Let R = {(a, b) : a = b 2} for all a, b ∈ N.
Show that R satisfies none of reflexivity, symmetry and transitivity.
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Let R = {(a, b) : a = b 2} for all a, b ∈ N.
Show that R satisfies none of reflexivity, symmetry and transitivity.
Let R = {(a, b) : a = b 2} for all a, b ∈ N.
Show that R satisfies none of reflexivity, symmetry and transitivity.
We have, R = {(a, b) : a = b2} relation defined on N.
Now,
We observe that, any element a ∈ N cannot be equal to its square except 1.
⇒ (a,a) ∉ R ∀ a ∈ N
For e.g. (2,2) ∉ R ∵ 2 ≠ 22
⇒ R is not reflexive.
Let (a,b) ∈ R ∀ a, b ∈ N
⇒ a = b2
But b cannot be equal to square of a if a is equal to square of b.
⇒ (b,a) ∉ R
For e.g., we observe that (4,2) ∈ R i.e 4 = 22 but 2 ≠ 42⇒ (2,4) ∉ R
⇒ R is not symmetric
Let (a,b) ∈ R and (b,c) ∈ R ∀ a, b,c ∈ N
⇒ a = b2 and b = c2
⇒ a ≠ c2
⇒ (a,c) ∉ R
For e.g., we observe that
(16,4) ∈ R ⇒ 16 = 42 and (4,2) ∈ R ⇒ 4 = 22
But 16 ≠ 22
⇒ (16,2) ∉ R
⇒ R is not transitive.
Thus, R is neither reflexive nor symmetric nor transitive.