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Home  >>  CBSE XII  >>  Math  >>  Relations and Functions
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Let n be a fixed positive integer.Define a relation R in Z as follows: a,b E\inE Z, aRb if and only if a-b is divisible by n. Show that R is an equivalence relation.

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Toolbox:
  • A relation R is an equivalnce relation if R is reflexive,symmetric and transitive
  • A relation R in a set A is called $\mathbf{ reflexive},$ if $(a,a) \in R\;$ for every $\; a\in\;A$
  • A relation R in a set A is called $\mathbf{symmetric}$, if $(a_1,a_2) \in R\;\Rightarrow\; (a_2,a_1)\in R \; for \;a_1,a_2 \in A$
  • A relation R in a set A is called $\mathbf{transitive},$ if $(a_1,a_2) \in R$ and $(a_2,a_3) \in R \; \Rightarrow \;(a_1,a_3)\in R$ for all$\; a_1,a_2,a_3 \in A$
Relation R on Z, n a fixed integer.
a R b if and only if a-b is divisible by n a R a is true since a-a=0 is divisible by n
Hence R is reflexive
$ a R b =>b R a$
Since a R b if and only if (a-b) is divisible by n implies b R a if and only if (b-a) is divisible by n
(b-a)=-(a-b) since R is defined in Z
If a-b is divisible by n;-(a-b) is also divisible by n$
Hence R is symmetric
$a R b \in R;b R c \in R$
a-b is divisible by n. and b-c is divisible by n
But $(a-c)=(a-b)+(b-c)$
Hence a-c is also divisible by n
Solution:Therefore R is reflexive,symmetric and transitive
Hence R is an equivalnce relation
answered Mar 3, 2013 by meena.p
edited Mar 26, 2013 by meena.p
 

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