My understanding is that students find absolute value to be challenging to learn or understand. Off the top of my head, I can come up with two possible reasons for this.

  1. Absolute value is a piecewise defined function. Piecewise defined functions are more difficult due to increased abstractness: they are not a simple formula, but include a conditional. (I do not know if this is true, but it sounds plausible enough.)

  2. Absolute value is difficult, because it combines algebra (changing sign) with a geometric interpretation as a distance of number. At least according to the thesis of Hähkiöniemi on derivative [1], it is challenging for students to change between perspectives in a fruitful way.

Is one of these the reason for the difficulties, or maybe it lies elsewhere? As always, answers using scientific literature are the most valuable and ones relying on explicit personal experiences are also fine.

[1] Hähkiöniemi, Markus. The role of representations in learning the derivative. No. 104 in Reports of university of Jyväskylä, department of mathematics and statistics. University of Jyväskylä, 2006. http://urn.fi/URN:ISBN:951-39-2639-7

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    $\begingroup$ I agree with @Jessica B, and I think another reason (which is possibly included in her reason if interpreted broadly enough) is that absolute value manipulations are different from the standard algebraic manipulations they're used to (distributive property, combining like terms, etc.). Indeed, the concepts in your second reason are often a way of getting students to better "visualize" absolute value manipulations and to help keep students from making incorrect manipulations or incorrect deductions involving absolute value manipulations. $\endgroup$ Commented Aug 16, 2019 at 13:01
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    $\begingroup$ The main difficulties I see are when students are asked to deal with absolute value inequalities. Inequalities are very hard for most students to deal with. $\endgroup$
    – Sue VanHattum
    Commented Aug 16, 2019 at 16:10
  • $\begingroup$ Because it's intricate. Yes, even just one or two "if then"s still makes something intricate! We are meat, not silicon. Just explaining a rule or set of rules is not adequate for us, if we have to remember some intricacies. Instead of tacitly searching for some lock-key explanation idea (what is the hurdle and how do we adroitly remove it), I recommend to drill, drill, drill. And then drill some more. This is the way to familiarity and to making struggling concepts routine. It's no different in music or sports. That's the way us meat people are. $\endgroup$
    – guest
    Commented Aug 18, 2019 at 18:00
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    $\begingroup$ I suspect most students have no trouble with the notion of the absolute value of a number, and that those who have trouble with the absolute value function do so because more generally they have trouble with the function concept (the same reason they struggle with the notion of the square root function). $\endgroup$
    – Dan Fox
    Commented Aug 21, 2019 at 19:11
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    $\begingroup$ Personally, I've not seen the "concept" of absolute value be difficult. The problem is in graphing, manipulation, and, as @JohnOmielan points out, inequalities. these are harder for students to wrap their heads around. I don't think the concept itself is hard, it's helping people figure out how to integrate that into other things they have learned, some of which may not have fully sunk in yet. $\endgroup$
    – johnnyb
    Commented Mar 17, 2023 at 21:17

11 Answers 11


(My answer is just a guess and not based on any formal research.)

I suspect the absolute value function may be difficult to understand because it involves "negative numbers that aren't negative." One way to define the absolute value of $x$ is:

$$|x|=\left\{\begin{array}{rl}-x, & x<0\\x, & 0\le x\end{array}\right.$$

I think the $-x$ confuses students, causing them to think that "sometimes the output of the function is negative."

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    $\begingroup$ Yes. In my experience, many students struggle to accept that for some values of x, it is true that |x| = -x. $\endgroup$
    – idmercer
    Commented Aug 22, 2019 at 20:05
  • $\begingroup$ This detail can be hard for students to grasp, but in many contexts the formal definition is entirely hand-waved away. I had one (weak) math major in a discrete math course a year ago, who I assume had done many absolute-value exercises in the past, who came to me after presenting the formal definition claiming "this is different from the absolute value we've been taught before", and surprised when I could show it produced the values he expected. $\endgroup$ Commented Aug 11, 2020 at 13:56
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    $\begingroup$ This is why I vastly prefer using the term "the opposite of $x$" to "negative $x$" for $-x$ in front of students, or "opposite $x$" for short. That removes a lot of the confusion what to do with $-x$ if $x<0$. $\endgroup$
    – Forklift17
    Commented Aug 12, 2020 at 19:23
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    $\begingroup$ @Forklift17 It turns out that many languages do read -3 as 'minus 3' rather than 'negative 3'; that is, they treat - as a unary operator. And nobody reads ±3 as 'positive negative 3'. $\endgroup$
    – ryang
    Commented Mar 15, 2023 at 17:40

My experience is that weak students latch onto absolute value of a number is always positive. They are fine working with constants. When you introduce a variable, it all falls apart. To these students it is clear that:

$$\lvert-x\rvert= x$$

After all the absolute value sign takes away the negative sign. If you want students to understand the absolute value of an algebraic expression then you MUST get them past this hump. All this is before you move onto word problems and inequalities.

  • 2
    $\begingroup$ This is the most concise description of the biggest problem, though Joel's answer also gets to this point fairly directly. $\endgroup$
    – kcrisman
    Commented Aug 7, 2020 at 14:30
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    $\begingroup$ @JoelReyesNoche thanks for catching my error $\endgroup$
    – Amy B
    Commented Aug 9, 2020 at 4:11
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    $\begingroup$ @JTP-ApologisetoMonica thanks for the edit $\endgroup$
    – Amy B
    Commented Aug 9, 2020 at 4:12
  • $\begingroup$ @kcrisman yes Joel's answer also addresses this point. He has a more formal definition of absolute value, which is often not introduced to high school students. I was thinking of how such students might fail in understanding. $\endgroup$
    – Amy B
    Commented Aug 10, 2020 at 7:56
  • $\begingroup$ Yes, that is exactly why I also upvoted your answer. $\endgroup$
    – kcrisman
    Commented Aug 12, 2020 at 15:13

In my experience, this is the first really clear example that students experience of dissonance between how something looks and what it is. One of the most common errors I see regarding absolute value - before bringing in variables, of course - looks like this:

$$|5 - 6| = 5 + 6 = 11$$

Whereas, of course, if I ask the student "what's $5 - 6$?" and then "okay, what's the absolute value of that?" they'll get the answer perfectly correct. The issue seems to be that they have a mental picture of "absolute values make negative signs positive", and they see a negative sign in the expression "$5 - 6$".

As experts, we understand that the absolute value operates on numbers, not on symbols; it cares whether the number it's given is negative, not whether the way the number is written includes a negative sign. But that's not an easy distinction to make if a student has grown up with the usual symbol-based approaches for problem-solving.

This is a similar issue to what often happens with simplifying fractions - one common mistake looks like this:

$$\frac{3 + 2}{6 + 2} = \frac{3}{6} = \frac12$$

A student who makes this mistake is thinking of cancellation as a symbolic action ("delete the same symbols from the top and bottom") instead of an arithmetic one ("multiplying the numerator and denominator by the same number is the same as multiplying by one").

Many students I've worked with get around this problem with absolute value by just memorizing the rule "simplify inside the absolute value first". That's correct, but it suddenly stops helping when variables come into play; that leads them to say things like this:

$$|-x| = x$$

The situation is made even worse when they're presented with the piecewise definition that other people have mentioned in the answers. My sense is that it's largely because they feel like they're being asked to accept a completely new meaning to a word that feels familiar; it's as if you were trying to convince them to accept that the word "elephant" now refers to a small bird. To the average student, "absolute value" means performing this very simple symbol-based process of removing negative signs; now you're telling them it's actually a complicated process of conditions and arithmetic!


I am writing this based on pure observation (e.g., entering year four of teaching this topic to secondary school students, and having co-taught a minicourse for teachers on absolute value functions$^\star$).

There are a lot of definitions/interpretations of absolute values:

  • the (abstract) axiomatic one;

  • the piecewise or "case-based" one (provided by Joel Reyes Noche);

  • the colloquial function one (erase any negative sign and return the result);

  • the equivalent function one of $x \mapsto \sqrt{x^2}$;

  • the geometric one (where $|a-b|$ is the distance on the number line between real numbers $a$ and $b$);

  • the positive difference interpretation (subtract the lesser number from the greater one, i.e., $|a-b|=\max(a,b)-\min(a,b)$ for real numbers $a$ and $b$);

  • the graphical one communicated by a $\mathsf{V}$-shaped curve;


A complication that has already arisen in the above collection of definitions/interpretations is its ambiguity around whether we are defining a function whose input is one number or two numbers; the one number version can, of course, be viewed as the two number version where (at least) one of the entries is zero. Nevertheless, I think that sometimes the topic is introduced with multiple approaches and without drawing this distinction; e.g., if you try to help a student get a grip on absolute value functions by asking the difference in age between them and a friend, then you are effectively asking about the two input interpretation, which could be confusing if you had just introduced it as, e.g., a piecewise function.

There may also be an issue of timing: When I delve into absolute value functions, equations, and inequalities, it is in an Algebra 2 course; in our present course sequence, this means that students have done Algebra 1 already, and then spent a year on Geometry. In other cases, you may have this topic broached in an Algebra 1 course; so, students are in the throes of matching graphical representations, geometric representations, symbolic representations, and so forth.

As to covering absolute values pedagogically, I think there is a great value in viewing more general absolute value functions of the form $x \mapsto a|x-h|+k$ for real parameters $a, h, k \in \mathbb{R}$ in terms of transformations, and, in addition, connecting this to quadratic functions written in vertex form, $x \mapsto a(x-h)^2 + k$, to emphasize similarities and differences (and to reinforce terminology: intercepts, roots, vertex, concavity, end behavior, domain, range).

$\star$: I would be remiss if I failed to link our (my colleague/department chair, Liz Brennan, and my) freely available materials from a minicourse taught through Math for America:

There are a lot of materials in there, which range from a question I asked on MathOverflow (MO 301514) to a misformulated problem (p. 9 #6 is impossible!) to problems that participants formulated during our final meeting (Desmos link).

The three-meeting course linked above is by no means exhaustive: In fact, I have been thinking more about absolute values and their role in defining equations as one endeavors to impose domain restrictions; for example, see the sequence of tweets that begins here.


At my school (a community college in California) the curriculum is set up so that students first take an algebra course in which essentially every function is a linear function, and only later do they do anything at all with nonlinear functions. Nonlinear functions are just more complicated, so they require more thought about the logic. Thinking about logic is harder than manipulating formulas according to recipes.

As an example, if I tell a student that $x^2=4$ and ask them to solve for $x$, the formula-manipulation approach would be "do the same thing to both sides," so they may get $x=2$. It requires an extra logical step to realize that there are two roots. One way to write this is $|x|=2$, but I'll see students do things like $x=|2|$, or $|x|=|2|$ (which is correct but kind of silly). Understanding why $|x|=2$ is the right choice requires some more logical thinking.

An example that comes up in physics is that students will memorize the fact that $a=-g$ for free fall, despite instruction to the effect that this depends on the coordinate system, and that they need to pick a coordinate system first. Some textbooks even encourage this. Picking a coordinate system requires an extra logical step, which is hard if you aren't used to thinking about math logically. I can tell them that $g$ is defined as $|a|$ so that we can put a value of $g$ in the book without reference to any coordinate system, but again, this requires some logical thinking about topics that they aren't used to thinking about, and have gotten the impression that they never have to think about.

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    $\begingroup$ "Picking a coordinate system requires an extra logical step, which is hard if you aren't used to thinking about math logically." — I think an extra complication is that in many pre-college environments the axes are drawn with two arrowheads, and it is assumed that values increase in the right or upwards direction; in this case arrowheads indicate that this is an infinite line. I believe that an alternative approach when an arrowhead indicates the direction in which value increases, and the infinite length of the axis is assumed just because it is not terminated with a point, is less ambiguous. $\endgroup$
    – Rusty Core
    Commented Aug 23, 2019 at 23:14

Absolute value is difficult for students because they have difficulty parsing and simplifying logical statements.

Some of the results of working with absolute value statements seem to actually hide the inner mechanics of how the logic of an absolute value statement play out. Because of their piecewise nature, you must understand logical statements to work deeply with absolute values. Students often develop a superficial, rote approach to these expressions because they are simpler than actually describing what is going on (rote is also sufficient for most exercises that I have encountered).

The best anecdote that I can offer was the following question, which was asked if me by a student in 11th grade: When you are solving the inequality abs(x-2)≤3, the solution is -1≤x≤5, which is an AND statement (-1≤x AND x≤5); but the solution process involves an OR statement (x<2 OR x≥2) to address both parts of the piecewise definition. Where does this change happen from OR to AND?

It is a question that really requires one to logically express the solution process clearly to answer (the answer is actually that the statement is always an OR statement that just happens to be equivalent to the AND one... (x<2 AND x≥-1) OR (x≥2 AND x<5), where the outermost OR is the continuation of the two distinct possibilities from the solution process).

All of the notions of the different interpretations of absolute values are tools to aid students' abilities to intuit answers to problems involving absolute values, but do little to actually let students rigorously understand them. To do so, you would have to actually pick one as the definition and relate the others rigorously to that. This is an exercise that almost never happens in classes.

In truth, I think that students find absolute values difficult because the subject is legitimately difficult! Otherwise, why would we bother having so many ways to think about how to interpret such statements? These interpretations are shortcuts around the rather tedious legwork of working directly with the logic (my personal chosen definition). I think that most teachers probably don't know absolute values as clearly as we would like them to.

I offer the following questions as food for thought on the difficulty of absolute values:

  1. When solving equations, as opposed to inequalities, it is more common in secondary school to find a countable number of solutions as opposed to an uncountable number, but the solution set for the equation abs(2x-2)-abs(x)=2-3x is [0,1]. Why can you get entire intervals of solutions to such an equation?
  2. How do you graph y=abs(x-abs(x-1))?

It is difficult because there are many things implicitly done behind the scenes:

  • Function definition with case analysis usually for the first time.

  • Solving problems, among which are equations and inequalities, usually involves case analysis for all variables and all their possible values and then checking if solution satisfies initial conditions, or in other words just doing Backtracking, but teachers don't do analogies showing general solving strategy, and for example comparing absolute value problem to some real life situation. They just do example and that's it.

  • and and or, De Morgan's laws etc statements when solving equations and inequalities, also doing set operations that teachers don't explain well.

  • Students don't have good algebra intuition - not explaining in details what happens with distance interpretation $|x-y|$ "What happens for example when $x$ is negative, $y$ is negative and other cases", or just seeing $-x$ that is positive.

  • Seeing other properties like reverse triangle inequality might be intimidating.

  • Graphing on $(x,y)$ plane and seeing some stroked lines for the first time and then doing mapping with algebra equation is a challenge too.

  • Convoluting even more for example adding some parameters and stating some questions about solutions without explaining overall problem solving strategy.

  • Resemblance with $x^2$ and other stuff that other commenters mentioned.


This answer is concerned with real numbers only. Adding to Dave Renfro's comment about absolute-value manipulations being—or at least feeling—different from usual algebraic manipulations, and Mark B's answer about the additional intricacy of the logic of absolute-value statements:

  1. The true statement $$\text{for each real }x,\:\:|x|=\pm x\tag{✔️}$$ looks like a mathematical identity, $$|x|\equiv\pm x,\tag{❌}$$ and misleadingly appears to suggest that $|x|$ and $\pm x$ are mutually substitutable.

    But of course $$x^2+3x=\color\red{|x|} \iff x^2+3x=\color\red{\pm x}\tag{❌}$$ is incorrect ($-2$ satisfies only the RHS equation).

    Yet this is correct: $$\color\red{|x|} = 5 \iff \color\red{±x} = 5.\tag{✔️}$$

  2. What's going on is that $|x|=\pm x$ is a faux equation and the first statement, which means $$\text{for each real }x,\:\:|x|\in \{x,-x\},\tag{✔️}$$ isn't actually an identity (a universally true equation).

    As a matter of fact, the result $$y=|x|\quad \iff\quad y=\pm x\:\:\:\text{and}\:\:\:y\ge0\tag{✔️}$$ is a quick derivation from the definition of absolute value, and the second ticked statement above is merely a particular case of it, rather than an instance of substitution-by-applying-an-identity.

  3. Thus, this is correct: \begin{align}&x^2+3x=|x| \\\iff{}&x^2+3x=\pm x \:\:\:\text{and}\:\:\:x^2+3x \ge0.\tag{✔️}\end{align}

    And this: \begin{align}&x^2+3x=|x| \\\iff{}&\big(x<0\:\:\:\text{and}\:\:\:x^2+3x=-x\big) \:\:\:\text{or}\:\:\:\big(x\ge0\:\:\:\text{and}\:\:\:x^2+3x=x\big).\tag{✔️}\end{align}

    And this: \begin{gather}x^2+3x=|x| \\\implies x^2+3x=x \:\:\:\text{or}\:\:\:x^2+3x=-x;\\ \text{eliminate the extraneous solution }-2.\tag{✔️}\end{gather} (The extraneous solutions created by the logic of an absolute-value equation, unlike those of a logarithmic or trigonometric equation, are generally not simply outside the equation's domain.)

  • $\begingroup$ I would not use the plus-minus sign in the manner you use it here. I would read it as there being two valid solution, one with plus and the other with minus, whereas you seem to be reading it as an or-operator. $\endgroup$
    – Tommi
    Commented Mar 17, 2023 at 7:36
  • $\begingroup$ @Tommi The implicitly universally quantified statement $$x^2=4⟺x=−2 \;\textbf{ or }\; x=2\tag1$$ says, “as $x$ varies over $\mathbb R$ (i.e., for each $x{\in}\mathbb R$), $x^2=4$ holds precisely when $x=-2$ or $x=2$"; in short: $$x^2=4 \text{ has solutions } {−}2 \;\textbf{ and }\; 2.\tag2$$ Statements (1) and (2) are equivalent to and consistent with each other. $\tag*{}$ To be clear: in $$f(x)=0⟺x=−2 \:\:\textbf{and }\:\: x=2⟺x\in\emptyset,$$ the solution set is empty; in $$g(x)=0⟺x=−2 \:\:\textbf{ or }\:\: x=2⟺x=\pm2⟺x\in\{-2,2\},$$ there are two solutions. $\endgroup$
    – ryang
    Commented Mar 17, 2023 at 17:40
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    $\begingroup$ You are correct, there is something interesting going on with my intuition about notation here. $\endgroup$
    – Tommi
    Commented Mar 22, 2023 at 7:54
  • $\begingroup$ @Tommi That misunderstanding is actually common and understandable; that said, "my son can only be Alan or Ben" (considering one object at a time) versus "I have 2 sons: Alan and Ben" (summarising) is a matter of framing. $\endgroup$
    – ryang
    Commented Mar 22, 2023 at 13:17

IMHO the main issue is that the absolute value doesn't satisfy any simple algebraic rules, so you cannot simplify the expressions including it. The computation itself is no more difficult than the computation with parentheses. If a student has more trouble with finding $f(x)=\Big\lvert\big\lvert\lvert x-3\rvert-5\lvert 3x-7\rvert\big\rvert-\lvert 6x+2\rvert\Big\rvert$ than with finding $g(x)=\Big(\big((x-3)-5(3x-7)\big)-(6x+2)\Big)$, for a given value of $x$, that merely means that they should put some effort into understanding the basic definitions and learning the order of operations. However, if one asks to solve the inequality $f(x)\le 10$ as opposed to solving $g(x)\le 10$, then the increase in the degree of difficulty is genuine not because of some fancy "psychological" reasons, but merely because it is a truly long and branching task, i.e., one can write a three line computer program that outputs the answer to the second question, but I wonder how much effort it will take you to write a piece of code that solves the first inequality correctly when I put arbitrary parameters instead of the numeric values into that rather simple expression. And if you have trouble teaching a machine to do something that requires just a routine algorithm, you'll certainly have trouble with students too, because at that level they are just imperfect computers.

  • $\begingroup$ That coincides with my experience when teaching absolute values. (By the way, I adjusted the size of the parantheses and the modulus a bit to make the formulas easier to read.) $\endgroup$ Commented Mar 17, 2023 at 17:43

These are great answers to a great question.

Although my answer might only be a rephrasing of what is already said, I think a main problem with absolute value is that it can be ambiguous (or have similar wording but different answers)..

For example, if I say the difference between $x$ and $4$ is $3$ then what is the answer?

  1. Is it $x-4 =3 \to 7$, using the "difference"?

  2. Is it $|x-4|=3$ and then it becomes $x-4=3 \to x= 7$ and $-(x-4)=3 \to x = 1$ using the absolute difference?

To me, the second option seems more likely if you think geometrically about placement on the number line, but more likely it's the first answer since the question didn't state "absolute difference"


Another thing I thought of for absolute value, is that it can be unnecessary like $|x^2|=x^2$, but then necessary with $|x| \neq x$


There is a lot of truth in many of the answers here, but they are all addressing this from a pure math standpoint. The logic of absolute value isn't hard, though - even if there are many examples here of weak prior knowledge that can cause problems.

Absolute value is hard for students to learn because it is almost always taught in extremely abstract and boring ways that just repel the vast majority of people from math while powerfully discouraging sense-making.

Here is an example from the ironically named "Math is Fun - Absolute Value" website

enter image description here

How fun and interesting is this? If we were to poll 1000 students, would 5 of them enjoy making meaning of this? Would even one see absolute value as useful?

I doubt it. The normal reaction almost assuredly would be something like "This is just more proof that math is just a bunch of weird rules and steps to obey. More rote crap to memorize." With this as a foundation, there is almost no way to make sense of any of the other very clear and logical points made in this thread.

In order to get students interested in the meaning of absolute value they need context in which absolute value thinking is useful.

You are a purchasing agent at a pharmacy. You need to order a large amount of a generic cancer drug and three different factories offer to sell it to you. You ask for a sample of their pills to check their quality. Each pill should have exactly 700mg of active ingredient, but no factory is perfect.

Factories A and B are brand new and not ready to produce enough pills for you. They send you pills just to get your feedback. Factories C and D are ready to sell to you now. They send a much larger sample of pills for you to analyze.

Your mission: (1) Recommend to your manager which factory to buy from. (2) Justify your answer. (3) Develop a written process for evaluating quality of all future pill samples.

Do great work because your patients' lives depends on it!

Pill Quality Case

Instructions are on the first sheet. Data is on the second sheet. Reflect explicitly to consolidate is on the third sheet.

Here's a little snippet of data.

enter image description here

First you can have them estimate the quality of each factory. B is clearly better than A.

Now, you can guide them through calculating, totaling, and averaging errors for each factory, in which case a pill being, say, 200mg over and another pill being 200mg under exactly cancels. And whaddaya know: Factory A is perfect and Factory B sucks! Math beats common sense! According to the total and mean error, we should go with Factory A. Maaaaaaybe grandma will scream in unnecessarily excruciating pain and the baby will go into overdose, but according to mean error calculations, that is OK. lolz

Then, the more outspoken ethical sticklers will say that they, in fact, don't want to torture grandma and kill a baby. (BUT, LIKE, MEAN ERROR CALCULATIONS YO, IT'S JUST MATH DUDE, GOTTA LISTEN TO DEM NUMBERS.)

Hmm. Maybe 200mg under and 200mg over should not cancel out when added up, because an overdosing baby doesn't really cancel out screaming grandma. So we could if we were very careful, always make sure to do:

  • Error = 700mg - (measured active ingredient) when the mg of active ingredient is less than 700mg


  • Error = (mg of active ingredient) - 700mg when the mg of active ingredient is greater than 700

And that way, we get no negative numbers. No more errors that offset each other.


We just have to be really careful on every single calculation. Every. Single. Pill.

That's pretty easy for Factories A and B.

Who wants to do all those calculations for the hundreds of pills from Factories C and D? Hands up for volunteers! Hands up please!

Uhh... anyone... anyone?

If only there were an easier way to get rid of those damn negative errors... Wouldn't that be awesome?

At this point, you can introduce absolute value (and clicking/dragging formulae in spreadsheets) and every student knows why it's necessary and what it means. You could go a little deeper here into this case, formally relate the two error formulae to absolute value, then gradually abstract away from this and towards all the other posts in this thread. It can now become obvious why $\lvert x-700 \rvert=400$ must have two solutions: An absolute error of 400mg can be caused by overage or underage.

This stuff can all make sense because absolute value has meaning in their minds.

Anyways, if you have any suggested updates for the case or other thoughts on how I teach absolute value, please reply. :-)

  • $\begingroup$ I like your example. For me, I consider distance as one fairly simple way to consider absolute value. Along a number line, the points at both $-2$ and $4$ are the same distance of $3$ from the point at $1$ since you need to move $3$ units in both cases to get to $1$, with $|-2 - 1| = |4 - 1| = 3$. In particular, distance doesn't care whether the other point is to the left or right of the point being compared to, just how many units it needs to move, with is given by the absolute value of the difference in values. $\endgroup$ Commented Aug 12, 2020 at 0:22
  • $\begingroup$ Thanks! Do you accompany your distance approach with some kind of context? I've never been able to get students interested in this without one. $\endgroup$ Commented Aug 12, 2020 at 6:51
  • $\begingroup$ You're welcome. When I tutored a $2$'nd year university engineering calculus course for a couple of years, I was asked several times about absolute value. This was over $30$ years ago, so I don't recall many details now of what I was specifically asked and how I responded, but I vaguely recall mentioning about its relation to distance at least once or twice, although I'm not sure of what sort of context I used then. Regardless, I agree giving appropriate context generally helps students to better understand mathematical concepts. $\endgroup$ Commented Aug 12, 2020 at 7:15

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