Prove $x<y \Rightarrow x+z < y+z$ with $x,y,z \in \mathbb{K}$ when $\mathbb{K}$ is an ordered field.

Question

Prove $x<y \Rightarrow x+z < y+z$ with $x,y,z \in \mathbb{K}$ when $\mathbb{K}$ is an ordered field.

My attempt:

$x<y \Leftrightarrow x+0<y+0 \Leftrightarrow x+(z+(-z))< y+ (z+(-z)$

$\Leftrightarrow x+(z+(-z))+z < y+(z+(-z))+z $

$\Leftrightarrow x+z < y+z \quad \Box$

Answer 1

By definition, $x<y$ means that $y-x\in\mathbb K^+$. But then, if $x<y$ and $z\in\mathbb K$,\begin{align}x+z<y+z&\iff(y+z)-(x+z)\in\mathbb K^+\\&\iff y-x\in\mathbb K^+,\end{align}which is true.

Answer 2

Your step $x + (z+(-z)) < y + (z+(-z)) \Rightarrow x + (z+(-z)) + z < y + (z+(-z)) + z$ is exactly what you are trying to prove.