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Decide if the following relation on X is transitive: X = Z, with relation a R b if −7 ≤ a − b ≤ 7

By adding

−7 ≤ a − b ≤ 7 and −7 ≤ b − c ≤ 7

by parts, I found

−14 ≤ a − c ≤ 14

, which gives a clue but is not enough since I need to prove (or disprove) that

−7 ≤ a − c ≤ 7... any ideas? Thanks in advance!

  • So you've noticed that the obvious method of proof doesn't work. Have you tried to disprove it? How might you do that? – user3482749 Nov 14 '18 at 23:45
  • @user3482749 well one obvious way is to try to reach a contradiction starting from −7 ≤ a − c ≤ 7... let me try that and I will come back with the results. Also, to be honest, my gut told me that this was true, that's why I did not try disproving it. –  Nov 14 '18 at 23:54
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    To disprove a thing, you only need to find one counterexample. – user3482749 Nov 15 '18 at 00:04

1 Answers1

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In order to show a relation $R$ is transitive, we must show $aRb$ and $bRc$ implies $aRc$.

Your relation $R$ is equivalent to $aRb$ if $|a - b| \leq 7$, which is not transitive.

Suppose $x = 10, y = 3, z = 0$. Then, we have $xRy$ because $|x - y| = 7 \leq 7$. Also, we have $yRz$ because $|y - z| = 3 \leq 7$. However, we do not have $xRz$ because $|x - z| = 10 > 7.$

Thus, we have a contadiction.

Ekesh Kumar
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