Let R be a relation on Z×Z defined by (a, b) R (c, d) if and only if ad-bc is divisible by 5. Then R is [2024]
(4)
Reflexive: For (a,b)R(a,b)
⇒ab-ab=0 is divisible by 5.
So,(a,b)R(a,b) ∀ a,b∈Z
∴ R is reflexive.
Symmetric : For (a,b)R(c,d), if ad-bc is divisible by 5.
Then, bc-ad is also divisible by 5.
So,(c,d)R(a,b) ∀ a,b,c,d∈Z
∴ R is symmetric.
Transitive : For (a,b)R(c,d)⇒ad-bc is divisible by 5 and (c,d)R(e,f)⇒cf-de is divisible by 5
Let ad-bc=5 k1 and cf-de=5 k2, where k1 and k2 are integers.
∴ adf-bcf=5k1f ...(i)
and cfb-deb=5k2b ...(ii)
Solving (i) and (ii), we get
adf-bcf+cfb-deb=5k1f+5k2b
⇒adf-deb=5(k1f+k2b)⇒d(af-be)=5(k1f+k2b)
af-be is not divisible by 5 for ∀ a,b,c,d,e,f∈Z
So, R is not transitive.
∴ R is Reflexive and symmetric but not transitive.