# Subtraction of inequality

I have this statement:

If $$a$$ belongs to the interval $$[- 4, - 1]$$ and $$b$$ belongs to the interval $$[- 2, 3]$$, what interval does it contain? all possible values ​​of $$(2a - b)$$?

I have developed it, but according to the guide my answer is incorrect and I would like to know why.

$$-4 \leq a \leq -1$$, multiply by $$2$$

$$-8 \leq 2a \leq -2$$

Now, the interval of $$b$$ is:

$$-2 \leq b \leq 3$$

I know the extreme values ​​of each interval, I will subtract the interval of $$b$$ :

$$-8 - -2 \leq 2a - b\leq -2 -3$$

$$-6 \leq 2a - b \leq -5$$, so my answer is: $$[-6,-5]$$

I would like to know why, my development is incorrect. Thanks in advance.

• Are you sure you gave us the full statement? Why do you multiply by $2$? Or is the second question: what are all possible values of $2a-b$ where $a\in[-4,-1]$ and $b\in[-2,3]$? – Clayton Jun 2 '19 at 1:06
• You’ve done the last thing wrong, because what you want to do is add the inequality for $-b$: the smallest that $2a-b$ can be will be when $a$ is as small as possible, and $b$ is as large as possible (not as small). So you get the smallest possible value of $2a-b$ when $a=-4$ and when $b=3$, not when $b=-2$. Similarly, the largest value of $2a-b$ will occur when? – Arturo Magidin Jun 2 '19 at 1:11
• That is, if $x\leq a\leq y$ and $z\leq b\leq w$, then $x+z\leq a+b\leq y+w$. But to subtract $b$, you need to add $-b$, and if you multiply by $-1$, you don’t get $-z\leq -b\leq -w$, you get $-w\leq -b \leq -z$. So adding $x\leq a\leq y$ to $-w\leq b\leq -z$ you get $x-w\leq a-b\leq y-z$. – Arturo Magidin Jun 2 '19 at 1:12
• Good explanation Arturo, the maximum value of $2a - b$ will be a = - 2, b = - 2. Thanks :D – Eduardo Sebastian Jun 2 '19 at 1:19

## 2 Answers

Max of $$2a=-2$$, min of $$b=-2$$, max of $$2a-b=0$$.

Min of $$2a=-8$$, max of $$b=3$$, min of $$2a-b=-11$$.

The error is assuming that $$A\le B$$ and $$C\le D$$ imply $$A-B\le C-D.$$ E.g. $$4\le 6$$ and $$2\le 5$$ but $$\neg (4-2\le 6-5).$$

On the other hand if $$E\le F$$ and $$G\le H$$ then $$\{x+y:x\in [E,F]\land y\in [G,H]\}=[E+G,F+H].$$

$$\{2a:a\in [-4,-1]\}=[-8,-2].$$

$$\{-b:b\in [-2,3]\}=[-3,2].$$

So $$\{2a-b: a\in [-4,-1]\land b\in [-3,2]\}=\{x+y:x\in [-8,-2]\land y\in [-3,2]\}=[-11,0].$$