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Hello Mathematics Community,

I have the relation:

$$R=\{(a,b)\in \mathbb{R} \times \mathbb{R} : |a-b|<1\}.$$

Is it true that this relation is not transitive, because:

Let $a=0, b=0$ and $c=0$:

$$ |0-0| < 1 \Rightarrow |0-0|<1$$

is true.

But if $a=b$ or $a,b \geq1$ and $c\geq1$ therefore this is not true. So the relation is not transitive.

sincerely, M.Hisoka

Juniven Acapulco
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M.Hisoka
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    It is not transitive, but you need to organize your argument. Give a concrete example, like $a=0$, $b=1/2$, $c=1$. Then, $|a-b|<1$ and $|b-c|<1$ but $|a-c|=1$. – olsen5 Oct 30 '17 at 12:09

1 Answers1

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As noted by olsen5 in a comment:

The relation is transitive, but your argument is a bit haphazard. To show a relation is not transitive, a simple counterexample suffices. Observe that $(0,1/2),(1/2,1) \in R$: this is because

$$\left| 0 - \frac 1 2 \right| = \left| \frac 1 2 - 1 \right| = \frac 1 2 < 1$$

However, $(0,1) \not \in R$ because $|0-1| = 1 \not < 1$.

Thus, in the framework of the variables prescribed, when we want to show that $(a,b),(b,c) \in R$ do not necessarily imply $(a,c) \in R$, take $a = 0, b = 1/2, c = 1$.

Mostly just posting this in order to get this answer out of the unanswered queue, and posting as Community Wiki since I have nothing further to offer.

PrincessEev
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