Question

Let R be a relation on NxN defined by
(a,b) R(c,d) => ad(b+c) = bc (a+d). Check whether
R is ( Reflexive
(i) Transitive
(iv) Equivalence
(in) Symmetric
​

Answer 1

Step-by-step explanation:

Let's do it step by step:

∀(a,b)∈N×N∗,ab=ab

So (a,b)R(a,b) and R is reflexive.

Assume now that (a,b)R(c,d) i.e ad=bc. Multiplication is commutative so we can write cb=da and this gives (c,d)R(a,b). The relation is symmetric

Now take (a,b)R(c,d) and (c,d)R(e,f) ; this means ad=bc and cf=de. Multiply the first equality by f≠0 to get afd=bcf and the second by b≠0 to get bcf=bed. So we have afd=bed and keeping in mind d≠0 we have af=be i.e (a,b)R(e,f) and the relation is transitive

It is therefore an equivalence relation

You need to prove that R is reflexive, symmetric and transitive.

I leave the first two to you. They are very straight forward.

Now suppose (a,b)≃(c,d) and (c,d)≃(e,f)

Then ad=bc and cf=de

Thus

(ad)(cf)=(bc)(de)

and by cancelling from both sides

af=be

Accordingly (a,b)≃(e,f) and R is transitive

Answer 2

Answer:

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Step-by-step explanation:

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