# My teacher describes absolute value confusingly: $|x|=\pm x,\quad \text{if}\enspace x>0$.

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?

• $|x|$ is always non-negative, but don't forget that $-x$ can be positive (that is, when $x$ is negative). – Lord Shark the Unknown Jul 24 '17 at 11:25
• 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
• "$|x| = -x$ 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
• 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
• 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

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

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

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

• i also showed that graph to my teacher. – user466742 Jul 24 '17 at 11:36
• Try to ask him to work some examples to see if his reasoning agree with the answers. – Crazy Jul 24 '17 at 11:44
• don't try and convince people not available to admit errors – G Cab Jul 24 '17 at 13:41
• @user466742 it seems like you are from India, is that correct? So many tution teachers in India dont understand absolute values and say and behave exactly like you told. – Kartik Jul 24 '17 at 16:18
• @Cracy: The statement $y$ positive should be corrected to $y$ non-negative. – Markus Scheuer Jul 25 '17 at 6:22

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.

$|x|=\pm x$ 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 $x<0$.

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)

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!

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?

• It is $$|x| = x, when x >= 0$$ $$|x| = -x, when x < 0$$. This is the way how modulus works. – AMAN Jul 26 '17 at 9:45

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?