Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 7 Maths Textbook Solution.

Answer: R is an equivalence relation.
Hint: To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
Explanation:
Given: R be a relation on A
$Set\: A\: o\! f\: order\: pair\: o\! f \: integers\: de\! f\! ined\: by\: (x, y) R(u, v)\: if\: xv=yu$
Reflexivity:
$Let (a, b) be\: an\: arbitrary\: element\: o\! f\: the\: set\: A.$
$Then, (a, b) \: \epsilon \: A$
$ab=ba$
$\left (a, b \right )\: R\: (a, b)$
Thus, R is reflexive on A.
Symmetry:
$\text{Let (x, y) and (u, v) }\: \epsilon \: A \: \text{ such that (x, y) R(u, v), Then}$
$xv=yu$
$\Rightarrow vx=uy$
$\Rightarrow uy=vx$
$\left ( u, v \right )R\: (x, y)$
So, R is symmetric on A.
Transitivity:
$Let\left ( x, y\right ), (u, v) and (p, q) \: \epsilon \: R\: such\: that\left ( x, y \right )R (u, v) and\left ( u, v \right ) R (p, q)$
$xv=yu \: \: \: \: \: \: \: \: ...(i)$
$uq=vp \: \: \: \: \: \: ...(ii)$
Multiply eq. (i) and (ii)
$xv\! \times \! uq=yu\! \times\! vp$
$xvuq = yuvp$
cancelling out vu it is common on both sides
$xq=yp$
$\left ( x, y\right ) R (p,q)$
So, R is transitive on A.
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation on A.

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