If R is a relation

Question

(C) Given $A = \{1, 2, 3, 4\}$ and $B = \{1, 3, 5\}$.
$R = \{(a, b) : a \in A, b \in B, a < b\} = \{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)

Vedclass · Accepted Answer

$\{(3, 3), (3, 5), (5, 3), (5, 5)\}$

Vedclass · Answer

(C) Given $A = \{1, 2, 3, 4\}$ and $B = \{1, 3, 5\}$.
$R = \{(a, b) : a \in A, b \in B, a < b\} = \{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)\}$.
The inverse relation $R^{-1} = \{(b, a) : (a, b) \in R\} = \{(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)\}$.
$R \circ R^{-1}$ is defined as $\{(x, z) : \exists y 	ext{ such that } (x, y) \in R^{-1} 	ext{ and } (y, z) \in R\}$.
For $(3, 1) \in R^{-1}$,we look for $y=1$ in $R$: $(1, 3) \in R, (1, 5) \in R \implies (3, 3), (3, 5) \in R \circ R^{-1}$.
For $(5, 1) \in R^{-1}$,we look for $y=1$ in $R$: $(1, 3) \in R, (1, 5) \in R \implies (5, 3), (5, 5) \in R \circ R^{-1}$.
For $(3, 2) \in R^{-1}$,we look for $y=2$ in $R$: $(2, 3) \in R, (2, 5) \in R \implies (3, 3), (3, 5) \in R \circ R^{-1}$.
For $(5, 2) \in R^{-1}$,we look for $y=2$ in $R$: $(2, 3) \in R, (2, 5) \in R \implies (5, 3), (5, 5) \in R \circ R^{-1}$.
For $(5, 3) \in R^{-1}$,we look for $y=3$ in $R$: $(3, 5) \in R \implies (5, 5) \in R \circ R^{-1}$.
For $(5, 4) \in R^{-1}$,we look for $y=4$ in $R$: $(4, 5) \in R \implies (5, 5) \in R \circ R^{-1}$.
Combining these,$R \circ R^{-1} = \{(3, 3), (3, 5), (5, 3), (5, 5)\}$.

If $R$ is a relation $ < $ from $A = \{1, 2, 3, 4\}$ to $B = \{1, 3, 5\}$ i.e.,$(a, b) \in R \iff a < b$,then $R \circ R^{-1}$ is

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