11
$\begingroup$

My fiancée is a teacher in a secondary school. She asked me a question connected to absolute value that I can't answer.

Let's consider the following problem $$ \lvert x - 2 \rvert < \lvert x + 4 \rvert. $$ The answer here is quite obvious, i.e. $x > -1, x \in \mathbb{R}$.

You can get the answer either by using a geometrical interpretation or dividing the real line in two points (i.e. 2 and -4) and consider some cases.

Her students solve this problem using the following method.

  1. They consider two inequalities $x - 2 < x + 4$ and $x - 2 > -(x + 4)$.
  2. They solve the inequalities, from the first one they get $x \in \mathbb{R}$ and from the second one $x > -1$.
  3. That's the point that I think is wrong, but I don't have any proper argument against. When the students have two intervals (i.e. two answers from the inequalities) they either take the union or the sum of them. They use the following rule: use the logical operator (between the solutions to the inequalities) which would lead neither to the real line nor to the empty set. In the described example sum would lead to $\mathbb{R}$ so they choose intersection instead.

The third point seems to be very wrong, nevertheless I can't find any decent argument which would prove that's it's not correct. Maybe it is? I would appreciate any counterexample or some argument for that.

Edit. Cases like $\lvert 2x \rvert < \lvert x \rvert$ are not the ones I'm looking for, because it's equivalent to $\lvert x \rvert < 0$ and now the third point does not appear.

$\endgroup$
9
  • $\begingroup$ $|x-2|=x-2$ only if $x\geq 2$. So the solution of the first inequality is $x\geq 2$ not $\mathbb{R}$. $\endgroup$ Commented May 27 at 6:41
  • 1
    $\begingroup$ Related: Can "doubt" sometimes mean "question"? $\endgroup$ Commented May 27 at 21:54
  • 1
    $\begingroup$ @Stef: Indeed. See my answer, which emphasizes the right attitude to learning mathematics. $\endgroup$
    – user21820
    Commented May 28 at 7:27
  • 2
    $\begingroup$ When you say “using geometrical interpretation”, do you mean by interpreting the problem along the lines of “𝑥 is nearer to 2 than it is to -4”?  (Of course, this method is less suitable for more complex problems — but I think it's useful to help build intuition.) $\endgroup$
    – gidds
    Commented May 28 at 10:06
  • 1
    $\begingroup$ You can always replace |x| <= |y| with x^2 <= y^2. In this case (x-2)^2 < (x+4)^2 leads to a simple solution x^2 - 4x + 4 < x^2 + 8x + 16 or 12x + 12 > 0, but that will not always be the case. $\endgroup$
    – gnasher729
    Commented May 28 at 15:43

9 Answers 9

13
$\begingroup$

Other answers have given examples where the heuristic (which essentially means a method that is designed to be often correct but may be wrong) the students used gives the wrong answer. But we need to much more strongly emphasize that in true mathematics everything must be justified. The correct question about the heuristic is not "Why is it wrong?" but rather "Why is it correct?", and if that question cannot be answered then the heuristic is not mathematically justified.

In particular, the right mathematical attitude is to clearly classify each statement as "justified" or "unjustified" (according to your current knowledge), and actively seek justification. Getting correct numerical answers is not mathematics. Getting understanding and correct justification is.

From this viewpoint, the students' method is wrong in its entirety, not just in one point!

  1. "They consider two inequalities ...". But why those two?
  2. "They solve the inequalities ...". For what?
  3. "When the students have two intervals (i.e. two answers from the inequalities) they ...". Already, they got these answers from previous steps that make no sense. So of course nothing you do with these answers make sense too.

Ultimately, as a teacher you should always focus on why is it correct not how to get the correct answer. Understanding comes from the former, not the latter.

$\endgroup$
9
$\begingroup$

Let's apply the students' method to two inequalities:

$|x-2|<|x-4|$

  1. $x-2<x-4$ and $x-2>-(x-4)$
  2. The first equation is equivalent to $-2<-4$, which has no solution; while the second one is equivalent to $x-2>4-x$, which gives $2x>6$, that is $x>3$.
  3. The heuristic used by the students leads to $(3,\infty)$ for solution set.

$|x-2|<|4-x|$

  1. $x-2<4-x$ and $x-2>-(4-x)$
  2. The first equation is equivalent to $2x<6$, that is $x<3$, while the second one is equivalent to $x-2>x-4$, that is $-2>-4$ for which every real number is a solution.
  3. The heuristic used by the students leads to $(-\infty,3)$ for solution set.

However, $|x-4|=|4-x|$ so, if the method were valid, it should give the same final result.

$\endgroup$
5
$\begingroup$

This "method" of "let's see what operation doesn't get us $\mathbb{R}$" which you describe seems to be a heuristic that fits solving an inequality like $|x+a| < |x+b|$ (for two constants $a$ and $b$), but not more general inequality. For example, consider the inequality $|x-1|>|x|-2$ - this is true for all x (entire $\mathbb{R}$) - you can't get rid of the solution $\mathbb{R}$ because that's the correct solution.

The reason why the "can't be all $\mathbb{R}$" heauristic does work for the inequalities of the form $|x+a| < |x+b|$ (in your example, $a=-2, b=4$), is because we know that not all numbers $x$ can ever solve such an equation: If $x$ is a very high positive number then it solves this inequality only if $a < b$, but if $x$ is a very high negative number, it only solves this inequality if $a > b$. In other words, the desired range is always a half-line, excluding either the very high positive half, or very negative half and can't possibly be the entire $\mathbb{R}$.

$\endgroup$
4
$\begingroup$

There are nine cases, depending on the signs of $x-2$ and $x+4$.
\begin{align*} x-2 &< 0, x+4 < 0 ,\text{ and}& -(x-2)&<-(x+4) \text{;} \\ x-2 &< 0, x+4 = 0 ,\text{ and}& -(x-2)&<0 \text{;} \\ x-2 &< 0, x+4 > 0 ,\text{ and}& -(x-2)&<x+4 \text{;} \\ x-2 &= 0, x+4 < 0 ,\text{ and}& 0&<-(x+4) \text{;} \\ x-2 &= 0, x+4 = 0 ,\text{ and}& 0&<0 \text{;} \\ x-2 &= 0, x+4 > 0 ,\text{ and}& 0&<x+4 \text{;} \\ x-2 &> 0, x+4 < 0 ,\text{ and}& x-2&<-(x+4) \text{;} \\ x-2 &> 0, x+4 = 0 ,\text{ and}& x-2&<0 \text{; or} \\ x-2 &> 0, x+4 > 0 ,\text{ and}& x-2&<x+4 \text{.} \end{align*}

It is straightforward to show that all cases except the third, sixth, and ninth have no solutions.

  • From the third, $$ x < 2, x > -4, \text{ and } x > -1 $$ giving $x \in (-1,2)$.
  • From the sixth, $$ x = 2, 2+4 > 0, \text{ and } 0 < 6 $$ giving $x \in \{2\}$.
  • Finally, from the ninth, $$ x > 2, x > -4, \text{ and } -2 < 4 $$ giving $x \in (2,\infty)$.

Then, $x \in (-1,2) \cup \{2\} \cup (2,\infty) = (-1,\infty)$. That is, $x > -1$.

  1. The students study part of the third case, but the choices of signs for the absolute values require $x < 2$ and $x > -4$, so the students have incorrectly discarded conditions in the third case. They should conclude $-1 < x < 2$, not $x \in \Bbb{R}$.

    The students study part of the ninth case, but the choice of signs for the absolute values require $x > 2$ and $x > -4$, so the students have incorrectly discarded conditions in the ninth case. They should conclude $x > 2$, not $x > -1$.

    The students do not study the sixth case, so cannot validly claim $x = 2$ is a solution.

  2. As explained, the student conclusions in cases three and nine are incorrect because they have discarded the inequalities representing the signs of the quantities $x-2$ and $x+4$ they used to write down the inequality they solve.

  3. If you choose to reason graphically, the graphs are a "V" with vertex at $x = 2$ and a "V" with vertex at $x = -4$. If you slide these Vs around (keeping their vertices distinct, otherwise they're the same expression and so an "$\leq$" or "$\geq$" would be satisfied by the whole line), you'll see that an inequality between them cannot be satisfied by the entire real line, even if we change the slopes of the lines making up the Vs, as in, for example, $|7x-14|<|3x+12|$.

So for [linear expression] [inequality] [linear expression], the heuristic that we should not get the entire line is valid. However, it is not generalizable. What do these students do with $$ -|x-1| < |x+1| \text{?} $$

$\endgroup$
4
  • 3
    $\begingroup$ Instead of considering 9 cases (the Cartesian product of considering where each absolute value is <0, =0, or >0), can you simplify by considering only 4 (where each absolute value is ≤0 or ≥0)? $\endgroup$
    – gidds
    Commented May 28 at 9:55
  • $\begingroup$ @gidds You shouldn't look at both $\ge 0$ and $\le 0$ because then you considered 0 twice, but yes, you can simplify to 4 cases by including 0 in one of the two cases. $\endgroup$
    – quarague
    Commented May 28 at 12:27
  • $\begingroup$ Considering 0 twice is no problem - as long as you are aware that you could get some solutions twice. The important thing is that you can solve any problems with k different absolute values by considering 2^k different cases. $\endgroup$
    – gnasher729
    Commented May 28 at 15:52
  • $\begingroup$ @gidds : Usually one can tell by inspection which case to lump the "${}=0$" case with. Sometimes there is enough complexity that it's not so obvious. As an example of all the moving parts and how they intermesh, I prefer to show everything. Optimization comes after correctness. $\endgroup$ Commented May 28 at 16:12
2
$\begingroup$

The fundamental way in which absolute values interact with inequalities are:

  • $|a|>b$ if and only if $a>b$ or $a<-b$;
  • $|a|<b$ if and only if $-b<a<b$, that is, if and only if $a<b$ and $-a<b$.

We apply these algebraic rules to the original problem to produce the following equivalent statements: $$ |x-2| < |x+4| $$ $$ x-2 < |x+4| \;\;\text{and}\;\; {-}(x-2) < |x+4| $$ $$ \bigl( x-2 < x+4 \;\;\text{or}\;\; x+4 < -(x-2) \bigr) \;\;\text{and}\;\; \bigl( -(x-2) < x+4 \;\;\text{or}\;\; x+4 < x-2 \bigr) $$ The first inequality $x-2<x+4$ is always true (hence the first "or" clause is always true) while the last inequality $x+4<x-2$ is always false; therefore this simplfies to just $$ -(x-2) < x+4 $$ $$ 2-4 < x+x $$ $$ -1 < x. $$ The point of this method is that it's strictly algebraic; it doesn't involve reasoning about what the answer might or might not look like.

$\endgroup$
4
  • $\begingroup$ I agree with everything in that answer, with the exception that I don't know what is the meaning of the word "fundamental" in the first sentence. The two rules that are listed are obviously true for anyone who is comfortable with absolute values, but I think they still require proof. $\endgroup$
    – Stef
    Commented May 28 at 8:30
  • $\begingroup$ Although the proof is easy, I don't think it's necessarily obvious for the students, so if the teacher just calls these rules "fundamental" when solving the exercises, it can make it feel like the teacher is saying "The rules you student use are wrong, only my rules are good" without justification, which sounds extremely arbitrary and arcane. Unfortunately, a lot of teachers do this kind of arcane teaching, and it's a complaint that comes up a lot when asking people who hate math "why do you hate math?" $\endgroup$
    – Stef
    Commented May 28 at 8:31
  • $\begingroup$ On a different note, I had a great teacher in 7-8-9th grades who encouraged students to think of absolute value in terms of "distance" as much as possible. So, $|x-a| < |x - b|$ can be read "$\operatorname{distance}(x, a) < \operatorname{distance}(x, b)$", which is equivalent to "$x$ is closer to $a$ than to $b$", which for anyone with a bit of geometry knowledge is equivalent to "$x$ is on $a$'s side of $(a+b)/2$". $\endgroup$
    – Stef
    Commented May 28 at 8:36
  • 1
    $\begingroup$ Agree on the distance interpretation being an important part of the topic. Yes, the absolute-value inequalities definitely require justification (although at an early level I think the number-line picture "proof" is sufficient). I guess by "fundamental" I mean that they should definitely be a part of any curriculum that expects students to use absolute values in an algebraic context. Omitting them is akin to not teaching students how to algebraically proceed from $x^2=C$. $\endgroup$ Commented May 28 at 16:39
2
$\begingroup$

This is the problem of teaching mathematics as a set of rules and the student is supposed to be able to guess what's the right one to apply.

In this case the “rule” the students apply by analogy is: $|a|<b$ is the same as $a<b$ and $a>-b$. This is correct, but in the case at hand $b$ is $|x+4|$ and not $x+4$.

What can one do after seeing that a wrong “rule” has been employed? Tell the students to apply the same method to $$ |x+4|<|x-2| $$ From $x+4<x-2$ we get the empty set. From $x+4>-(x-2)$ we get $2x>-2$ and so $x>-1$.

What logical operator would they use now? Nothing will provide the actual solution set. If they found $x>-1$ as the solutions for the initial inequality, they must find $x<-1$ as the solutions for the new one.

Conclusion: the students' method worked by pure chance and it's like computing $64/16$ by “striking the $6$'s”.

Inequalities with absolute values (taken with parsimony) can be very instructive because they show that there is generally no “automatic rule” to apply in every situation: they require thinking!

Well, there is a “mechanical” way: divide the reals into pieces, in this case $(-\infty,-4)$, $[-4,2)$, $[2,\infty)$. A division points can go in either interval it bounds (even both, but it's better if we do a partition). Thus we can study three sets of inequalities and take the union of the solution sets.

Thus we have, in the particular case of $|x-2|<|x+4|$, $$ \mathrm{(A)}\ \begin{cases} x<-4 \\[6px] -(x-2)<-(x+4) \end{cases} \qquad \mathrm{(B)}\ \begin{cases} -4\le x<2 \\[6px] -(x-2)<x+4 \end{cases} \qquad \mathrm{(C)}\ \begin{cases} x\ge 2 \\[6px] x-2<x+4 \end{cases} $$ that become $$ \mathrm{(A)}\ \begin{cases} x<-4 \\[6px] 2<-4 \end{cases} \qquad \mathrm{(B)}\ \begin{cases} -4\le x<2 \\[6px] x>-1 \end{cases} \qquad \mathrm{(C)}\ \begin{cases} x\ge 2 \\[6px] 2<4 \end{cases} $$

  • Solution set for (A) is the empty set
  • Solution set for (B) is $-1<x<2$
  • Solution set for (C) is $x\ge2$

Take the union and we're done.

Now show the same for the other inequality and see what goes wrong with the “naïve” method.

However, for the particular inequality there's a smarter method: square both sides, because an inequality between nonnegative numbers is equivalent to the inequality between their squares. We get $$ (x-2)^2<(x+4)^2 $$ hence $$ (x+4)^2-(x-2)^2>0 $$ hence $$ (x+4-x+2)(x+4+x-2)>0 $$ and finally $$ 12(x+1)>0 $$ Note that if we start from $|x+4|<|x-2|$ the same steps bring to $$ -12(x+1)>0 $$ which is expected, isn't it?

This method cannot be applied to, say, $|x|<-1$, because the two sides aren't nonnegative. But it works for more complicated situations such as $|x-2|<|2x+3|$ which leads, after squaring and factoring, to $(x+5)(3x+1)>0$, which is elementary.

$\endgroup$
0
$\begingroup$

This is an answer which might be more appropriate for Math Educators SE, but it was asked here, so I'll address it here.

Students love to have procedures for dealing with things. They want some algorithm that they can apply to a problem without thinking. I think that good mathematics education should train them out of that. Instead of looking to apply tricks that they don't understand, they should be required to explain their work, in terms of whatever theorems and definitions have been provided to them them. In this case, there are just two basic ideas which are relevant:

  • Additive Cancelation: For any $a,b,c\in \mathbb{R}$, $$ a < b \iff a+c < b+c. $$ This is usually taken as an axiom of the real numbers, and doesn't need to be justified, though it does need to be understood. This is the axiom which, in effect, says that any real number can be added to, or subtracted from, both sides of an inequality.

  • Multiplicative Cancelation: For any $a,b,c,\in\mathbb{R}$, $$ a<b \iff ac < bc $$ and $c > 0$. Again, this is an axiom to be understood—it can be interpreted as saying that both sides of an inequality can be multiplied or divided by any positive number.

  • Absolute Values: By definition, if $x \in \mathbb{R}$, then $$ |x| = \begin{cases} x & \text{if $x\ge 0$, and} \\ -x & \text{if $x < 0$.} \end{cases}$$ It is possible to give other definitions of the absolute value (e.g. I have seen students define $|x| = \sqrt{x^2}$) but, in general, I think that this is the most common definition in elementary settings.

It is possible to solve the original inequality, $$ |x-2| < |x+4|,$$ by invoking only these "facts" (the axiom and the definition). To start, neither of the above cited axiom for real numbers (nor any of the axioms which I haven't cited) give us any idea how to handle absolute values when combined with inequalities or equalities. As such, the first order of business should be to handle those. From the definition of the absolute value, there are four possible cases:

  1. $x-2 \ge 0$ and $x+4 \ge 0$,
  2. $x-2 > 0$ and $x+4 < 0$,
  3. $x-2 < 0$ and $x+4 \ge 0$, or
  4. $x-2 < 0$ and $x+4 < 0$.

Notice the "or". Any value of $x$ which satisfies the conditions given, and which also solves the original inequality will solve the original inequality. In fancy language, this means that the goal is to find a set of solutions to the inequality in each case, and then to take the union of those sets.

Within each case, additive cancelation can be used to deduce solutions.

  • Case 1: If $x-2 \ge 0$ and $x+4 \ge 0$ then, by definition of the absolute value, $$ |x-2| = x-2 \qquad\text{and}\qquad |x+4| = x+4. $$ Hence the original inequality becomes \begin{align} |x-2| < |x+4| &\iff x-2 < x+4 \\ &\iff x+(-x) < 2+(4) && \text{(additive cancelation)} \\ &\iff 0 < 6. \end{align} Since $0< 6$ is a true statement, any value of $x$ will solve the original inequality. But wait! Don't forget that this is only the first case. The conditions of the first case are that $x-2\ge 0$ and that $x+4 > 0$. So it must be the case that $x\ge 2$. Therefore any $x \in [2,\infty)$ will solve the original inequality.

  • Case 2: If $x-2 \ge 0$ and $x+4 < 0$ then, by definition of the absolute value, $$ |x-2| = x-4 \qquad\text{and}\qquad |x+4| = -(x+4). $$ Hence the original inequality becomes \begin{align} |x-2| < |x+4| &\iff x-2 < -x-4 \\ &\iff x+(x) < -4+(2) && \text{(additive cancelation)} \\ &\iff 2x < -2 \\ &\iff x < -1. && \text{(multiplicative cancelation)} \end{align} Hence if $x$ satisfies the conditions of this case, then it must be that $x<-1$. Since $x-2 \ge 0$ and $x+4 < 0$, it follows that $$x>2\qquad\text{and}\qquad x<-4, $$ which is impossible. Oops! There are no solutions in this case.

  • Case 3: If $x-2 < 0$ and $x+4 \ge 0$ then, by definition of the absolute value, $$ |x-2| = -(x-2) \qquad\text{and}\qquad |x+4| = x+4. $$ Hence the original inequality becomes \begin{align} |x-2| < |x+4| &\iff -x+2 < x+4 \\ &\iff 2+(-4) < x+(x) && \text{(additive cancelation)} \\ &\iff -2 < 2x \\ &\iff -1 < x. && \text{(multiplicative cancelation)} \end{align} Hence if $x$ satisfies the conditions of this case, then it must be that $x>-1$. Since $x-2<0$ and $x+4\ge 0$, it follows that $-4 \le x < 2$. Therefore any $x \in (-1,2)$ solves the original inequality.

  • Case 4: If $x-2 < 0$ and $x+4 < 0$ then, by definition of the absolute value, $$ |x-2| = -(x-2) \qquad\text{and}\qquad |x+4| = -(x+4). $$ Hence the original inequality becomes \begin{align} |x-2| < |x+4| &\iff -x+2 < -x-4 \\ &\iff -x+(x) < -4+(-2) && \text{(additive cancelation)} \\ &\iff 0<-6. \\ \end{align} This is nonsense, hence this case yields no solutions.

Reviewing the work above, only cases 1 and 3 give any solutions. In case 1, any $x\in[2,\infty)$ solves the inequality; while in case 4 solutions are $x \in (-1,2)$. Since either of these may hold, the set of solutions to the original inequality is the union of those two intervals, i.e. $$ x \in (-1,\infty). $$

TL;DR

I think the the problem with the original solution given in the question is that it attempts to apply some "rule" without having any understanding of what that rule says, where it comes from, or how it is justified. It appears as though the author is trying to take a shortcut without really knowing how that shortcut works.

The solution I have provided is not nice. It is tedious, and takes time to work through. On the other hand, it only relies on knowing a couple of things (additive and multiplicative cancelation, and the definition of the absolute value). Moreover, these few things are essential axioms and definitions, which students need to know anyway. Students need to learn to work with elementary tools before they start learning shortcuts. Once they are comfortable with elementary tools, then you can show them "tricks" (or, better yet, have them derive those tricks and prove that they work).

$\endgroup$
0
$\begingroup$

Note Step 1 matches exclusively the combination of inequalities corresponding to $x+4>0$. As it happens, this is consistent with the solution, but that isn't a safe expectation.

Step 2 copies over the problem from step one, failing to consider $|x+4|<0$ without justifying leaving it out.

Step 3: Each of the three assumptions about the absolute value of $x+4$ implies 2 different inequalities which must be solved simultaneously. This means you always take the intersection within each specific case. Finally, then (step 4) take the union of what remains between each case.


Demo below.

Using a compound, three-way inequality could help simplify things.

$|x-2|<|x+4|$

$\implies -|x+4|<x-2 < |x+4|$

Getting rid of one absolute value expression. Then break up the other absolute value into cases allows complete removal of absolute value.

Suppose $x+4=0 $. Then there are no satisfactory $x$ since $|x-2| \ne 0$.

Suppose $x+4<0$.

$x+4<x-2<-x-4$. We have a contradiction since $x+4>x-2$ for all $x$ and both inequalities must be satisfied..

Suppose $x+4>0$.

$-x-4<x-2<x+4$

$0<2x+2<2x+8$

$0<x+1<x+4$

$x+4>x+1 \forall x \in \mathbb R$

$x+1>0\implies x>-1$ which gives us all satisfactory $x$.


$\endgroup$
0
$\begingroup$

The zeroth step is wrong

The easiest way I think one can explain this is that they have an unproven initial claim that $$\mid x-2 \mid<\mid x+4 \mid \iff x-2<x+4 \text{ and } x-2<-(x+4)$$ This is a logical leap that is not justified. What they're dismissing is all the possibilities where (x-2) is negative (or non-positive if you want to really take a look at zero), which turns out not to be relevant in this particular situation, but this fact hasn't been shown. The rest of their process would work as long as they didn't omit the other possibilities.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .