Lesson 2 of 10 · 9 min
Reflexive, symmetric and transitive
NCERT §1.2
In the stands, three friends compare notes: 'is taller than', 'sits next to', 'is in the same house as'. Some of these links work both ways, some pass along a chain, and one of them links everybody to themselves.
The lesson in notes
In short
R on A is reflexive if (a, a) ∈ R for every a ∈ A: every element is related to itself. One missing diagonal pair is enough to break it.
R is symmetric if (a₁, a₂) ∈ R always forces (a₂, a₁) ∈ R: every pair comes with its mirror image.
R is transitive if (a₁, a₂) ∈ R and (a₂, a₃) ∈ R together force (a₁, a₃) ∈ R: a two-step chain always has its one-step shortcut.
Perpendicularity on the lines of a plane is symmetric, since L₁ ⊥ L₂ gives L₂ ⊥ L₁. It is not reflexive, since no line is perpendicular to itself, and not transitive, since L₁ ⊥ L₂ and L₂ ⊥ L₃ make L₁ parallel to L₃.
On {1, 2, 3}, R = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 2)} is reflexive; it is not symmetric, because (1, 3) ∈ R but (3, 1) ∉ R, and not transitive, because (1, 3) and (3, 2) lie in R but (1, 2) does not.
On the real numbers, a R b ⇔ a ≤ b is reflexive (a ≤ a) and transitive (a ≤ b ≤ c gives a ≤ c) but not symmetric: 1 ≤ 2 while 2 ≤ 1 is false.
To show a property holds, argue for all elements; to show it fails, one counterexample pair is enough.