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.

  • $\begingroup$ 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 $b\in[-2,3]$? $\endgroup$ – Clayton Jun 2 '19 at 1:06
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    $\begingroup$ 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? $\endgroup$ – Arturo Magidin Jun 2 '19 at 1:11
  • $\begingroup$ 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 $x-w\leq a-b\leq y-z$. $\endgroup$ – Arturo Magidin Jun 2 '19 at 1:12
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    $\begingroup$ Good explanation Arturo, the maximum value of $2a - b$ will be a = - 2, b = - 2. Thanks :D $\endgroup$ – Eduardo Sebastian Jun 2 '19 at 1:19

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


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