Let R = $\{ (n+4,n) \mid n \in \Bbb Z^+\}$, Find $R^2$

Question

I found that this relation is not transitive, does this mean that $R^2$ does not exists?

Any help is appreciated thanks!

Answer 1

The relation $R$ on the set $\mathbb{Z}^+$ is given by $$ R \colon= \left\{ \, (n+4, n) \colon n \in \mathbb{Z}^+ \, \right\}. $$ This $R$ can also be written as $$ R = \big\{ \, (5, 1), (6, 2), (7, 3), (8, 4), \ldots \big\} $$ Thus the relation $R^2 \colon= R\circ R$ is given by $$ R^2 \colon= \left\{ \, (a, b) \in \mathbb{Z}^+ \times \mathbb{Z}^+ \colon \exists \, c \in \mathbb{Z}^+ \mbox{ for which } (a, c) \in R, (c, b) \in R \, \right\}. $$ That is, $$ \begin{align} R^2 &= \left\{ \, (a, b) \in \mathbb{Z}^+\times \mathbb{Z}^+ \colon \exists \, c \in \mathbb{Z}^+ \mbox{ for which } c+4 = a, b+4 = c \, \right\} \\ &= \left\{ \, (a, b) \in \mathbb{Z}^+ \times \mathbb{Z}^+ \colon \exists \, c \in \mathbb{Z}^+ \mbox{ for which } a-4 = c = b+4 \, \right\} \\ &= \left\{ \, (a, b) \in \mathbb{Z}^+ \times \mathbb{Z}^+ \colon a-4 = b+4 \, \right\} \\ &= \left\{ \, (a, b) \in \mathbb{Z}^+ \times \mathbb{Z}^+ \colon a = b+8 \, \right\} \\ &= \left\{ \, (b+8, b) \colon b\in \mathbb{Z}^+ \, \right\}. \end{align} $$

Using the same logic, we can show that, for any $k \in \mathbb{Z}$, we have $$ R^k = \left\{ \, (b+4k, b) \colon b \in \mathbb{Z}^+ \, \right\}. $$

And, then $$ R^* = \bigcup_{k=1}^\infty R^k. $$

Hope this helps.

Answer 2

For $n,m\in\mathbb Z^{+}$ we have:$$nRm\iff n=m+4$$

Then by definition of $R^2$ for $n,m\in\mathbb Z^{+}$ we have: $$nR^2m\iff n=k+4\text{ and }k=m+4\text{ for some }k\in\mathbb Z^{+}$$

This shows immediately that: $$nR^2m\implies n=m+8$$

Conversely if $n,m\in\mathbb Z^{+}$ with $n=m+8$ then $k:=m+4\in\mathbb Z^+$ and this with $n=k+4$.

So we conclude that: $$nR^2m\iff n=m+8$$

Or equivalently that:$$R^2=\{(n+8,n)\mid n\in\mathbb Z^+\}$$