Question

Show that the relation \( S = \{ (a, b) : a \leq b^3,\ a, b \in \mathbb{R} \} \) is neither reflexive, nor symmetric, nor transitive.

Answer 1

1. Reflexivity:
A relation \( S \) on \( \mathbb{R} \) is reflexive if \( (a, a) \in S \) for all \( a \in \mathbb{R} \), i.e., \( a \leq a^3 \).
Consider \( a = \frac{1}{2} \). Then: \[ a = \frac{1}{2},\quad a^3 = \left( \frac{1}{2} \right)^3 = \frac{1}{8} \] \[ \text{So, } a \leq a^3 \text{ becomes } \frac{1}{2} \leq \frac{1}{8}, \text{ which is false}. \] Hence, \( (a, a) \notin S \). Therefore, \( S \) is not reflexive.
2. Symmetry:
A relation \( S \) is symmetric if \( (a, b) \in S \Rightarrow (b, a) \in S \).
Take \( a = 1,\ b = 2 \). Then: \[ a \leq b^3 \Rightarrow 1 \leq 8 \Rightarrow (1, 2) \in S. \] But: \[ b \leq a^3 \Rightarrow 2 \leq 1 \Rightarrow \text{False} \Rightarrow (2, 1) \notin S. \] Hence, \( S \) is not symmetric.
3. Transitivity:
A relation \( S \) is transitive if \( (a, b) \in S \) and \( (b, c) \in S \Rightarrow (a, c) \in S \).
Let \( a = 0,\ b = 1,\ c = 2 \). Then: \[ a \leq b^3 \Rightarrow 0 \leq 1 \Rightarrow (0, 1) \in S, \] \[ b \leq c^3 \Rightarrow 1 \leq 8 \Rightarrow (1, 2) \in S. \] But: \[ a \leq c^3 \Rightarrow 0 \leq 8 \Rightarrow (0, 2) \in S. \] This example does not disprove transitivity. Let's try another one.
Let \( a = 8,\ b = 2,\ c = 1.5 \). Then: \[ a \leq b^3 \Rightarrow 8 \leq 8 \Rightarrow (8, 2) \in S, \] \[ b \leq c^3 \Rightarrow 2 \leq (1.5)^3 = 3.375 \Rightarrow (2, 1.5) \in S. \] But: \[ a \leq c^3 \Rightarrow 8 \leq 3.375 \Rightarrow \text{False} \Rightarrow (8, 1.5) \notin S. \] Hence, \( S \) is not transitive.
Conclusion:
The relation \( S = \{ (a, b) : a \leq b^3 \} \) is neither reflexive, nor symmetric, nor transitive.