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Name: Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 14 Maths Textbook Solution.
Uploaded: Sun, 2021-07-04 23:06
Description: Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 14 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 14 Maths Textbook Solution.

Answers (1)

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 $Z\times Z_{0}$