If a natural number \(x\) is smaller than another natural number \(y\), then the inequality is preserved, if and only if we multiply it by a natural number \(z\neq 0\)1, formally:
\[x < y\Longleftrightarrow \begin{cases} z x < z y&\text{or}\\ x z < y z. \end{cases}\]
The same can be stated about the order relations smaller or greater "\( \le \)", greater or equal "\( \ge \)", and greater "\( > \)":
\[x \le y\Longleftrightarrow \begin{cases} z x\le z y&\text{or}\\ x z\le y z. \end{cases}\]
\[x \ge y\Longleftrightarrow \begin{cases} z x \ge z y&\text{or}\\ x z \ge y z. \end{cases}\]
\[x > y\Longleftrightarrow \begin{cases} z x > z y&\text{or}\\ x z > y z. \end{cases}\]
Proofs: 1
Please note that the proposition would be wrong, if we allowed \(z=0\), e.g. \(3 < 5\), but \(3\cdot 0= 5\cdot 0\). ↩
