Misconceptions in Solving Inequalities...
The pitfall of solving inequalities...
A few days ago I came across an inequality problem asking for the range of values of 2a-b, where
$$1 < a + b < 2\tag{1}$$$$-2 < a - b < -1\tag{2}$$
The teacher happily added the two expressions together, getting $-1<2a<1$; then multiplied the second expression by (-1), getting $1 < b - a < 2$, and added this to (1), getting $2 < 2 b< 4 \Rightarrow 1 < b < 2$; then multiplied this expression by (-1), and added it to $-1<2a<1$. So the result was obvious: $-3<2a-b<0$. Readers, do you think there's a problem with this approach? more
First, the teacher's method is rigorous in form but not tight in substance, and the result itself isn't wrong. But we can notice that no matter what, we can never come up with an example that achieves $2a-b=-2.99$. This shows that the range has actually been enlarged. Why is that? Let's start with a very simple example:
$$\begin{aligned} 1 < a < 2 \\ 1 < a < 3\end{aligned}$$
If we add the two expressions together, we get $1 < a < 2.5$. Clearly, this result enlarges the range of the first expression, so this kind of manipulation is not valid.
So how should we do it properly to keep things tight? Let's go back to the original problem. Let a+b=x, a-b=y, so that
$$\begin{aligned}1 < x < 2 \\ -2 < y < -1 \end{aligned}$$
and solving the equations
$$\begin{aligned}a+b=x \\ a-b=y\end{aligned}$$
we get $a=\frac{x+y}{2},b=\frac{x-y}{2}\Rightarrow 2a-b=\frac{x+3y}{2}$
It's easy to see that the larger (or smaller) x and y are, the larger (or smaller) this expression becomes. Taking $y=1,x=2$ gives the maximum, and taking $x=1 , y= -2$ gives the minimum. Hence $(2a-b) \in (-5/2,-1/2)$. Now we can see that the answer $ -3 < 2a - b < 0$ obtained earlier was indeed too wide!
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