A relation R on the set $A = \{1, 2, 3, \ldots, 13, 14\}$ defined as $R = \{(x,y): 3x - y = 0\}$ is

The set $A = \{1, 2, 3, \ldots, 13, 14\}$

The relation $R = \{(x,y): 3x - y = 0\}$

This means $y = 3x$

A pair $(x,y)$ is in relation $R$ only if the second number is exactly 3 times the first number.

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Finding which pairs from set $A$ satisfy $y = 3x$:

When $x = 1$: $y = 3(1) = 3$ → $(1,3) \in R$

When $x = 2$: $y = 3(2) = 6$ → $(2,6) \in R$

When $x = 3$: $y = 3(3) = 9$ → $(3,9) \in R$

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

When $x = 5$: $y = 3(5) = 15$ (15 is not in set $A$)

Therefore, $R = \{(1,3), (2,6), (3,9), (4,12)\}$

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

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

$3x - x = 0$

$2x = 0$

$x = 0$

Since $0 \notin A$, no element like $(1,1), (2,2), (3,3), \ldots$ is in $R$.

The relation is not reflexive.

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A relation is symmetric if $(x,y) \in R$ implies $(y,x) \in R$.

Consider $(1,3) \in R$. For symmetry, $(3,1)$ should also be in $R$.

Check: $3(3) - 1 = 9 - 1 = 8 \neq 0$

So $(3,1) \notin R$.

The relation is not symmetric.

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A relation is transitive if $(x,y) \in R$ and $(y,z) \in R$ implies $(x,z) \in R$.

Consider $(1,3) \in R$ and $(3,9) \in R$. For transitivity, $(1,9)$ should also be in $R$.

Check: $3(1) - 9 = 3 - 9 = -6 \neq 0$

So $(1,9) \notin R$.

The relation is not transitive.

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The relation $R$ is neither reflexive nor symmetric nor transitive.