Prove $x<y \land z>0 \Rightarrow x\cdot z < y \cdot z$ for all$x,y,z\in \mathbb{K}$

Question

is it possible to prove it like that:

$\begin{gather*} x\cdot z < y\cdot z \quad | \cdot z^{-1} \\ x\cdot \underbrace{(z \cdot z^{-1})}_{\overset{}=1} \overset{}< y \cdot \underbrace{(z \cdot z^{-1})}_{\overset{}=1} \\ 1\cdot x \overset{}< 1\cdot y \\ x \overset{}< y \quad \Box \end{gather*} $

Thanks in advance!

Answer 1

If $x<y$ then $0<y-x$. Since $0<z$. Then, by multiplicative axiom, you have $0<(y-x)(z)$. So, $0<yz-xz$. Hence $xz<yz$.