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!

  • $\begingroup$ 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. $\endgroup$ – JacobCheverie May 3 at 15:28
  • $\begingroup$ $\mathbb{K}$ is an ordered field. $\endgroup$ – Analysis May 3 at 15:39
  • 1
    $\begingroup$ 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. $\endgroup$ – Julian Mejia May 3 at 16:52

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 $xz<yz$.


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