Prove that if $\alpha < x < \beta $ and $\alpha < y < \beta $ then $ \left | x-y \right |< \beta -\alpha $. Interpret the result geometrically as a statement about the interval $\left ( \alpha ,\beta \right )$.
I know that it's true but I don't know where to start...
UPDATE:
Suppose that $$x\geq y$$ $$x <\beta$$ $$x-y<\beta-y $$
$$\alpha <y $$ $$\alpha +\beta -y<\beta$$ $$\beta-y<\beta-\alpha$$ $$\Rightarrow \left | x-y \right |<\beta -\alpha $$
$ \vert x-y\vert:= \begin{cases} -(x-y)&\text{if}\,\ x-y < 0\\ x-y&\text{if}\,\ x-y\geq 0.\\ \end{cases} $
i.e.,
$ \vert x-y\vert:= \begin{cases} -x+y&\text{if}\,\ x < y\\ x-y&\text{if}\,\ x\geq y.\\ \end{cases} $
We consider the two cases, $x<y,$ and $x\geq y,$ separately.
Suppose that $x<y.$
Adding $-x$ to both sides of the assumption $ y < \beta$ gives: $ y-x < \beta - x.$
Adding $(\beta-x-\alpha)$ to both sides of the assumption $\alpha < x$ gives: $ \beta - x < \beta - \alpha. $
We have $ y-x < \beta - x$ and $ \beta - x < \beta - \alpha. $ It follows that $y-x < \beta - \alpha.$
Since $x<y,\ $ it follows that $\vert x - y \vert = -x + y = y - x < \beta - \alpha.$
This proves that $ \left | x-y \right |< \beta -\alpha $ for the case $x<y.$
Now try to imitate this proof to prove the original statement for the other case, $x\geq y.$