1.

Let `N`denote the set of all natural numbers and R be the relation on `NxN`defined by `(a , b)R(c , d) a d(b+c)=b c(a+d)dot`Check whether R is an equivalence relation on `NxNdot`

Answer» Reflexive
Since, `(a,b)R(a,b)iffab(b+a)=ba(a+b),AAa,binN` is true.
Hence, R is reflexive.
Symmetric (a, b) R (c, d)
`iff ad(b+c)=bc(a+d)`
`iff bc(a+d)=ad(b+c)`
`iff cb(d+a)=da(c+b)`
`iff (c,d)R(a,b)`
Hence, R is symmetric.
Transitive
Since, `(a,b)R(c,d)iffad(b+c)=bc(a+d)`
`iff (b+c)/(bc)=(a+d)/(ad)`
`if (1)/(c)+(1)/(b)=(1)/(d)+(1)/(a)`
`iff (1)/(a)-(1)/(b)=(1)/(c)-(1)/(d)`
`therefore (a,b)R(c,d)iff(1)/(a)-(1)/(b)=(1)/(c)-(1)/(d)" ... (i)"`
and similarly (c,d) R (e,f) `iff (1)/(c)-(1)/(d)=(1)/(e)-(1)/(f)" ... (ii)"`
From Eqs. (i) and (ii),
`(a,b)R(c,d)and(c,d)R(e,f)iff(1)/(a)-(1)/(b)=(1)/(e)-(1)/(f)iff(a,b)R(e,f)`
So, R is transitive. Hence, R is an equivalence relation.


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