TITLE: Multiplying the inequalities gives a weird result which is obviously wrong
QUESTION [6 upvotes]: Here's a chain of conclusions which leads to the wrong outcome but I don't understand at what point I made a mistake.
$$\begin{align}
1 < a < 2 &\quad\Rightarrow\quad 1 < \,a^2\, < 4 \\[4pt]
1 < a < 2 &\quad\Rightarrow\quad \frac12 < \frac1a < 1
\end{align}$$
Now I want to multiply the two sets of inequations that I have to the right of the '$\Rightarrow$' so here we have:
$$\frac12 < \frac{a^2}a < 4 \quad\Leftrightarrow\quad \frac12 < a < 4$$
Obviously that's wrong, because $1 < a < 2$, but I don't understand where I went wrong, leading to a false conclusion.
I appreciate any help.

REPLY [3 votes]: When you find yourself scratching your head over a result like this, it pays sometimes to redo the problem using only variables
$$\begin{align}
A < x < B &\quad\Rightarrow\quad A^2 < \,x^2\, < B^2 \\[4pt]
A < x < B &\quad\Rightarrow\quad \frac1B < \frac1x < \frac1A
\end{align}$$
multiply them together and you get
$$\begin{align}
 \frac{A^2}B < \,x\, < \frac{B^2}A \\[4pt]
\end{align}$$
or
$$\begin{align}
 A\frac{A}B < \,x\, < B\frac{B}A \\[4pt]
\end{align}$$
Now, since you know that $A<B$ (since $A<x<B$), that means that $\frac{A}{B} < 1$.
That means that $A\frac{A}B < A$ and $B\frac{B}A > B$, and your final inequality will always be weaker than your original inequality.