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

[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.