Proposition 2.2.8 If \(a\) is positive and \(b\) is a natural number, then \(a+b\) is positive (and hence \(b+a\) is also, by Proposition 2.2.4.
\(Proof~by~Tao\). We use induction on \(b\). If \(b=0\), then \(a+b=a+0=a\), which is positive, so this proves the base case Now suppose inductively that \(a+b\) is positive. Then \(a+(b\pp)=(a+b)\pp\), which cannot be zero by Axiom 2.3, and is hence positive. This closes the induction. \(\square\)