Exercise 1.1 · Q6: {(1, 2), (2, 1)} on {1, 2, 3}

Basics you need

Solution

1. Draw the relation

   Draw each element of $A$ as a dot and each pair $(a,b)\in R$ as an arrow from $a$ to $b$⁠.

   With only two pairs the picture shows everything at once: an arrow can be reversed (symmetric), no dot has a loop (not reflexive), and going $1 \to 2 \to 1$ would need a loop at 1 (not transitive). The next steps check each property carefully.
   In an arrow diagram, a pair $(a,a)$ appears as a loop at $a$⁠.

2. Symmetric: reverse each pair

   We must check that for every pair $(a,b)$ in $R$⁠, the reversed pair $(b,a)$ is also in $R$⁠. There are only two pairs:

   Pair in $R$	Reversed pair	Reversed pair in $R$⁠?
   $(1,2)$	$(2,1)$	yes
   $(2,1)$	$(1,2)$	yes

   Every pair passes, so $R$ is symmetric.

   The symmetric condition only has to be checked for pairs that are in $R$⁠: it says "if $(a,b)\in R$ then …". Pairs not in $R$⁠, such as $(1,3)$⁠, impose no condition. Since $R$ has two pairs, two checks make a complete proof.
   Symmetric: $(a_1,a_2)\in R \Rightarrow (a_2,a_1)\in R$ (Definition 3(ii), Section 1.2).

3. Reflexive? No: $(1,1)$ is missing

   Reflexive needs all of $(1,1)$⁠, $(2,2)$⁠, $(3,3)$ in $R$⁠. But $R$ contains only $(1,2)$ and $(2,1)$⁠, so none of these is in $R$⁠. One is enough: $(1,1)\notin R$⁠. Hence $R$ is not reflexive.

   Reflexive is a "for every $a\in A$" statement, so a single $a$ with $(a,a)\notin R$ disproves it. Here the failure is total: no element is related to itself.
   Reflexive: $(a,a)\in R$ for every $a\in A$ (Definition 3(i)).

4. Transitive? No: the chain $1 \to 2 \to 1$ needs $(1,1)$

   List every chain: a pair $(a_1,a_2)\in R$ followed by a pair $(a_2,a_3)\in R$ that starts where the first one ends.

   First pair $(a_1,a_2)$	Second pair $(a_2,a_3)$	Transitivity needs $(a_1,a_3)$	In $R$⁠?
   $(1,2)$	$(2,1)$	$(1,1)$	no
   $(2,1)$	$(1,2)$	$(2,2)$	no

   Take the first row: $(1,2)\in R$ and $(2,1)\in R$⁠, so transitivity would require $(1,1)\in R$⁠, which is false. So $R$ is not transitive.

   Many students look for three different numbers, find none, and wrongly decide transitivity holds. But the definition says "for all $a_1, a_2, a_3 \in A$", and nothing stops $a_3$ from being equal to $a_1$⁠. Here $a_1 = 1$⁠, $a_2 = 2$⁠, $a_3 = 1$⁠. The table lists every possible chain, so we are not missing any case; one failing chain is already enough.
   Transitive: $(a_1,a_2)\in R$ and $(a_2,a_3)\in R \Rightarrow (a_1,a_3)\in R$ (Definition 3(iii)). A symmetric relation containing $(a,b)$ with $a\ne b$ can only be transitive if it also contains $(a,a)$ and $(b,b)$⁠.

5. Conclude

   $R$ is symmetric (Step 2), not reflexive (Step 3) and not transitive (Step 4). So $R$ is symmetric but neither reflexive nor transitive, as required.

   The same relation is a ready-made answer to Exercise 1.1 Q10(i), which asks for exactly this combination of properties.

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