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Homework Help: Is this true: |x|=+-x?

  1. Jun 20, 2007 #1
    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.
  2. jcsd
  3. Jun 20, 2007 #2

    D H

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    What happens if x is negative?
  4. Jun 20, 2007 #3
    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.]
  5. Jun 20, 2007 #4
    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 [tex]x^{\frac{1}{2} [/tex] 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.
    Last edited: Jun 20, 2007
  6. Jun 20, 2007 #5
    Thanks morson.

    [By the way, how do you insert mathematical symbols into posts like you just did?]
  7. Jun 20, 2007 #6


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    No, only one of the signs is allowed, depending on the sign of x. If they wrote [itex]|x| = \pm x [/itex], 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
    [itex]|x| = x \, \rm{if} x \geq 0 [/itex]


    [itex] |x| = -x \, \rm{if} x <0 [/itex]

    (the equality case could have been put on either of those or on both since zero does not care about the sign)
    Last edited: Jun 21, 2007
  8. Jun 20, 2007 #7
    Umm ... is nrqed disagreeing with morson, or did I just misunderstand someone?
  9. Jun 20, 2007 #8

    D H

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    Per your original post, your CliffsNotes say that [tex](x^2)^{1/2} = |x|[/tex] in one place and [tex](x^2)^{1/2} = \pm x[/tex] in another. First, look back at these notes. Does it say that [tex]\sqrt{x^2} = |x|[/itex] rather than [tex](x^2)^{1/2} = |x|[/tex]?

    As morson indicated, [tex]\sqrt x[/tex] 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 [tex]x^{1/2}[/tex]. As you indicated, the CliffNotes are a bit unreliable.
  10. Jun 20, 2007 #9
    First of all, it does say "square root of x", not (x^2)^1/2, I just haven't figured out yet how to write that with a square root sign!

    So according to what you'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 = "+-(square root of y)" [ahhh!! Somebody tell me how to get math symbols in here!] instead of just "square root of y", since although the first way is technically correct, it is not understood without prefacing the square-root symbol with a "+-". Is that correct?
  11. Jun 20, 2007 #10
    Click on the square root symbol generated by someone else and that will give you the code. Here it was \sqrt x
  12. Jun 20, 2007 #11
    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

    |±x| = √(±x2) = +x but
    |±x| ≠ (√±x)2 ≠ ±x
    Last edited: Jun 20, 2007
  13. Jun 20, 2007 #12
    I have always disagreed with the mathematical definition of the square root being single-valued.
    But we have to live with it. :smile:

    Anyone knows who was the "genius" who came up with that anyway?
  14. Jun 20, 2007 #13
    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.
  15. Jun 20, 2007 #14
    √-x = -x is obviously false.

    But at any rate, you cannot prove or disprove the workings of an operator. It is simply a definition.
  16. Jun 20, 2007 #15
    My fault. I thought you were talking about the negative square root; while you meant √(x2) = ±x and not, +x or -x.
  17. Jun 21, 2007 #16


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    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.
  18. Jun 21, 2007 #17
    modulus x is always positive yes, but you can replace it with a +-x in calculations i think.
  19. Jun 21, 2007 #18
    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.
  20. Jun 21, 2007 #19
    I don't see what so funny here, the inverse of y = x2 is not a function.

    And you think that is a valid argument?
  21. Jun 21, 2007 #20
    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?
  22. Jun 21, 2007 #21
    I am not arguing for a change.
  23. Jun 21, 2007 #22
    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.
  24. Jun 21, 2007 #23
    Indeed, and now please demonstrate the logic of that restriction.
  25. Jun 22, 2007 #24
    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.
  26. Jun 22, 2007 #25


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    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.

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

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

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

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

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


    [tex]\sqrt{-x ^ 2}[/tex] is not defined for [tex]x \neq 0[/tex]

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


    We know that |x| will always return the positive value, so, we have:
    [tex]|x| = \left\{ \begin{array}{l} x , \ \ \ x \geq 0 \\ -x , \ \ \ x < 0 \end{array} \right.[/tex]
    To make it short, we can write:
    [tex]|x| = \pm x[/tex], i.e depends on x, whether x is negative or not, a "-", or a "+" sign can be chosen.
    Last edited: Jun 22, 2007
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