First slide

Introduction to Relations (XI) on Sets

Question

Solution set of $x \equiv 3 (\mod 7), p \in Z,$ is given by

Easy

A

$\{3\}$
B

$\{7 p - 3 : p \in Z\}$
C

$\{7 p + 3 : p \in Z\}$
D

None of these

Solution

$x = 3 (\mod 7) \Rightarrow x - 3 = 7 p, (p \in z)$
$\Rightarrow x = 7 p + 3, p \in z i . e ., \{7 p + 3 : p \in z\} .$

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