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. \(\square\)
\(Proof~by~Tao\). Suppose for the sake of contradiction that \(a\neq0\) or \(b\neq0\). If \(a\neq0\) then \(a\) is positive, and hence \(a+b\) is positive by Propostion 2.2.8, a contradiction. Similarly if \(b\neq0\) then \(b\) is positive, and again \(a+b\) is positive by Proposition 2.2.8, a contradiction. Thus \(a\) and \(b\) must both be zero. \(\square\)
\(\neq\) { Proposition 2.2.8: \(a,b\in\mathbb{N}.~b\neq0\implies a+b=b+a\neq0\) }
\(\neq\) { Proposition 2.2.8: \(a,b\in\mathbb{N}.~a\neq0\implies a+b\neq0\) }