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So I have the following relation $R \subset \mathbb{Z} \times \mathbb{Z}$: $a\:R \:b$ $\Leftrightarrow$ $a \leq b+1$. I did not have problem proving the relation is reflexive and giving a counterexample the relation is not symetric. Still I have the impression the relation is transitive but I cannot prove this. How do I clearly prove that if $a \leq b+1$ and $b \leq c+1$ then $a \leq c+1$ ? Thanks. :)

Andrés E. Caicedo
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Cos
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2 Answers2

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What if $a=1$, $b=0$, $c=-1$?

Both $aRb$ and $bRc$ hold, but $aRc$ doesn't. Hence $R$ is not transitive.

vadim123
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You have an infinite number of counterexamples. Choose some $c$. Now let $a=c+2$ and $b=c+1$.

Then you have $a=c+2\leq b+1=c+2$ and $b=c+1\leq c+1$ but $a=c+2>c+1$.

Too see why there is room for non-transitivity you can do this:

From $a\leq b+1$ and $b\leq c+1$ you obtain $a-b\leq 1$ and $b-c \leq 1$ and when you add these two you obtain $a-b+b-c=a-c\leq 2$ which is equivalent to $a\leq c+2$.