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which are within the same interval is equal or less to the difference of the marginal values of the interval?

I have the following inequalities: $$a \le x \le b$$ $$a \le y \le b$$

How to prove that: $$|x - y| \le b - a $$

I tried subtracting y from the first inequality to get:

$$a - y \le x - y \le b - y$$ but don't know how to conclude: $$| x - y | \le b - a$$ ?

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2 Answers 2

up vote 3 down vote accepted

Hint: Multiply the second inequality with $-1$ and then add the two. Can you continue this reasoning? Also note that $$\left|x\right|\le y\iff -y\le x\le y$$ for $y\ge 0$

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1  
So if I understood it correctly: $$ a \le x \le b $$ $$ a \le y \le b $$ $$\implies$$ $$ a \le x \le b $$ $$ -a \ge -y \ge -b $$ $$\implies$$ $$ a \le x \le b $$ $$ -b \le -y \le -a $$ $$ \implies a - b \le x - y \le b - a $$ $$ a - b = - (b - a) $$ and by using $$ -p \le q \le p \implies |q| \le p $$ I get $$ |x - y | \le b - a $$ is that correct? –  Shirohige Dec 28 '12 at 16:07
    
@Shirohige You are 100% correct. –  Nameless Dec 28 '12 at 17:47

Hint Prove that

$$x-y \leq b-a \, \mbox{and} \, y-x \leq b-a \,.$$

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