1. **Problem:** Check whether the relation $S = \{(a,b) : a \leq b^3\}$ on $\mathbb{R}$ is reflexive, symmetric, or transitive. 2. **Reflexivity:** A relation $R$ on a set is reflexive if for every $a$ in the set, $(a,a) \in R$. For $S$, check if $a \leq a^3$ for all real $a$. - For $a=1$, $1 \leq 1^3 = 1$ is true. - For $a=0$, $0 \leq 0$ is true. - For $a=\frac{1}{2}$, $\frac{1}{2} \leq \left(\frac{1}{2}\right)^3 = \frac{1}{8}$ is false. Since it fails for some $a$, $S$ is **not reflexive**. 3. **Symmetry:** A relation $R$ is symmetric if whenever $(a,b) \in R$, then $(b,a) \in R$. Assume $(a,b) \in S$, so $a \leq b^3$. Check if $b \leq a^3$ must hold. Counterexample: Let $a=1$, $b=2$. - $1 \leq 2^3 = 8$ true. - But $2 \leq 1^3 = 1$ false. So $S$ is **not symmetric**. 4. **Transitivity:** A relation $R$ is transitive if whenever $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$. Given $a \leq b^3$ and $b \leq c^3$, does $a \leq c^3$ hold? Counterexample: Let $a=1$, $b=1$, $c=0$. - $1 \leq 1^3 = 1$ true. - $1 \leq 0^3 = 0$ false, so $(b,c) \notin S$; need both true. Try $a=0$, $b=1$, $c=1$: - $0 \leq 1^3=1$ true. - $1 \leq 1^3=1$ true. - Check $0 \leq 1^3=1$ true. Try $a=1$, $b=0$, $c=1$: - $1 \leq 0^3=0$ false, so no. Try $a=-1$, $b=0$, $c=1$: - $-1 \leq 0^3=0$ true. - $0 \leq 1^3=1$ true. - Check $-1 \leq 1^3=1$ true. Try $a=2$, $b=1$, $c=1$: - $2 \leq 1^3=1$ false, no. Try $a=1$, $b=1$, $c=2$: - $1 \leq 1^3=1$ true. - $1 \leq 2^3=8$ true. - Check $1 \leq 2^3=8$ true. No counterexample found easily, but consider $a=2$, $b=1.5$, $c=1$: - $2 \leq (1.5)^3=3.375$ true. - $1.5 \leq 1^3=1$ false, so no. Try $a=1$, $b=0.5$, $c=0.5$: - $1 \leq (0.5)^3=0.125$ false. Try $a=0$, $b=-1$, $c=0$: - $0 \leq (-1)^3=-1$ false. Since no counterexample found where both premises hold but conclusion fails, $S$ is **transitive**. **Summary:** - Reflexive: No - Symmetric: No - Transitive: Yes Final answer: $S$ is not reflexive, not symmetric, but transitive.