Question

Show that the relation R in the set R of real numbers, defined as

R = {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive.

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Answer 1

Answer:

R = {(a, b): a ≤ b2}

It can be observed that: (see the attachment)

∴R is not reflexive.

Now, (1, 4) ∈ R as 1 < 42

But, 4 is not less than 12.

∴(4, 1) ∉ R

∴R is not symmetric.

Now,

(3, 2), (2, 1.5) ∈ R

(as 3 < 22 = 4 and 2 < (1.5)2 = 2.25)

But, 3 > (1.5)2 = 2.25

∴(3, 1.5) ∉ R

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

Answer 2

$\huge\star\underline{\mathtt\orange{❥a} \mathfrak\blue{n } \mathbb\purple{ Տ} \mathbb\pink{wer}}\star\:$

$R = \{(a, b): a \leq b\} \\ Clearly (a, a) \in R \\ [as a = a] \\ \therefore R is reflexive. \\ Now, (2, 4) \in R (as 2 < 4) \\ But, (4, 2) \notin R as 4 is greater than 2 . \\ \therefore R is not symmetric. \\ Now, let (a, b), (b, c) \in R. \\ Then, a \leq b and b \leq c \\ \implies a \leq c\\ \implies (a, c) \in R \\ \therefore R is transitive. Hence R is reflexive and transitive but not symmetric$