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

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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Nathalie Redfern

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, $$B \not{\subseteq} A$$