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Answer: R is an equivalence relation on$Z\times Z_{0}$
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
$Given: Z\: be\: set\: o\! f \: integers\: and\: Z_{0}\: be\: the\: set\: o\! f \: non-zero \: integers.$
$R=\left \{ ((a, b),( c, d)):ad=bc \right \}\: be\: a\: relation\: on\: Z\times Z_{0}$
Explanation:
Reflexivity:
$(a, b) \: \epsilon \: Z\times Z_{0}$
$ab=ba$
$(a, b), (a, b)) \: \epsilon \: R$
R is reflexive
Symmetric:
$Let\left ( (a, b), (c, d)\right ) \: \epsilon \: R$
$ad=bc$
$cb=da$
$\left ((c,d), (a,b) \right )\epsilon \: R$
R is symmetric
Transitive:
$Let ((a, b), (c, d)) \epsilon \: R\: and\: (c, d), (e, f)) \epsilon \: R$
$ad=bc\: and\: cf=de$
$\frac{a}{b}=\frac{c}{d} \: \:\: \: \: \: ...(i)\: and \: \frac{c}{d} =\frac{e}{f}\: \: \: \: \: ....(ii)$
Using (ii) in (i), we get
$\frac{a}{b}=\frac{e}{f}$
$a\! f=bc$
$((a, b),(e, f)) \epsilon \: R$
R is transitive
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on$Z\times Z_{0}$

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