a < x < b and a < y < b. How to prove that $\left| {x - y} \right| < b - a$?

Hint: We know that $\left| {x - y} \right|$ has two values: $x - y$ and $ - x + y$ . Therefore, we need to find the range for both these terms. For this, we will first multiply one of the equalities with $ - 1$ and add the other inequality to it and then similar for the other one. Finally, this will lead us to prove the given condition.

Complete step by step answer:
We will first find the range of $x - y$. For this,
We are given that $a < x < b$ and $a < y < b$.
Now, we will multiply the second inequality $a < y < b$ with $ - 1$. After doing this, we get
$ - a > - y > - b \Rightarrow - b < - y < - a$
Now, we will add this to the first inequality.
$\left( {a < x < b} \right) + \left( { - b < - y < - a} \right) = a - b < x - y < b - a$
Now, we will find the range of $ - x + y$.
For this, we will multiply the first inequality $a < x < b$ with $ - 1$. After doing this, we get
$ - a > - x > - b \Rightarrow - b < - x < - a$
Now, we will add this to the second inequality.
$\left( { - b < - x < - a} \right) + \left( {a < y < b} \right) = a - b < - x + y < b - a$
Thus, we have determined that $a - b < x - y < b - a$ and $a - b < - x + y < b - a$.
From this, we can say that both the values $x - y$ and $ - x + y$ are less than $b - a$. Therefore, the valye of their mode will also be less than $b - a$ which is $\left| {x - y} \right| < b - a$.
Hence, it is proved that $\left| {x - y} \right| < b - a$.

Note: We have proved the required condition mathematically in this question. However, we can also logically understand and solve the same question. We are given that $a < x < b$ and $a < y < b$.This means that there are two numbers on a number line \[\left( {a,b} \right)\] and another two numbers between them $x$ and $y$. Therefore, it is clear that the difference between $b$ and $a$ will always be greater than the difference of any two numbers which are between them.