Show that the relation R on the set Z of all integers,given by  R={(a,b):2 divides(a-b)} is an equivalence relation.

 

 

 

 
 
 
 
 

Answers (1)

   (i) Reflexive:  \left ( a,a \right )\epsilon R           given: R= \left \{ \left ( a,b \right ):2\: divides\: a-b \right \}
  \frac{a-a}{2}= \frac{0}{2}= 0
\therefore 2\: divides\: a-a\: \: \therefore \left ( a,a \right )\epsilon R\Rightarrow R \: is \: reflexive
(ii) symnatic:  \left ( a,b \right )\epsilon R\: \: \Rightarrow \: \left ( b,a \right )\epsilon R
\frac{a-b}{2}= p\: \: \: p\epsilon Z
a-b= 2p\: \: \: \: \Rightarrow b-a= 2\left ( -p \right )\: \: -p\epsilon Z
\frac{b-a}{2}= -p
2 divides      b-a \Rightarrow \: \left ( b,a \right )\epsilon R
         \therefore R \: is\: symnatic
(iii) Transitive: \left ( a,b \right )\epsilon R,\: \: \left ( b,c \right )\epsilon R\: \: \Rightarrow \left ( a,c \right )\epsilon R
\frac{a-b}{2}= p---(1)             \frac{b-c}{2}= q            p,q\epsilon Z
Add (1)& (2)
\frac{a-b}{2}+\frac{b-c}{2}= p+q
\frac{a-b+b-c}{2}= p+q              \Rightarrow \frac{a-c}{2}= p+q
2 divides  a-c                                \therefore p+q\epsilon Z
             \therefore \left ( a,c \right )\epsilon R
\Rightarrow  R is transitive
Hence R is equivalence relations

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