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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.

Andrés E. Caicedo
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Zippy The Pinhead
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    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
  • Help me to understand why it is not right to read an implicit union. – Zippy The Pinhead Nov 30 '12 at 18:48
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    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
  • Thanks for the quick reply. I suppose that when I consider something of the form a>x>b, I am first noting whether a>b or b>a. In the former case, I read an implicit intersection, and when b>a, an implicit union. I should have made that explicit. – Zippy The Pinhead Nov 30 '12 at 19:16
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    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
  • There are other instances where we work with inequalities on a conditional basis. For example, if a>b, ac>bc only when c>0. But I can't think of how to explain why a>x>b is an intersection, regardless. I would be so grateful if you would help me connect the dots, because to be direct, treating it as an implicit union when b>a seems to work so well. – Zippy The Pinhead Dec 01 '12 at 01:23
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    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 01 '12 at 01:35
  • I really appreciate your comments, many thanks. I will continue to think about this. – Zippy The Pinhead Dec 01 '12 at 01:55
  • One more question, please: I feel compelled to ask, regarding your generalization: is not a piecewise function something which changes what it means depending on the values of the parameters? – Zippy The Pinhead Dec 01 '12 at 02:25
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    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 01 '12 at 02:40
  • There is a typo in your last comment. If you please, fix it, so I can see the complete comment. Thanks in advance... – Zippy The Pinhead Dec 01 '12 at 03:06
  • "It means square $x$ if $x\lt a$, and $\dots$.'We need to distinguish between a computer program, which (usually) does not change, and what it does given certain inputs, which of course is input-dependent. – André Nicolas Dec 01 '12 at 03:08

2 Answers2

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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.

If you have something complicated like:

$-2>-3<5>2$, 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 $-2>2$?

Alex R.
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  • Regarding the example in your last paragraph, I agree it makes perfect sense if read in a pairwise fashion, but I would be reluctant to leap to the conclusion that it also means -2>2, because the direction is not consistent throughout. What I mean is, a>b>c>d certainly means a>d, for example. But the chain is broken when one of the symbols is reversed, no? – Zippy The Pinhead Nov 30 '12 at 18:57
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    The fact that you have to ask whether the chain is broken when one of the symbols is reversed implies there is ambiguity involved. I will say that if there is to be a universal standard, then it should be read pairwise to avoid any ambiguity. – Alex R. Nov 30 '12 at 19:01
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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 $-2$.

Michael Hardy
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  • I saw your comment, but see that Alex already stated an example of what I was trying to convey, and did so much better than I! So, my answer has been deleted, accordingly. – amWhy Nov 30 '12 at 18:39