Let R be 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 $B = \{1, 3, 5\}$.
$R = \{(a, b) : a \in A, b \in B, a < b\} = \{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)\}$.
$R^{-1} = \{(b, a) : (a, b) \in R\} = \{(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)\}$.
Now,$R \circ R^{-1}$ is a relation on $B$ defined as $R \circ R^{-1} = \{(x, z) : \exists y \in A 	ext{ such that } (x, y) \in R^{-1} 	ext{ and } (y, z) \in R\}$.
This is equivalent to $\{(x, z) : \exists y \in A 	ext{ such that } (y, x) \in R 	ext{ and } (y, z) \in R\}$.
For $x, z \in B$,we check pairs $(y, x) \in R$ and $(y, z) \in R$:
If $x=3, z=3$: $y=1$ works since $(1, 3) \in R$. So $(3, 3) \in R \circ R^{-1}$.
If $x=3, z=5$: $y=1$ works since $(1, 3) \in R$ and $(1, 5) \in R$. So $(3, 5) \in R \circ R^{-1}$.
If $x=5, z=3$: $y=1$ works since $(1, 5) \in R$ and $(1, 3) \in R$. So $(5, 3) \in R \circ R^{-1}$.
If $x=5, z=5$: $y=1$ works since $(1, 5) \in R$. So $(5, 5) \in R \circ R^{-1}$.
Thus,$R \circ R^{-1} = \{(3, 3), (3, 5), (5, 3), (5, 5)\}$.

Let $R$ be a relation '$ < $' from $A$ to $B$,where $A = \{1, 2, 3, 4\}$ and $B = \{1, 3, 5\}$ such that $(a, b) \in R \iff a < b$. Then $R \circ R^{-1}$ is:

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