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

is it possible to prove it like that:

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

• What is $\mathbb{K}$? It looks nice as long as you have inverses. Also, you started with what you wanted to prove and ended at your conditional. Typically, we would do it the other way. – JacobCheverie May 3 at 15:28
• $\mathbb{K}$ is an ordered field. – Analysis May 3 at 15:39
• How do you know that multiplying by $z^{-1}$ mantain your inequality sign? You are using the proposition to prove the proposition, you can't do that. – Julian Mejia May 3 at 16:52
If $$x then $$0. Since $$0. Then, by multiplicative axiom, you have $$0<(y-x)(z)$$. So, $$0. Hence $$xz.