For each of the following sets A,B prove or disprove whether A \subseteq B and B

ringearV 2021-08-20 Answered
For each of the following sets A,B prove or disprove whether \(\displaystyle{A}\subseteq{B}\) and \(\displaystyle{B}\subseteq{A}\)
a) \(\displaystyle{A}={\left\lbrace{x}\in{Z}:\exists_{{{y}\in{z}}}{x}={4}{y}+{1}\right\rbrace}\)
\(\displaystyle{B}={\left\lbrace{x}\in{Z}:\exists_{{{y}\in{z}}}{x}={8}{y}-{7}\right\rbrace}\)

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Expert Answer

Nathalie Redfern
Answered 2021-08-21 Author has 5858 answers

Step 1
Given: \(\displaystyle{A}={\left\lbrace{x}\in{Z}:\exists_{{{y}\in{z}}}{x}={4}{y}+{1}\right\rbrace}\)
\(\displaystyle{B}={\left\lbrace{x}\in{Z}:\exists_{{{y}\in{z}}}{x}={8}{y}-{7}\right\rbrace}\)
To prove or disprove: \(\displaystyle{A}\subseteq{B},{B}\subseteq{A}\)
Step 2
Counter Example:
Reason: Let \(\displaystyle{5}\in{A}\)
\(\displaystyle{5}={4}{\left({1}\right)}+{1}\)
But \(\displaystyle{5}={8}{y}-{7}\)
\(\displaystyle\Rightarrow{\frac{{{3}}}{{{2}}}}={y}\)
\(\Rightarrow \frac{3}{2} \not{\in} Z\)
\(\Rightarrow 5 \not{\in} B\)
Therefore, \(A \not{\subseteq} B\)
Step 3
Counter Example:
Reason: Let \(\displaystyle{40}\in{B}\)
\(\displaystyle{40}={8}{\left({5}\right)}-{7}\)
But \(\displaystyle{40}={4}{y}+{1}\)
\(\Rightarrow \frac{39}{4}=y\)
\(\Rightarrow \frac{39}{4} \not{\in} Z\)
\(\Rightarrow \not{\in} A\)
Therefore, \(B \not{\subseteq} A\)

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