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He says (direct quote): "In higher mathematics the absolute value of a number, $|x|$, is equal to positive and negative $x$, if $x$ is a positive number."

Then he wrote: $|x|=\pm x,\quad \text{if}\enspace x>0 $.

I think he misunderstood the definition of absolute value, or did I? From what I understand, absolute value of a number is the distance of a number from zero, so it is always positive. Am I wrong?

Michael Rozenberg
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user466742
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    $|x|$ is always non-negative, but don't forget that $-x$ can be positive (that is, when $x$ is negative). – Angina Seng Jul 24 '17 at 11:25
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    Seems to me like he's trying to condense the following into a single line:$$|x| = \cases{x & if $x > 0$\-x & if $x\leq 0$}$$(This is true, by the way). It is the case that $|x|$ is either $+x$ or $-x$ (thus $\pm x$), depending on whether $x > 0$, so all the important elements are there, but it's just a confused mess in my opinion. – Arthur Jul 24 '17 at 11:25
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    The concept you have for absolute value is right, however and is a good way of thinking about it – Shuri2060 Jul 24 '17 at 11:27
  • He also said that $|x|=-x$ if x is a negative integer. He also made an example: $\sqrt{2^2}=|2|=\pm 2$ . Which made things even worse. – user466742 Jul 24 '17 at 11:28
  • you right. but consider the point $(-x,0)$, the distance between the point $(-x,0)$ and $(0,0)$ is $x$ unit, because of this we write $|x|=\pm x$. – Amin235 Jul 24 '17 at 11:29
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    "$|x| = -x$ if $x$ is negative" is a true statement. For instance, $|-5| = -(-5)$. Adding a sign is how we "remove" the sign from a negative number when we don't have typographical access to the sign itself, such as with "$x$" when $x$ is negative. – Arthur Jul 24 '17 at 11:29
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    I'll point out that if you're trying to solve the equation $|x|=a$, then the solution would be $x=\pm a$ if $a\ge0$ which is perhaps what the last thing refers to? – Shuri2060 Jul 24 '17 at 11:32
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    The $\sqrt{2^2}=\pm 2$ example pretty much shows that this teacher, if he is as you present him, knows less than nothing about "higher mathematics." There are way too many of these guys out there. They minored in math, majored in PE, and were hired to coach volleyball and teach 5 sections of algebra 1 on the side. – B. Goddard Jul 24 '17 at 13:00
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    In practice, I think it would be reasonably common to see $\sqrt{2^2} = \pm 2$ written in the context of an elementary algebra or arithmetic class. Of course this doesn't agree with the standard definition of the square root function. I suppose it's possible the teacher has mixed up two interpretations of $\sqrt{\cdot}$: the one where $\sqrt{x^2} = \lvert x\rvert$ (standard), and the very different and nonstandard one where $\sqrt{x^2} = \pm x$. – David Z Jul 24 '17 at 13:33
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    Don't you agree that is is more likely that you misunderstood your teacher, than that he has a different definition of absolute value to all other mathematicians? – jwg Jul 24 '17 at 14:06
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    @DavidZ "in the context of elementary...." But the teacher specifically said, "in higher mathematics." If our picture of this guy is accurate, he's blustering to cover his ignorance. – B. Goddard Jul 24 '17 at 14:21
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    @B.Goddard I'm not talking about what this teacher is saying about higher mathematics. I'm saying that, in general, I think it is reasonably common to see $\sqrt{x^2} = \pm x$ in the context of elementary-level math education. Perhaps this teacher has picked up that usage of the square root symbol the same way many other teachers do, and just failed to realize that the same symbol is being used for different purposes. In any case I do agree that he's probably trying to cover his ignorance. – David Z Jul 24 '17 at 15:01
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    How about a title which is less click-baity and more about the actual content of the question? – Asaf Karagila Jul 24 '17 at 17:50
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    "Higher mathematics"? – anomaly Jul 24 '17 at 18:29
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    In old books about conic sections- more than a century old- I have seen $$\tan \theta = \pm\dfrac{(m+M)}{(1- Mm)}$$ standing for $$\tan \theta = |\frac{(m+M)}{(1- Mm)}|$$. – user8277998 Jul 24 '17 at 19:16
  • Those words are very confusing indeed, to say the least. – Pedro A Jul 25 '17 at 00:05
  • @Arthur Condensing the definition into one line correctly is easy. $|x| = x\mathrm{sgn}(x)$. – Deepak Jul 25 '17 at 05:33
  • @Deepak Most expressions can be condensed into one line if you hide everything behind other definitions. In your expression there is no explicit mention of neither $\pm$ nor "if $x>0$". My point was, if you want those two things to appear in your definition, then it is best to do it in two lines. – Arthur Jul 25 '17 at 06:18
  • $\lvert\pm x\rvert= x$ for all $x\geq0$ is surely what he meant. – Bonnaduck Jul 26 '17 at 03:50
  • I think you know what absolute value means. I recommend that you pretend to agree with the teacher and feed back to him whatever he wants until the course is over and you have received your grade. Then avoid him in the future.

    On the other hand, I remember some of the most terrible math teachers imaginable. Years later, I found that they wrote the best books on certain subjects. You never know.

     – richard1941 Jul 27 '17 at 15:21
  • $$|x|=±x$$ is a true statement, not just for positive $x,$ but for every real $x.$ – ryang Mar 16 '23 at 15:24

7 Answers7

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Mathematical definition for absolute value of $x$.

$$y=|x|$$

$$|x|=\begin{cases}x &, x\geq 0\\-(x) &, x < 0\end{cases}$$

The geometrical definition for absolute value of x is

enter image description here

Notice that they will always make the value of y non-negative.

Just to elaborate...

$$x=-1$$

Absolute value of $-1$ is then $1$.

$$x=3$$

Absolute value of $3$ is still $3$.

$$x=0$$

Absolute value of $0$ is still $0$.

What is

$$y=|x-1|$$

$$|x-1|=\begin{cases}(x-1) &, x-1 \geq 0\\-(x-1) &, (x-1) < 0\end{cases}$$

For $x-1$

$$x-1 \geq 0$$

$$x \geq 1$$

For $-(x-1)$

$$(x-1) < 0$$

Which is

$$ x<1$$

The new notation becomes

$$|x-1|=\begin{cases}(x-1) &, x \geq 1\\-(x-1) &, x <1\end{cases}$$

Crazy
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I think you are right, and your teacher - though I would assume he knows the definition - was too sloppy when he introduced the absolute value.

You are right, the absolute value of a number is its distance to the point $0$ on the real line, so the correct definition would be $$ |x| := \begin{cases} x, \quad & \text{if } x \geq 0 \\ -x, \quad & \text{if } x<0. \end{cases} $$ The definition $|x| = \pm x$ yields not even a function, because this would have two $y$-values for every $x$-value and this is certainly not what he wants. I guess he might have wanted to say it is $\pm x$, depending on whether $x$ is positive or negative, in the way I wrote it above.

Maybe to have an additional example: $|5| = 5$, since $5 > 0$, so you do not have to change anything. But $|-3| = -(-3) = 3$, since $-3 < 0$, so the absolute value function tells you to multiply with $(-1)$, i.e. to cancel the minus sign.

And, to round things up, as pointed out above in the comments, if you want to solve an equation involving an absolute value, you typically get two solutions. Let us say you are looking for all $x$ that satisfy the equation $$ |x| = 4. $$ Then there are two solutions, namely $x = \pm 4$, but note that the absolute value itself is NOT $\pm 4$. It is just $4$, in either case.

Andre
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$|x|=\pm x$ it's wrong! For example, $|1|=\pm1$ it's absurd.

$|x|$ it's the distance between origin and the point $x$ at $x$ axis.

From this definition we obtain:

$|x|=x$ for $x\geq0$ and $|x|=-x$ for $x<0$.

Michael Rozenberg
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What your teacher might've meant by:

In higher mathematics, the absolute value of a number, |x|, is equal to positive and negative x if x is a positive number.

Is that in |x|, the variable x can be ±x.

              |-3|=|3|=3

The last: "if x is a positive number" is incorrect. It must've been a result of misinterpretation. So, we usually end up using this in places like this:

                  |3|<x

                 -3<x<3


As, x can be both 3,-3 simultaneously to give an absolute value of 3, which the inequality doesn't demand. (x should be smaller than absolute value of 3)

Alpha Mineron
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Your teacher seems to think that $|x| = +x $ and $-x $ at the same time. That's not true, as you correctly have understood (unless $x=0$). Instead we have $|x| =+x $ or $-x $ depending on sign of $x $.

Point your teacher to this discussion to show him the truth, the whole truth and nothing but the truth!

md2perpe
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Hmm... I'm a little confused by some of the comments here.

My understanding of Absolute value is that it is the distance FROM 0.

Meaning, that unless x has a value of 0, the Absolute value of x is ALWAYS positive...

|x| = |x| Because Absolute value depicts only the distance from 0, not direction, and there is no way to graphically depict a number stripped of it's +/-. So the value of x is irrelevant: Absolute value of a number is the whole number value (+) of the variable.

If x = -5 then |x| = 5

If x = +5 then |x| = 5

If x = -12 then |x| = 12

If x = +12 then |x| = 12

DISTANCE from 0 is the depiction, not direction. So while saying that |(+/-)x| = x would be true, but trying to say -x = |x| = +x would be false. After all, in absolute value, ALL values are positive, because they are an ABSOLUTE value as a distance from 0, and as THAT is never depicted in direction, on a graph Absolute value would be depicted as more a vertical line, as long as to the left was (+) numbers, and to the right (-) numbers (or vise-versa)...

That would be a proper understanding of Absolute value, no?

Asuka Jr.
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  • It is $$ |x| = x, when x >= 0 $$ $$ |x| = -x, when x < 0 $$. This is the way how modulus works. – AMAN Jul 26 '17 at 09:45
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I cannot comment, so I have to add this down here:

The last: "if x is a positive number" is incorrect. It must've been a result of misinterpretation. So, we usually end up using this in places like this:

              |3|<x

             -3<x<3

As, x can be both 3,-3 simultaneously to give an absolute value of 3, which the inequity doesn't demand. (x should be smaller than absolute value of 3)

This is one answer I'm not understanding. The two 'comments' are contridictory. In the first, |3| < x (Absolute value of 3 is less than x) is pretty straightforward. But is then followed by: -3 < x < 3 (negative 3 is less than x which is less than 3). Huh? Unless these are two completely independent statements, they are contradictory. In the first, x is greater than the absolute value of 3. Which is 3. So if the second comment is intended to be a corollary, shouldn't the value be stated as: -3 < x > 3? And as such is completely redundant, as the first establishes that the value of x is GREATER than the absolute value of 3, which is 3. Thus: x>3 done.

Did I miss something? Absolute value HAS no sign, and in fact DENIES the application OF a sign...

Then, to make it even worse for me, the poster then adds under that 'As, x can be both 3,-3 simultaneously'... What? No, x MUST be a value of LESS than 3 if the second inequity IS intended to be a second statement, AND MUST be GREATER than -3 for the same reason. Thus in the second, x = (-3,3) meaning that the value of x is any number (whole or not) that is greater than -3, and less than 3.

Now if the second IS intended as a corollary to the first, it's totally messed up, as it should then be: -3 < 3 < x and is again, redundant, because any number larger than 3 (|3|) is AUTOMATICALLY larger than -3... thus the most complete answer would be x = [3,+∞)... Or again, did I miss something?

Asuka Jr.
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