Propostion 2.3.4 (Distributive law). For any natural numbers \(a,b,c\), we have \(a(b+c)=ab+ac\) and \((b+c)a=ba+ca\).
\(Proof~by~Tao\). Since multiplication is commutative we only need to show the first identity \(a(b+c)=ab+ac\). We keep \(a\) and \(b\) fixed, and use induction on \(c\). Let's prove the base case \(c=0\), i.e., \(a(b+0)=ab+a0\). The left-hand side is \(ab\), while the right-hand side is \(ab+0=ab\), so we are done with the base case. Now let us suppose inductively that \(a(b+c)=ab+ac\), and let us prove that \(a(b+(c\pp))=ab+a(c\pp)\). The left hand ise is \(a((b+c)\pp) = a(b+c)+a\), while the right-hand side is \(ab+ac+a=a(b+c)+a\) by the induction hypothesis, and so we can close the induction. \(\square\)