Function - Test of Transitivity Relation R in the set N of natural numbers defined as
R = $\{(x, y): y = x + 5 $and $x < 4\}$
We can make set : (1,6)(2,7)(3,8)
Is this a transitive function please guide..
 A: Transitivity requires that 
$$(x, y), (y, z) \in R \implies (x, z) \in R$$
where $(x, y) \in R$ means $x$ is related to $y$, under the relation defined by $R$.

Counter example: for $R$ on the set of integers:
Let $x = -4,\; y = 1,\; z = 6$
Then $y = x + 5$ and $x\lt 4$, so $(x, y) = (-4, 1) \in R$
And $z = y + 5$ and $y \lt 4$, so $(y, z) = (1, 6) \in R$.
But we do not have that $(x, z) = (-4, 6) \in R$. Why not?
Hence, since $(-4, 1), (1, 6) \in R$, but $(-4, 6) \notin R$. So, $R$ on the set of integers fails to be transitive.

In contrast:
On the set of natural numbers, since we will never have $x, y, z \in \mathbb N$ such that both $(x, y)\in R$ and $(y, z) \in R$ (and we will never have reflexivity nor symmetry), the relation is vacuously transitive. That is, it cannot fail to be transitive on the set of natural numbers. 
A: Transitivity is a property of relations. You don't need it to be a function. Transitivity means that $xRy, yRz \implies xRz.$ So it's transitive if $y=x+5, x<4, z=y+5, y<4 \implies z=x+5, x<4.$
