UPSC NDA I 2015 Mathematics Paper

Concept:The relation $R$ on $\mathbb{Z}$ defined by $aRb \iff (a-b)$ is divisible by $5$ is an equivalence relation. Its equivalence classes are the residue classes modulo $5$.Explanation:Check the three properties of an equivalence relation:Reflexive: For any $a \in \mathbb{Z}$, $a-a = 0$ is divisible by $5$. So $aRa$ holds.Symmetric: If $aRb$, then $a-b$ is divisible by $5$. Then $b-a = -(a-b)$ is also divisible by $5$, so $bRa$ holds.Transitive: If $aRb$ and $bRc$, then $a-b$ and $b-c$ are each divisible by $5$. Their sum $(a-b)+(b-c)=a-c$ is also divisible by $5$, so $aRc$ holds.Thus $R$ is an equivalence relation.An equivalence relation partitions the set into disjoint equivalence classes. Here, $aRb$ means $a \equiv b \pmod{5}$, so the equivalence classes are:$\{5k\}, \{5k+1\}, \{5k+2\}, \{5k+3\}, \{5k+4\}$ for $k \in \mathbb{Z}$.There are exactly five such classes, and they are pairwise disjoint and cover all integers. Therefore statement 1 is correct.Statement 2 is a fundamental property of equivalence classes: any two distinct equivalence classes are disjoint; if they are not disjoint they are equal. Hence statement 2 is also correct.Answer:Both statements 1 and 2 are correct. The correct option is C.