Find the singleton sets and the pair of disjoint sets from the following 1. A = {x |x and x4 − 4 =0} and B = {y | y and x4 − 16 =0 =0}.

Solution:

A = {x |x and x4 − 4 =0} and B = {y | y and y4 − 16 =0 =0}

$x^4 − 4 =0 \\ \Rightarrow x^4 =4 \\ \Rightarrow x^2=\pm2 \\ \Rightarrow x=\pm\sqrt{\pm2} \\ \Rightarrow x=\sqrt{2},-\sqrt{2},\sqrt{2}i,-\sqrt{2}i$

And,

$y^4-16=0 \\\Rightarrow y^4=16 \\\Rightarrow y^2=\pm 4 \\\Rightarrow y=\pm \sqrt{\pm4} \\\Rightarrow y=2, -2, 2i, -2i$

So, $A=\{\sqrt{2},-\sqrt{2},\sqrt{2}i,-\sqrt{2}i \}, B=\{2,-2,2i,-2i\}$

None is the singleton set (having just one element).

$\because A\cap B=\phi$

So, they are disjoint sets.