Mathematical Analysis
Exercise 1-2
2026-01-04
\(\cdot\) is right-distributive over \(+\). Prove that for all \(x, y, z \in \symbb{R}\) we have \((x + y)z = xz + yz\).
Axiom 1.1 part 9 from Schröder (2008, 2) only requires that multiplication be left-distributive over addition. However, using commutativity of multiplication and left-distributivity, \[\begin{align*} (x + y) \cdot z &= z \cdot (x + y)\\[0.75em] &= z \cdot x + z \cdot y\\[0.75em] &= x \cdot z + y \cdot z, \end{align*}\] whence it follows that multiplication is right-distributive over addition as well.