Let A=1,2,3.

Question

Let $A=\{1,2,3\}$. The number of relations on $A$, containing $(1,2)$ and $(2,3)$, which are reflexive and transitive but not symmetric, is \_\_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

Reflexivity forces $(1,1),(2,2),(3,3)$ into $R$. Transitivity applied to $(1,2)$ and $(2,3)$ forces $(1,3)$ into $R$. This gives the minimal relation
$$R_0 = \{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$$
The remaining possible pairs are $(2,1), (3,1), (3,2)$. Checking each way of extending $R_0$ while preserving transitivity:

- Adding nothing: $R_0$ itself is transitive and not symmetric (e.g. $(1,2)\in R_0$ but $(2,1)\notin R_0$). Valid.
- Adding only $(2,1)$: still transitive (check all new chains), and not symmetric since $(2,3)$ has no reverse. Valid.
- Adding only $(3,2)$: still transitive, not symmetric since $(1,2)$ has no reverse. Valid.
- Adding only $(3,1)$: breaks transitivity, since $(3,1),(1,2)\Rightarrow(3,2)$ is required but absent. Invalid.
- Adding any two of these three pairs together either breaks transitivity (a missing forced pair) or forces full symmetry. All invalid.

So exactly 3 relations satisfy the conditions.

Solutions · 1