1. **Problem:** Prove that the relation $R$ on $\mathbb{R} \times \mathbb{R}$ defined by $(a,b) R (x,y)$ iff $a \leq x$ and $b \leq y$ is a partial ordering. 2. **Recall:** A relation $R$ is a partial order if it is reflexive, antisymmetric, and transitive. 3. **Reflexivity:** For any $(a,b) \in \mathbb{R} \times \mathbb{R}$, we have $a \leq a$ and $b \leq b$, so $(a,b) R (a,b)$. 4. **Antisymmetry:** Suppose $(a,b) R (x,y)$ and $(x,y) R (a,b)$, i.e., $a \leq x$, $b \leq y$, and $x \leq a$, $y \leq b$. Combining these, $a = x$ and $b = y$, so $(a,b) = (x,y)$. 5. **Transitivity:** Suppose $(a,b) R (x,y)$ and $(x,y) R (m,n)$, i.e., $a \leq x$, $b \leq y$, and $x \leq m$, $y \leq n$. By transitivity of $\leq$ on $\mathbb{R}$, $a \leq m$ and $b \leq n$, so $(a,b) R (m,n)$. 6. Since $R$ is reflexive, antisymmetric, and transitive, $R$ is a partial order on $\mathbb{R} \times \mathbb{R}$. **Final answer:** $R$ is a partial ordering on $\mathbb{R} \times \mathbb{R}$.