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Let $X = \{0, 1, 2, ... , 10\}$, Define the relation $R$ on $X$ by, for all $a, b \in X$, $aRb$ if and only if $a + b = 10$.

Is $R$ reflexive? symmetric, transitive? Give reasons.


Here are my answers, please see if I made any mistakes?

$R$ is not reflexive, because there exists $a \in X$ such that $a$ does not relate $a$. For example, let $a=1$, $1+1=2$ which is not equal to $10$.

$R$ is symmetric, because $a + b = b + a$. sum of integers are symmetric. So $4R6$ and $6R4$.

$R$ is not transitive, because there are $a, b, c$ integers such that $aRb$ $bRc$ but $a$ does NOT relate $c$. Let $a = 4, b = 6$ and $c = 4$. Then, $4R6$ and $6R4$ but $4$ does NOT relate $4$.

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  • $\begingroup$ Everything looks good, but there's a typo in the last sentence: you probably meant "Then $4R6$ and $6R4$..." $\endgroup$ – coldnumber Aug 16 '15 at 19:41
  • $\begingroup$ @coldnumber Thanks - fixed that. My doubt is about the last part, because a=c, so is that still not transitive? $\endgroup$ – user3282081 Aug 16 '15 at 19:43
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    $\begingroup$ Your example works even if $a=c$. Even if we keep the elements abstract, $aRb, bRa \implies aRa$ for a transitive relation (note that this doesn't imply reflexivity because you may not always have such pairs). $\endgroup$ – coldnumber Aug 16 '15 at 19:44
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Your overarching ideas are correct, but there are some minor mistakes along the way. So some minor clarifications and comments:

  • For reflexivity, your work is fine. If $R$ were reflexive, then we would have $1R1$, but $1+1 \ne 10$, so that doesn't hold.

  • For symmetry, you have the right idea, but you should bear in mind that a simple example does not mean that it holds all of the time. (For instance, that $5R5$ holds doesn't mean the relation as a whole is reflexive!) What you need to see is that symmetry holds if, whenever $aRb$ holds, then $bRa$ holds. This follows from commutativity of addition:

$$aRb \iff a+b = 10 \iff b+a = 10 \iff bRa$$

  • For transitivity, you have the right idea: a counterexample is always a nice means of disproof when possible. If it were transitive, then we would have $4R6$ and $6R4$ implies $4R4$, but $4+4=8 \ne 10$, so we don't have transitivity.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.

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