# Backwards Compound Inequalities?

Some textbooks I've seen declare inequalities such as $-2>x>2$ to have no solution, or to be ill-defined, which I disagree with. I'm curious to know if anyone else thinks the same.

Inequalities can always be written two ways. For example, $x>2$ is the same as $2<x$. So far as I understand, the same applies to compound inequalities; for example, everyone would regard $-3<x<3$ to be well-defined, and it can be written "backwards" as $3>x>-3$.

When someone interprets $-3<x<3$, upon reflection, it is understood that there is an implicit intersection behind the scenes, as it can be read out-loud as "$-3<x$ and $x<3$." And when they interpret $3>x>-3$, it is the "backwards" version of $-3<x<3$. Both are two different, compact ways of expressing {$x<3$} $\cap$ {$x>-3$}.

So when I look at an inequality such as $-2>x>2$, I take it to mean there is an implicit union behind the scenes. In other words, $-2>x>2$ and $2<x<-2$ both refer to the same thing, namely {$x<-2$} $\cup$ {$x>2$}. Were I to read $-2>x>2$ out-loud, I would read it as "$-2>x$ or $x>2$."

Am I crazy, or is there something wrong with this interpretation?

It seems to offer some advantages. For example, it makes the solution of certain absolute value inequalities very easy and natural.

• You are right about the implicit "intersection," though I think of it as logical "and" (usual notation $\land$). But then for the inequalities running the other way, intead of saying intersection, which would be right, you switch to union, which is not right. – André Nicolas Nov 30 '12 at 18:43
• In the case $3\gt x \gt -3$ you recognized that we are dealing with an intersection. Now the expression $-2\gt x\gt 2$ has exactly the same shape as the previous one, the numbers are a bit different. So if the first is an intersection, so is the second. It so happens that the intersection of the two sets $\{x: -2\gt x\}$ and $\{x: x\gt 2\}$ is empty, but that's irrelevant to how one interprets the logic of the situation. – André Nicolas Nov 30 '12 at 18:55
• It is intersection independently of the relationship between $a$ and $b$. Of course, if $a\le b$ it is not very interesting, but it still is an intersection. – André Nicolas Nov 30 '12 at 19:19
• It looks as if I will not succeed. You are comfortable with interpreting $1\lt x\lt 5$ as an intersection, (or conjunction). You want the meaning of a formula $F(x, a,b)$ that involves the parameters $x$, $a$, and $b$ to depend on the values of these parameters. That is not the way things are ordinarily done, and it is unlikely, at least in the short run, that you will alter standard mathematical practice. – André Nicolas Dec 1 '12 at 1:35
• The meaning of the expression $f(x)=x^2$ if $x\lt a$, and $f(x)=x^3$ if $x\gt a$ is independent of $a$. It means square $x$ if $x\t a$, $\dots$. Of course the function changes. Similarly, the meaning of $a \lt x\lt b$ does not change, but the actual set being described does. – André Nicolas Dec 1 '12 at 2:40

You can write inequalities any way you want if you think of the elements satisfying the inequality as members of a set combined with a truth table. There isn't any ambiguity in writing $-2>x>2$ or $-2>x<5$ as long as you read it left to right, or right to left in a PAIRWISE fashion: $-2>x$ and $x>2$ or $-2>x$ and $x<5$ respectively.. If you the write $\{x<-2\}\cap \{x>2\}$ you will realize that this intersection is the empty set, that there are no $x$ which satisfy the inequality. On the the other hand, if you write $-2>2$, this can be interpreted as a true or false statement, in this case being false.
$-2>-3<5>2$, again it will be unambiguous if you read it left to right or right to left, in a pairwise fashion. In other words, $-2>-3$, $-3<5$, $5>2$. It WILL be ambigious otherwise, because for example, are you saying $-2>2$ as well as $-2>2$?
I wouldn't say the system of inequalities $2<x<-2$ is "ill defined", but certainly it has no solutions: there is no number that is both bigger than $2$ and less than $-2$.