Question

Let n denote the set of all natural numbers and r be the relation on nn defined as (a,b)r((c,d) iff ad(b+c)=bc(a+d) .Show that r is an equivalence relation.

Answer 1

Answer:

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Need to show reflexive, symmetric and transitive.

*Reflexive*

Putting (c,d) = (a,b) in the definition of r:

(a,b) r (a,b)  <=>  ab(b+a) = ba(a+b)

Since ab(b+a) = ba(a+b) is always true (by commutativity of multiplication and of addition), also (a,b) r (a,b) is always true.

Therefore r is reflexive.

*Symmetric*

Swapping (a,b) and (c,d) in the definition:

(c,d) r (a,b)

<=> cb(d+a) = da(c+b)

<=> da(c+b) = cb(d+a)       ( swapped sides )

<=> ad(b+c) = bc(a+d)       (by commutativity of multiplication and of addition)

<=> (a,b) r (c,d)

Therefore r is symmetric.

*Transitive*

(a,b) r (e,f)   and   (e,f) r (c,d)

=> af(b+e) = be(a+f)   and   ed(f+c) = fc(e+d)

=> abf + aef = abe + bef   and   def + cde = cef + cdf

=> ab ( f - e ) = ( b - a ) ef   and   cd ( f - e ) = ( d - c ) ef

=> ab / ( b - a )  =  ef / ( f - e )  =  cd / ( d - c )

=> ab ( d - c ) = cd ( b - a )

=> abd - abc = bcd - acd

=> abd + acd = abc + bcd

=> ad ( b + c ) = bc ( a + d )

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

Therefore r is transitive.

*Equivalence relation*

Since r is reflexive, symmetric and transitive, r is an equivalence relation.