Question

Show that the relation R on the set N×N
(a,b)R(c,d) if and only if a+d = b+c is an equivalence relation

Answer 1

let N={1,2,........,9}
Toolbox:

• 1.R is an equivalance relation if R is
• a) reflexive ie (a,b)∈N×N (a,b)R(a,b)
• b) symmetric ie (a,b)R(c,d)=>(c,d)R(a,b) (a,b)(c,a)∈N×N
• c) transitive ie (a,b)R(c,d);(c,d)R(e,f)=>(a,b)R(e,f) (a,b)(c,d),(e,f)∈N×N

A={1,2,3......9}A={1,2,3......9}

R in N×N

(a,b)R(c,d) if (a,b)(c,d)∈N×N

a+b=b+c

Consider (a,b)R(a,b) (a,b)∈N×N

a+b=b+a

Hence R is reflexive

Consider (a,b)R(c,d)given by (a,b)(c,d)∈N×N

a+d=b+c=>c+b=d+a

=>(c,d)R(a,b)

Hence R is symmetric

Let (a,b)R(c,d)and(c,d)R(e,f)

(a,b),(c,d),(e,f),∈N×N

a+b=b+c and c+f=d+e

a+b=b+c

=>a−c=b−d........(1)
c+f=d+e,,,,,,,,,,,,,,,(2)

adding (1) and (2)

a−c+c+f=b−d+d+e
a+f=b+e

=>(a,b)R(e,f)

R is transitive

R is an equivalnce relation