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 \u2212 4 =0\n\\\\ \\Rightarrow x^4 =4\n\\\\ \\Rightarrow x^2=\\pm2\n\\\\ \\Rightarrow x=\\pm\\sqrt{\\pm2}\n\\\\ \\Rightarrow x=\\sqrt{2},-\\sqrt{2},\\sqrt{2}i,-\\sqrt{2}i"
And,
"y^4-16=0\n\\\\\\Rightarrow y^4=16\n\\\\\\Rightarrow y^2=\\pm 4\n\\\\\\Rightarrow y=\\pm \\sqrt{\\pm4}\n\\\\\\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.
Comments
Leave a comment