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I Definition of absolute value

  1. Mar 15, 2017 #1
    If we define ##|a| = \sqrt{a^2}##, then why can't we do something like ##\sqrt{a^2} = (\sqrt{a})^2 = a##? Or equivalently ##\sqrt{a^2} = (a^2)^{1/2} = a^{2/2} = a##? Isn't this a contradiction?

    Also, how would this relate to showing that ##\sqrt{|a|} = |\sqrt{a}|## is true or false?
     
  2. jcsd
  3. Mar 15, 2017 #2
    Hi Mr Davis:
    When you take a square-root the result may be + or -. Therefore the definition of √a2 should be (+/-) a.

    Regarding
    I am not sure how this topic is taught this century, but I would say the answer is neither "true" or "false", but rather the answer is "incomplete", since it is also correct that
    √|a| = -|√a|.​

    Hope this helps.

    Regards,
    Buzz
     
  4. Mar 15, 2017 #3

    FactChecker

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    The ##\sqrt{ a }## symbol with no sign in front of it is understood by convention to be the positive value. If you want the negative value, you must put a minus sign in front. It could have been done other ways, but it would have no advantage or would be ambiguous.
    Many of your other questions are ignoring that the square root of a negative number is a problem. So you can not say that √(a2) = (√ a )2 = a if a is negative.
     
  5. Mar 15, 2017 #4
    Hi @FactChecker:

    Would you agree that for a < 0 it is OK to say:
    (√ a )2 = - √(a2) ?​

    Regards,
    Buzz
     
  6. Mar 15, 2017 #5

    FactChecker

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    Depends on whether complex numbers are allowed. Staying within the reals, I would say that the left side is undefined.
     
  7. Mar 15, 2017 #6
    Hi @FactChecker:

    Another question occurred to me.

    What does the notation (+/-)√a means by current conventions?

    Is is ambiguous? Or does it mean both + and - values are possible as correct interpretations? Or does it mean that either one or the other value, or both values may be a correct interpretation?

    Regards,
    Buzz
     
  8. Mar 15, 2017 #7

    FactChecker

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    Good question. In the places I have seen it, I always interpreted it as "try both options and use any and all that work". But I don't know if that is always how it should always be interpreted.
     
  9. Mar 15, 2017 #8

    Mark44

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    I agree with FactChecker. By tradition, the square root symbol represents the positive square root of a number. (I am considering only the square root of a nonnegative real number.)
    ##\sqrt{(-4)^2} = |-4| = 4##.

    It is not ##\pm 4##.
     
  10. Mar 15, 2017 #9

    Stephen Tashi

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    That is not correct. Even people who define the phrase "square root" so it allows some numbers to have two different square roots (such as the current Wikipedia article https://en.wikipedia.org/wiki/Square_root ) also define the notation "##\sqrt{}##" to mean the principal square root. Many math texts disagree with the Wikipedia definition and define the (verbal) phrase "square root" so that a number has at most one square root.

    Computing ##\sqrt{a^2}## is a different task than solving the equation ##x^2 = a^2##.

    The solutions to ##x^2 = a^2## are members of the set ##\{-a, a\}## and this fact may be abbreviated by the notation ##x = \pm a ##.
     
  11. Mar 15, 2017 #10
    I am still confused. The manipulation ##\sqrt{x^2} = x## seems valid by the rules of exponents, just as the manipulation ##(x^3)^{1/3} = x## is valid. Under what conditions is this valid or invalid, and does it also depend on the underlying field (real or complex)?
     
  12. Mar 15, 2017 #11

    FactChecker

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    Don't worry about complex numbers. The problem is when x is negative. ##\sqrt{x^2}## is always positive.
     
  13. Mar 15, 2017 #12
    So when exactly does the exponent rule ##(x^a)^{1/a} = x^{a/a} = x## hold true? We established that it doesn't when a = 2, but it does when a = 3.
     
  14. Mar 15, 2017 #13

    FactChecker

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    It is the √ symbol that causes the problem. I believe that it implies the positive value. When you use general exponents, you are free to interpret as you need, but when there are multiple roots, you should be clear about which roots you mean.
     
  15. Mar 15, 2017 #14

    Stephen Tashi

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    What rules of exponents are you referring to? If ##x = -2## then ##\sqrt{x^2} = -x##.

    What do you mean by "this"? Are you asking about the validity of ##\sqrt{x^2} = x## or about the validity of ##(x^3)^{1/3} = x## or about the general thought that ##(x^a)^b = x^{ab}##?

    ##\sqrt{x^2} = x## is valid for those ##x \ge 0##.

    In the complex domain, a non-zero number has n distinct n-th roots. Since a non-zero ##x## has 3 distinct cube roots, the notation ##(x^3)^{1/3} = x## is ambiguous unless we define which cube root is to be taken. If we use the convention that ##b^{1/3}## shall mean the real number (i.e. the one with zero imaginary part) that is a cube root of ##b## then ##(x^3)^{1/3}## is not valid as a generality for all complex numbers ##x##. It is valid for those ##x## in the subset of the complex numbers that consists of the real numbers (i.e. those with no imaginary part).

    In the domain of real numbers, ##(x^3)^{1/3} = x ## is valid. However, in the domain of real numbers, ##(x^a)^b = x^{ab}## is not valid as a generality for all ##x,a,b##. For example, with ##x = -1, a = 2/3, b = 1/a = 3/2## we have ##x^a = ( (-1)^2) ^{1/3} = 1^{1/3} = 1## and ## (x^a)^b = (1)^{3/2} = 1## but ##x^{ab} = x^1 = (-1)^1 = -1 ##.

    In the domain of real numbers, we can give sufficient conditions for it to hold true, but it would be harder to describe all possible instances when it does hold true. It is sufficient that ##a## be an odd integer
     
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