Relation R on the set $A = \{1, 2, ..., 15\}$ defined as $R = \{(x, y): y - 4x = 0\}$ is

Set $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}$

Relation $R = \{(x, y): y - 4x = 0\}$

The condition $y - 4x = 0$ gives $y = 4x$

$(x, y) \in R$ when $y = 4x$ and both $x, y \in A$.

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Finding all pairs in $R$:

$x = 1$: $y = 4(1) = 4$ (4 is in $A$) → $(1, 4) \in R$

$x = 2$: $y = 4(2) = 8$ (8 is in $A$) → $(2, 8) \in R$

$x = 3$: $y = 4(3) = 12$ (12 is in $A$) → $(3, 12) \in R$

$x = 4$: $y = 4(4) = 16$ (16 is NOT in $A$)

For $x \geq 5$: $y = 4x \geq 20$, which exceeds 15

Therefore $R = \{(1, 4), (2, 8), (3, 12)\}$

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A relation is reflexive if $(a, a) \in R$ for all elements $a \in A$.

For $(a, a)$ to be in $R$:

$a = 4a$

$3a = 0$

$a = 0$

Since $0 \notin A$, no element satisfies this condition.

Checking: $(1, 1) \notin R$ since $1 \neq 4(1) = 4$

$R$ is NOT reflexive.

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A relation is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$.

We have $(1, 4) \in R$.

For $(4, 1)$ to be in $R$: $1 = 4(4) = 16$, which is false.

Therefore $(4, 1) \notin R$.

$R$ is NOT symmetric.

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The relation $R$ is neither reflexive nor symmetric.