हिंदी

Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b^2} is neither reflexive nor symmetric nor transitive.

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प्रश्न

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.

योग
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उत्तर

(i) Reflexive:

R = {(a, b) : a ≤ b2}

Let a ∈ R

a ≤ a2 which is false.

(a, a) ∉ R

∴ R is not reflexive.

(ii) Symmetric:

Let a, b ∈ R and (a, b) ∈ R

a ≤ b2 and b ≤ a2, which is false.

(a, b) ∈ R, but (b, a) ∉ R

∴ R is not symmetric.

(iii) Transitive:

Let a, b, c ∈ R

Consider (a, b) ∈ R, (b, c) ∈ R

a ≤ b2 and b ≤ c2

a ≤ c2 is false.

(a, c) ∉ R

∴ R is not transitive.

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

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अध्याय 1: Relations and Functions - EXERCISE 1.1 [पृष्ठ ५]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 1 Relations and Functions
EXERCISE 1.1 | Q 2. | पृष्ठ ५

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