Which of the following are examples of the singleton set?
(i) $\left\{x: x \in Z, x^{2}=4\right\}$.
(ii) $\{x: x \in Z, x+5=0\}$.
(iii) $\{x: x \in Z,|x|=1\}$.
(iv) $\left\{x: x \in N, x^{2}=16\right\}$.
(v) $\{x: x$ is an even prime number $\}$
(i) Integers $=\ldots-3,-2,-1,0,1,2,3, \ldots$
Given equation:
$x^{2}=4$
$\Rightarrow x=\sqrt{4}$
$\Rightarrow x=\pm 2$
If $x=-2$, then $x^{2}=(-2)^{2}=4$
If $x=2$, then $x^{2}=(2)^{2}=4$
So, there are two elements in a set.
$\therefore$ It is not a singleton set.
(ii) Integers $=-6,-5,-4,-3,-2,-1,0,1,2,3,4, \ldots$
Given equations:
$x+5=0$
$\Rightarrow x+5-5=0-5$
$\Rightarrow x=-5$
So, there is only 1 element in a given set.
$\therefore$ It is a singleton set.
(iii) Integers $=\ldots,-2,-1,0,1,2, \ldots$
Given equation: $|x|=1$
If $x=-1$, then $|x|=|-1|=1$
If $x=1$, then $|x|=|1|=1$
So, there are 2 elements in a given set
$\therefore$ It is not a singleton set.
(iv) Natural Numbers = 1, 2, 3, …
Given equation:
$x^{2}=16$
$\Rightarrow x=\sqrt{16}$
$\Rightarrow x=\pm 4$
$\Rightarrow x=-4,4$
but $x=-4$ not possible because $x \in N$
So, there is only 1 element in a set.
∴ It is a singleton set.
(v) Prime number = 2, 3, 5, 7, 11, …
Even Prime number = 2
∴ It is a singleton set.