From the sets given below, select equal sets:
$A=\{2,4,8,12\}, B=\{1,2,3,4\}, C=\{4,8,12,14\}, D=\{3,1,4,2\}$
$E=\{-1,1\}, F=\{0, a\}, G=\{1,-1\}, H=\{0,1\}$
$A=\{2,4,8,12\} ; B=\{1,2,3,4\} ; C=\{4,8,12,14\}$
$D=\{3,1,4,2\} ; E=\{-1,1\} ; F=\{0, a\}$
$G=\{1,-1\} ; H=\{0,1\}$
It can be seen that
$8 \in A, 8 \notin B, 8 \notin D, 8 \notin E, 8 \notin F, 8 \notin G, 8 \notin H$
$\Rightarrow A \neq B, A \neq D, A \neq E, A \neq F, A \neq G, A \neq H$
Also, $2 \in A, 2 \notin C$
$\therefore A \neq C$
$3 \in B, 3 \notin C, 3 \notin E, 3 \notin F, 3 \notin G, 3 \notin H$
$\therefore B \neq C, B \neq E, B \neq F, B \neq G, B \neq H$
$12 \in C, 12 \notin D, 12 \notin E, 12 \notin F, 12 \notin G, 12 \notin H$
$\therefore C \neq D, C \neq E, C \neq F, C \neq G, C \neq H$
$4 \in D, 4 \notin E, 4 \notin F, 4 \notin G, 4 \notin H$
$\therefore D \neq E, D \neq F, D \neq G, D \neq H$
Similarly, $E \neq F, E \neq G, E \neq H$
$F \neq G, F \neq H, G \neq H$
The order in which the elements of a set are listed is not significant.
$\therefore B=D$ and $E=G$
Hence, among the given sets, $B = D$ and $E = G$.
Two finite sets have $m$ and $n$ elements. The total number of subsets of the first set is $56$ more than the total number of subsets of the second set. The values of $m$ and $n$ are
Match each of the set on the left described in the roster form with the same set on the right described in the set-builder form:
$(i)$ $\{ P,R,I,N,C,A,L\} $ | $(a)$ $\{ x:x$ is a positive integer and is adivisor of $18\} $ |
$(ii)$ $\{ \,0\,\} $ | $(b)$ $\{ x:x$ is an integer and ${x^2} - 9 = 0\} $ |
$(iii)$ $\{ 1,2,3,6,9,18\} $ | $(c)$ $\{ x:x$ is an integer and $x + 1 = 1\} $ |
$(iv)$ $\{ 3, - 3\} $ | $(d)$ $\{ x:x$ is aletter of the word $PRINCIPAL\} $ |
List all the elements of the following sers :
$F = \{ x:x$ is a consonant in the Englishalphabet which precedes $k\} $
Let $A=\{1,2,\{3,4\}, 5\} .$ Which of the following statements are incorrect and why ?
$\varnothing \subset A$
Write the following as intervals :
$\{ x:x \in R, - 4\, < \,x\, \le \,6\} $