Investigating whether a given relation is reflexive, symmetric, and transitive

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 $$4$$.

• Everything looks good, but there's a typo in the last sentence: you probably meant "Then $4R6$ and $6R4$..." – coldnumber Aug 16 '15 at 19:41
• @coldnumber Thanks - fixed that. My doubt is about the last part, because a=c, so is that still not transitive? – user3282081 Aug 16 '15 at 19:43
• 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). – coldnumber Aug 16 '15 at 19:44

• 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.