A binary relation $R$ over a set $X$ is transitive if:
$$\forall a,b,c \in X.(aRb\wedge bRc) \Rightarrow aRc$$
Let me define a particular relation $R$ over $\mathbb{N} = \{0, 1, 2,\dots\}$:
$$aRb := \exists n\in \mathbb{N}. a = b + n$$
Clearly, $R$ is reflexive. However, $R$ seems to be "more" than transitive. In particular, if I know that
$$aRb \wedge bRc$$
then clearly I know $aRc$. To be exact, if I know that
$$(\exists n_1\in \mathbb{N}. a = b + n_1)\ \wedge\ (\exists n_2\in \mathbb{N}. b = c + n_2)$$
then I know:
$$\exists n_3\in \mathbb{N}. a = c + n_3$$
However, I know precisely which $n_3$, namely $n_1 + n_2$. Is there a mathematical notion for this kind of relation?
Many thanks, and happy holidays!
Martin
Your relation is equivalent to $\geq$ over the natural numbers. Indeed, if $a,b\in\Bbb N_0$, where $\Bbb N_0=\{0,1,\ldots\}$, then $$aRb\implies a=b+n\implies a\geq b.$$ On the other hand, if $a\geq b$ for $a,b\in\Bbb N$, then there exists $n\in\Bbb N_0$ such that $a=b+n$, or in your notation, $aRb$.
One question you have is if anything has changed because you know $n_3$ precisely; I claim that you have equally as much information using $\geq$ over the natural numbers. From above, we know the relation and $\geq$ are equivalent, so assume $a\geq b$ and $b\geq c$. This means there exists and $n_1\in\Bbb N$ such that $b=c+n_1$. Additionally, there exists an $n_2$ such that $a=b+n_2$. Substituting in for $b$, we have $$a=b+n_2=c+\underbrace{n_1+n_2}_{n_3}$$
So even in this circumstance, we have as much information regarding $n_3$ as in the relation $R$'s case.