,....} \right\} - \left\{ {5,6,7,8,9....} \right\}\]
\[ \Rightarrow \left\{ {1,2,3,4.5,.....} \right\}\]
\[ \Rightarrow \left\{ {x:x \le 4.5} \right\}\]

Note: In these types of questions we should first write all the given sets as defined and then do its complement that is we should not include the given sets from the universal set in the answers. This is one of the best ways to solve these kinds of problems.