Given,
(x, y) R x + 2y = 8 where x ∈ N and y ∈ N
x + 2y= 8
x = 8 – 2y
Putting the values y = 1, 2, 3,…… till x ∈ N
When, y = 1, x = 8 – 2(1) = 8 – 2 = 6
When, y = 2, x = 8 – 2(2) = 8 – 4 = 4
When, y = 3, x = 8 – 2(3) = 8 – 6 = 2
When, y = 4, x = 8 – 2(4) = 8 – 8 = 0
Now, y cannot hold value 4 because x = 0 for y = 4 which is not a natural number.
∴ R = {(2, 3), (4, 2), (6, 1)}
R‑1 = {(3, 2), (2, 4), (1, 6)}
Answered by Sakshi | 1 year 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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