Understanding Absolute Value in Algebra: |x|=+-x Explained

[alenglander]

I am using CliffsNotes QuickReview Algebra I (which by the way I find to be fantastic, except that it is very prone to typos) to review the algebra that I haven't learned in almost a decade. It says there that (x^2)^1/2 = |x|, but it also says that (x^2)^1/2 = +-x. But isn't it true that |x| > 0? So how can |x| = +-x? Someone please help me understand this. Thanx.

 

[D H]

What happens if x is negative?

 

[alenglander]

If x is negative then |x| = -x (for example, say x = -3, so |-3| = -(-3) = 3), because the absolute value of a number is always a positive number. But +-x refers to the number you get as a result of putting either a + or a - in front of x. So if x = -3, +-x could equal +(-3) OR -(-3). Only ONE of these is a positive number (-(-3)).

So to use the example of -3, my question is that |-3| = 3, but +-(-3) could equal 3 OR -3, and since 3 obviously does not equal -3, how can you say that |-3| = +-3?

[Sorry if that's a little convoluted. I tried to be clear, but I may not have succeeded.]

 

[morson]

The answer to your thread's question is no, if you're considering nonzero x, and if you mean "plus AND minus." Some consider the symbol √x as the principle value of the square roots of x (ie: only the positive square root). Some consider x^{\frac{1}{2} to represent BOTH the square roots of x, but I have come to think of them as the same. So, |x| ≠ +-x, if x is nonzero.

 

[alenglander]

Thanks morson.

[By the way, how do you insert mathematical symbols into posts like you just did?]

 

[nrqed]

If x is negative then |x| = -x (for example, say x = -3, so |-3| = -(-3) = 3), because the absolute value of a number is always a positive number. But +-x refers to the number you get as a result of putting either a + or a - in front of x. So if x = -3, +-x could equal +(-3) OR -(-3). Only ONE of these is a positive number (-(-3)).

So to use the example of -3, my question is that |-3| = 3, but +-(-3) could equal 3 OR -3, and since 3 obviously does not equal -3, how can you say that |-3| = +-3?

[Sorry if that's a little convoluted. I tried to be clear, but I may not have succeeded.]

No, only one of the signs is allowed, depending on the sign of x. If they wrote |x| = \pm x, it was a shorthand for the more completed definition:


OOPS: EDIT: should have been a larger or equal than... I thought that leq was larger or equal but it is less or equal... It's corrected below
|x| = x \, \rm{if} x \geq 0

and

|x| = -x \, \rm{if} x <0

(the equality case could have been put on either of those or on both since zero does not care about the sign)

 

[alenglander]

The answer to your thread's question is no, if you're considering nonzero x, and if you mean "plus AND minus." Some consider the symbol √x as the principle value of the square roots of x (ie: only the positive square root). Some consider x^{\frac{1}{2} to represent BOTH the square roots of x, but I have come to think of them as the same. So, |x| ≠ +-x, if x is nonzero.

No, only one of the signs is allowed, depending on the sign of x. If they wrote |x| = \pm x, it was a shorthand for the more completed definition:

|x| = x \, \rm{if} x \leq 0

and

|x| = -x \, \rm{if} x <0

(the equality case could have been put on either of those or on both since zero does not care about the sign)

Umm ... is nrqed disagreeing with morson, or did I just misunderstand someone?

 

[D H]

Per your original post, your CliffsNotes say that (x^2)^{1/2} = |x| in one place and (x^2)^{1/2} = \pm x in another. First, look back at these notes. Does it say that \sqrt{x^2} = |x|[/itex] rather than (x^2)^{1/2} = |x|?<br /> <br /> As morson indicated, \sqrt x universally indicates the the principal (i.e., positive) root. If you want to allow negative and positive roots, it is better to use the form x^{1/2}. As you indicated, the CliffNotes are a bit unreliable.

 

[alenglander]

Per your original post, your CliffsNotes say that (x^2)^{1/2} = |x| in one place and (x^2)^{1/2} = \pm x in another. First, look back at these notes. Does it say that \sqrt{x^2} = |x|[/itex] rather than (x^2)^{1/2} = |x|?<br /> <br /> As morson indicated, \sqrt x universally indicates the the principal (i.e., positive) root. If you want to allow negative and positive roots, it is better to use the form x^{1/2}. As you indicated, the CliffNotes are a bit unreliable.

<br /> <br /> First of all, it does say &quot;square root of x&quot;, not (x^2)^1/2, I just haven&#039;t figured out yet how to write that with a square root sign!<br /> <br /> So according to what you&#039;re saying: If I have an equation, say x^2 = y, and I want to solve for x, which means taking the square of both sides, then I need to specify x = &quot;+-(square root of y)&quot; [ahhh! Somebody tell me how to get math symbols in here!] instead of just &quot;square root of y&quot;, since although the first way is technically correct, it is not understood without prefacing the square-root symbol with a &quot;+-&quot;. Is that correct?

 

[Vagrant]

Click on the square root symbol generated by someone else and that will give you the code. Here it was \sqrt x

 

[cshum00]

This is bad though.
|±x| ≠ ±x

The absolute value is defined to always ONLY give positive values.
|±x| = +x

But mathematically, the way to do it is to square it and the take the square root. For |±x| = +x
(±x)² = +x² but +x² ≠ +x so
√(+x²) = +x

So we can define that,
|±x| = √(±x²)

Cliff Notes is wrong that,
|±x| = ±x

But the thing is that, √(±x²) = ±x because
√(±x²) = (√±x)² and
(√±x)² = ((±x)²)^1/2 = (±x)¹

However,
|±x| = √(±x²) = +x but
|±x| ≠ (√±x)² ≠ ±x

 

[MeJennifer]

I have always disagreed with the mathematical definition of the square root being single-valued.
But we have to live with it.

Anyone knows who was the "genius" who came up with that anyway?

 

[cshum00]

I have always disagreed with the mathematical definition of the square root being single-valued.
But we have to live with it.

Anyone knows who was the "genius" who came up with that anyway?

Well, i also disagree with that but the problem is that we are still unable to prove that √-x = -x. Until you find a prof, we have to stick with it.

 

[MeJennifer]

Well, i also disagree with that but the problem is that we are still unable to prove that √-x = -x. Until you find a prof, we have to stick with it.

√-x = -x is obviously false.

But at any rate, you cannot prove or disprove the workings of an operator. It is simply a definition.

 

[cshum00]

My fault. I thought you were talking about the negative square root; while you meant √(x²) = ±x and not, +x or -x.

 

[nrqed]

Umm ... is nrqed disagreeing with morson, or did I just misunderstand someone?

Sorry. My mistake. I corrected it now. I used the latex "leq" thinking it was larger or equal to but it is for less or equal to. Corrected now. Sorry.

 

[Neutralino]

modulus x is always positive yes, but you can replace it with a +-x in calculations i think.

 

[ZioX]

I have always disagreed with the mathematical definition of the square root being single-valued.
But we have to live with it.

Anyone knows who was the "genius" who came up with that anyway?

Some people like functions to be well defined...



Of course, you could always map it to a 2-tuple, which is what I think you were hinting at. The graphing complications that would naturally arise would make it confusing for a lot of people.

 

[MeJennifer]

Some people like functions to be well defined...

I don't see what so funny here, the inverse of y = x² is not a function.

The graphing complications that would naturally arise would make it confusing for a lot of people.

And you think that is a valid argument?

 

[ZioX]

Well, I'm not even really sure what you were suggesting. Are you saying sqrt(x) should map to a 2-tuple, namely (sqrt(x),-sqrt(x))?

Besides, what is your argument for even changing it in the first place?

 

[MeJennifer]

Besides, what is your argument for even changing it in the first place?

I am not arguing for a change.

 

[ZioX]

Why do you disagree then?

Besides, the curve y=x^2 is invertible on the positive side, which is where we restrict our domain for sqrt(x) in the first place.

 

[MeJennifer]

Besides, the curve y=x^2 is invertible on the positive side, which is where we restrict our domain for sqrt(x) in the first place.

Indeed, and now please demonstrate the logic of that restriction.

 

[ZioX]

We could arbitrarily choose our domain to be the negative real axis and get the same results. The whole point is to get a functional inverse. It's the same deal with inverses in trigonometry.

 

[VietDao29]

Indeed, and now please demonstrate the logic of that restriction.

Well, for one function to have an inverse, it must be a 1-to-1 function. So, we just choose the domain, where it satisfy the requirement. R⁺, or R^- would do it.

This is bad though.
|±x| ≠ ±x

The absolute value is defined to always ONLY give positive values.
|±x| = +x

Errr... What if x is negative? Say x = -1.

So, we have:
|-1| = +(-1) = -1?

But mathematically, the way to do it is to square it and the take the square root. For |±x| = +x
(±x)2 = +x2 but +x2 ≠ +x

What if x = 0? And ,btw, how can you go from this to the following claim?

...so
√(+x2) = +x

Again, what if x = -1?
\sqrt{(-1) ^ 2} = \sqrt{1} = +(-1) = -1?

So we can define that,
|±x| = √(±x2)

-x² is a negative number for x \neq 0, you cannot have: \sqrt{-x ^ 2} , \ \ \ x \neq 0. It's not even defined in the reals.

Cliff Notes is wrong that,
|±x| = ±x

Really?

But the thing is that, √(±x2) = ±x because
√(±x2) = (√±x)2 and
(√±x)2 = ((±x)2)1/2 = (±x)1

However,
|±x| = √(±x2) = +x but

\sqrt{-x ^ 2} is not defined for x \neq 0

|±x| ≠ (√±x)2 ≠ ±x

Are you sure: |\pm x| \neq ( \sqrt{\pm x} ) ^ 2?

-------------------------

@OP:
We know that |x| will always return the positive value, so, we have:
|x| = \left\{ \begin{array}{l} x , \ \ \ x \geq 0 \\ -x , \ \ \ x &lt; 0 \end{array} \right.
To make it short, we can write:
|x| = \pm x, i.e depends on x, whether x is negative or not, a "-", or a "+" sign can be chosen.

 

[MeJennifer]

Well, for one function to have an inverse, it must be a 1-to-1 function. So, we just choose the domain, where it satisfy the requirement. R⁺, or R^- would do it.

Begging the question why a function supposedly "must" have an inverse function. It is simply the case that some functions do not have an inverse function. I don't see the problem with that.

Anyway it appears we can agree to disagree, I do not think it logical at all to define the square root as an operator that returns a single value, while others here think it is as logical as apple pie. Bottom line is that we have to accept the definition as is.

 

[cshum00]

What if x = 0? And ,btw, how can you go from this to the following claim?

Again, what if x = -1?
\sqrt{(-1) ^ 2} = \sqrt{1} = +(-1) = -1?

That is a good one. But then, are implying that |x| ≠ √(x²)?

-x² is a negative number for x \neq 0, you cannot have: \sqrt{-x ^ 2} , \ \ \ x \neq 0. It's not even defined in the reals.

\sqrt{-x ^ 2} is not defined for x \neq 0

Are you sure: |\pm x| \neq ( \sqrt{\pm x} ) ^ 2?

If you take the square root first, it is not defined. But if you play around with the exponents first, it is defined. It is like calculus problems where it is not defined if you substitute limits immediately.

We know that |x| will always return the positive value, so, we have:
|x| = \left\{ \begin{array}{l} x , \ \ \ x \geq 0 \\ -x , \ \ \ x &lt; 0 \end{array} \right.
To make it short, we can write:
|x| = \pm x, i.e depends on x, whether x is negative or not, a "-", or a "+" sign can be chosen.

I still don't like the |x| = ±x notation because it is contradicts to what the absolute value does. Once x is outside the absolute value symbol, it means that the absolute value is already taken and show only positive values.

Now, as the example you showed before, √(x²) is not a good representation of |x| because it also gives a -x; which is the same as not taking any absolute value. Then what should be do? Tell the mathematicians to consider removing this bad representation?

 

[K.J.Healey]

I still don't like the |x| = ±x notation because it is contradicts to what the absolute value does. Once x is outside the absolute value symbol, it means that the absolute value is already taken and show only positive values.

Just because x is outside the |x| doesn't mean its positive at all. We're not using operator algebra like quantum.
|x|=x+4 is possible because x != |x|.


Ok, here's how you have to think of it, "x" can be positive or negative, we don't know.
Given |x| = ±x
The left had side can only be positive right?
But the right had side COULD be either, but the ± makes SURE its positive.
If x = -2, then the "±" becomes a "-". If x=2, then it becomes a "+".
If can, and MUST BE, either the + or the -.
Basically you have to say :
If x is positive, |x| = x
if x is negative, |x| = -x
Therefor at all times |x| = ±x

get it?

 

[malawi_glenn]

Also \sqrt{x^{2}} = |x|

It is a big difference in just writing the
\sqrt{... }
It implies that we choose only the positive value.

But the equation:
x ^{2} = 9
has two solutions:
x = \pm \sqrt{9}

 

[cshum00]

Also \sqrt{x^{2}} = |x|

It is a big difference in just writing the
\sqrt{... }
It implies that we choose only the positive value.

But the equation:
x ^{2} = 9
has two solutions:
x = \pm \sqrt{9}

No, i don't think so. Only in square root of natural numbers implies to choose the positive result. But absolute values, we deal with negative numbers too. Although we can set the absolute value to get only the positive number of the square root but then why the whole fuss of squaring it in the first place?

For example:
if a is positive
|-a| = +a by excluding the negative sign
but if |x| = √(x²)
√((-a)²) = √(a²) = ±a = +a by picking the positive one
That is why i say it is not a good representation.

Just because x is outside the |x| doesn't mean its positive at all. We're not using operator algebra like quantum.
|x|=x+4 is possible because x != |x|.Ok, here's how you have to think of it, "x" can be positive or negative, we don't know.
Given |x| = ±x
The left had side can only be positive right?
But the right had side COULD be either, but the ± makes SURE its positive.
If x = -2, then the "±" becomes a "-". If x=2, then it becomes a "+".
If can, and MUST BE, either the + or the -.
Basically you have to say :
If x is positive, |x| = x
if x is negative, |x| = -x
Therefor at all times |x| = ±x

get it?

I see. True, but the notation still easily confuse people.
I prefer,
if x = ±a
|x| = +a

 

[VietDao29]

For example:
if a is positive
|-a| = +a by excluding the negative sign
but if |x| = √(x2)
√((-a)2) = √(a2) = ±a = +a by picking the positive one
That is why i say it is not a good representation.

Yes, if a is positive, then this is correct. But, what if a is negative? Shouldn't it be |-a| = -a? So, to find a general formula, we use:
|a| = \pm a, i.e, depends on the sign of a, a plus sign (+), or a minus sign (-) can be wisely chosen so that the final result we get is positive.

I see. True, but the notation still easily confuse people.
I prefer,
if x = ±a
|x| = +a

Noooo. =.="

The notation is fine. You seem to be confuse between +1, and +a.
+1 means that it is positive, to distinguish it from -1, which is a negative number.

However, the second case is completely different, +a does not mean that it is positive.
It can also be negative, say, if a = -3, then +a = +(-3) = -3, a negative number, whereas, -a = -(-3) = +3, a positive number.
So +a does not mean that it's positive.

 

[turdferguson]

|x| is either positive or negative, but not both at the same time. When taking the square root of both sides of an equation, you must take cases and usually come up with 2 possible solutions, a positive and negative answer

 

[cshum00]

lol (XD)
When i said = +a i meant the final answer and after substituting it and all.
In other words, |-3| = +3.

 

[laurelelizabeth]

Maybe what they are trying to say is that |x| is what is a "piecewise funcion", in other words made up of more than one piece

If x is greater than or equal to 0, then y=x
If x is less than 0, then y=-x

If you graph that you get that V shape :)

 

[HallsofIvy]

No, i don't think so. Only in square root of natural numbers implies to choose the positive result. But absolute values, we deal with negative numbers too. Although we can set the absolute value to get only the positive number of the square root but then why the whole fuss of squaring it in the first place?

No, you are wrong. the square root of any real number, a, is, by definition, the positive real number x such that x²= a.

For example:
if a is positive
|-a| = +a by excluding the negative sign
but if |x| = √(x²)
√((-a)²) = √(a²) = ±a = +a by picking the positive one
That is why i say it is not a good representation.

Once again, no. whether a is positive or negative, \sqrt{(-a)^2}= \sqrt{a^2}= |a|. The squareroot function has only one value.

I see. True, but the notation still easily confuse people.
I prefer,
if x = ±a
|x| = +a

That's nonsense. If a= -4 that would say "if x= ±(-4), then |a|= -4.

 

[CompuChip]

Anyone knows who was the "genius" who came up with that anyway?

Probably a genius to whom functions were one-to-one and non-injective functions weren't functions at all, or something like that

 

[HallsofIvy]

Probably a genius to whom functions were one-to-one and non-injective functions weren't functions at all, or something like that

No, "one-to-one" has nothing to do with being a function. f(x)= x² is not one-to-one but is a function. It has everything to do with being "well defined"- for a given value of x, you get a specific, "well defined" value of f(x).

I've always thought of it as similar to what we demand of experiments- if two people do it in exactly the same way, they get the same result.