CUET UG 2025 — Mathematics Algebra

The set is $\{1, 2, 3\}${1,2,3} and the relation is $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 3)\}$R={(1,1),(2,2),(3,3),(1,2),(1,3),(2,3)}.

A relation is reflexive if every element is related to itself. This requires $(1,1)$(1,1), $(2,2)$(2,2), and $(3,3)$(3,3) to be present.

$(1,1)$(1,1) is present

$(2,2)$(2,2) is present

$(3,3)$(3,3) is present

$R$R is reflexive.

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A relation is symmetric if whenever $(a,b)$(a,b) exists, $(b,a)$(b,a) must also exist.

$R$R contains $(1,2)$(1,2) but not $(2,1)$(2,1)

$R$R contains $(1,3)$(1,3) but not $(3,1)$(3,1)

$R$R is not symmetric.

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A relation is transitive if whenever $(a,b)$(a,b) and $(b,c)$(b,c) exist, $(a,c)$(a,c) must also exist.

$(1,2)$(1,2) and $(2,3)$(2,3) exist, need $(1,3)$(1,3) - present

$(1,2)$(1,2) and $(2,2)$(2,2) exist, need $(1,2)$(1,2) - present

$(1,1)$(1,1) and $(1,2)$(1,2) exist, need $(1,2)$(1,2) - present

$(1,1)$(1,1) and $(1,3)$(1,3) exist, need $(1,3)$(1,3) - present

$(2,2)$(2,2) and $(2,3)$(2,3) exist, need $(2,3)$(2,3) - present

$(2,3)$(2,3) and $(3,3)$(3,3) exist, need $(2,3)$(2,3) - present

All required pairs exist.

$R$R is transitive.

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Therefore, $R$R is reflexive and transitive.