Given relation f is defined as

f = {(ab, a + b): a, b ∈ Z}

We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B.

As 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f

i.e., (12, 8), (12, –8) ∈ f

It’s clearly seen that, the same first element, 12 corresponds to two different images (8 and –8).

Therefore, the relation f is not a function.

Answered by Pragya Singh | 11 months agoLet R = {(a, b) : a, b, ϵ N and a < b}.Show that R is a binary relation on N, which is neither reflexive nor symmetric. Show that R is transitive.

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**(i) **Write R in roster form.

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