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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

Martin Adam
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  • How do you define $\Bbb N$? If you don’t include $0$, your relation is not reflexive. $R$ is otherwise simply a different notation for $>$ (or $\geq$ if you include $0$). – Clayton Dec 20 '18 at 18:00
  • I added my intended definition of $\mathbb{N}$: it does include $0$. You're right, this is essentially the $\geq$ relation. However, I was wondering whether being able to precisely determine $n_3$ "changed" anything? Maybe I'm just overthinking this. – Martin Adam Dec 20 '18 at 18:09

1 Answers1

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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.

Clayton
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