Divisibility

Theorem 1. If \(a\), \(b\) and \(c\) are integers such that \(a|b\) and \(a|c\), then \(a|(b+c)\).

Proof. We assume that \(a\), \(b\) and \(c\) are integers such that \(a|b\) and \(a|c\), and will show that \(a|(b+c)\). Since \(a|b\), the definition of divisibility means that \(b = xa\) for some integer \(x\). Similarly \(c = ya\) for some integer \(y\). Now

\begin{align*} b + c &= xa + ya \\ &= (x+y)a \end{align*}

So we have shown that \(b+c\) is an integer multiple of \(a\), since \(x + y\) is an integer by closure under addition. Thus we have shown that if \(a\), \(b\) and \(c\) are integers such that \(a|b\) and \(a|c\), then \(a|(b+c)\). □