How to prove that if $x,y\in(a,b)$, then $|x-y|<b-a$?

Question

Let $(a, b)$ be the open interval $\left\{z\in\mathbb{R} : a < z < b\right\}$.

Write the theorem "If $x,y\in(a,b)$ then $|x − y| < b − a$" in logic form, and then prove the theorem.

Answer 1

Let $x,y \in (a,b)$. By definition, $$ a<x<b , a<y<b$$ so $$ -b< -x<-a, a<y<b.$$ $$-b+a<y-x<b-a \mbox{ similary } -b+a<x-y<b-a$$ so $$\left | x-y \right| < b-a.$$

Answer 2

Here is an alternative form of the first answer, with some suggestions on how this proof could be designed.$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\Ref}[1]{\text{(#1)}} \newcommand{\then}{\Rightarrow} \newcommand{\when}{\Leftarrow} \newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}} \newcommand{\abs}[1]{\lvert #1 \rvert} $

You know something about $\;x\;$ and $\;y\;$ separately, and you have to prove something about the combination of the two. Therefore seem easiest to start with the conclusion $\;\abs{x-y} < b-a\;$, and try and split this into a part about $\;x\;$ and a part about $\;y\;$.

Therefore we calculate: $$\calc \abs{x-y} < b-a \op\equiv\hint{basic property of $\;\abs{\cdot}\;$} x-y < b-a \;\land\; y-x < b-a \op\when\hints{arithmetic: $\;p<q \land r<s \;\then\; p+r<q+s\;$, twice} \hint{-- this seems the simplest way to separate $\;x\;$ and $\;y\;$} x < b \land -y < -a \;\land\; y < b \land -x <-a \op\equiv\hint{arithmetic, reorder} a < x < b \;\land\; a < y < b \op\equiv\hint{definition of open interval} x \in (a,b) \;\land\; y \in (a,b) \endcalc$$