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A proof for sqrt(x^2)=|x|

  1. May 29, 2009 #1

    ShayanJ

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    Hi
    Can someone give me a proof for sqrt(x^2)=|x|?
    I mean why sqrt(x^2)!=+x,-x?
    thanks
     
  2. jcsd
  3. May 29, 2009 #2

    HallsofIvy

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    [itex]\sqrt{x}[/itex], like any function, is "single valued". That's part of the definition of function: for any specific value of x, f(x) must be a single value.

    It is true that, for example, [itex]x^2= 4[/itex] has two roots: x= -2 and x= 2. But [itex]\sqrt{4}[/itex] is specifically defined as "the non-negative number whose square is 4". More generally [itex]\sqrt{a}[/itex] is defined as "the non-negative number whose square is a".

    In fact, the reason why we have to write the solutions to [itex]x^2= a[/itex] as "[itex]\pm\sqrt{a}[/itex]" is because [itex]\sqrt{a}[/itex] alone does NOT include "[itex]\pm[/itex]".
     
  4. May 29, 2009 #3

    jambaugh

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    Remember that the square function
    [tex]f(x)=x^2[/tex]
    is not invertible over its whole domain. (Its graph doesn't pass the horizontal line test.) This because there is not a unique solution for x to the equation:
    [tex] y = x^2[/tex]
    To be invertible the inverse relation must be a function i.e. yield a single unique value.

    In order to "make" our square function invertible we must restrict the domain:

    [tex]f_+(x) = x^2\quad (x\ge 0)[/tex]
    (i.e. if x is negative the function is undefined)

    In this case we can then define an inverse function
    [tex] f_+^{-1}(y) = \sqrt{y}[/tex]

    Note we could have chosen to restrict the domain differently:

    [tex] f_-(x) = x^2 \quad (x \le 0)[/tex]

    This then would have inverse:
    [tex] f_-^{-1}(y)= - \sqrt{y}[/tex]

    The square root function gives the "principle root". Since there are two solutions to the equation x^2 = 4, namely 2 and -2 then in order that sqrt be a function (have a unique value) we must pick one of the two solutions. We adopt the convention of choosing the positive solution and call this the "principle" solution. We could have "gone the other way" but more often the solution we want is the positive one.

    As for your proof you should break the problem down into cases, x = 0, x>0, x < 0. Show that the value output is the same as |x| in each case.
     
  5. May 29, 2009 #4
    How about

    [tex]|z|=|a+bi|=\sqrt{a^2+b^2}\implies |a|=\sqrt{a^2}[/tex]
     
  6. May 30, 2009 #5
    It's more of a definition than anything you can prove.
     
  7. May 30, 2009 #6

    ShayanJ

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    Hey qntty that is right only if the set of real numbers were a subset of the set of complex numbers,but its not.
    And to you tibarn.What if I tell you that the sentence "There are infinite number of prime numbers.",has a proof?
     
  8. May 30, 2009 #7

    matt grime

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    Shyan, the set of real numbers *is* a subset of the complex numbers.

    I've no idea what you mean by your last sentence, Shyan. It is the definition of the square root function (or perhaps convention if you prefer) that the square root of a positive number r is the positive one of the two possible solutions to x^2=r. We could form a perfectly good version of mathematics where it is the negative one. It would be perverse and silly but perfectly possible. In that sense you can't 'prove' it in any non-trivial way
     
  9. May 30, 2009 #8

    ShayanJ

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    yeah.I was making a big mistake and I understood it when I was writing a tough answer(I thought it is) for you matt grime.So I think qntty's answer is the ultimate one(but thanks to all posters)and the thread is finished.
    Thank every one for the answers
     
  10. May 30, 2009 #9
    Shyan, let me ask you a question. How do you prove that |x| = x and not |x| = +x,-x ?

    Because it is DEFINED that way. Same thing goes for the square root. "qntty" has not proved anything.
     
  11. May 30, 2009 #10

    ShayanJ

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    first:|x|=x if x>=0,-x if x<0
    second:Yes he did.
    sqrt(x^2) is defined to be |x|.but I didn't know why it is.qntty told the reason.So he PROVED it.
    In fact he passed the way that the mathematicians has passed to reach to the definition of sqrt(x^2) in front of our eyes.And I think that is one kind of PROVING.
     
    Last edited: May 30, 2009
  12. May 30, 2009 #11
    "first:|x|=x if x>=0,-x if x<0"

    NO NO NO. You are using the definition and that is no good according to your own standards. PROVE it to me!
     
  13. May 30, 2009 #12

    matt grime

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    qntty didn't prove anything - he just used the fact that sqrt(x) returns the positive root. Which is what you claim you want proving....
     
  14. May 30, 2009 #13

    ShayanJ

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    Yes I used the definition but what was the thing that you used?
    if it was a definition too:
    then there are just two candidates for the definition of |x|.yours and mine.So if I prove that yours is wrong,then mine will be true.And that's exactly what I'm going to do.
    Imagine x=-2.as your definition says,|-2|=-2,but we all know that a modulus won't be equal to a negative number.So your definition is wrong and mine is true.
     
  15. May 30, 2009 #14
    Now prove that modulus is always positive.
     
  16. May 30, 2009 #15

    ShayanJ

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    Ok.|a| means the distance between the 0 point and +a or -a.a distance can not be negative.So modulus is always positive or zero.
    Is it finished?
     
  17. May 30, 2009 #16
    This question is a matter of definition.

    If I ask "What is sqrt(25)?", the answer is +5.

    If I ask "Which real number x has the property that x^2 = 25?", the answer is there are two: +5 and -5.

    sqrt(25) is defined to be a function, hence, single-valued.The positive root is taken, as far as I can tell, because writing it takes less time (the + sign can be assumed on positive numbers...)
     
  18. May 31, 2009 #17

    matt grime

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    But you've defined the modulus function to be the distance. You didn't need to do that; I can define |x| to be negative and then claim that distance is -|x|. As I said before it is entirely perverse to do so, but would also have been legitimate.

    You are not seeking a proof that is anything other than the vacuous one. You might perhaps want a justification for it. The functions |x| and sqrt(x^2) are identical on the real numbers, and that is an immediate consequence of the definition of each. It is not a deep fact, and it requires nothing that really merits the word 'proof'.
     
  19. May 31, 2009 #18

    ShayanJ

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    Well I think in the chain of proofs,we reach somewhere that we have to accept sth without proof and also sth must be set as a law.Like for example:axioms of euclid.
    And that's because we can fight for ever and get no result.so I think you are right at the end.and sqrt(x^2)=|x| is sth like an axiom or maybe a law.
    Ok
    thanks to all.
     
  20. May 31, 2009 #19
    sqrt(x^2) equals x if x>0 and -x otherwise.
    |x| equals x if x>0 and -x otherwise.
    Therefore, sqrt(x^2) = |x|.
    Pedantic, yes, but this is a definition, not a deep fact.
     
  21. May 31, 2009 #20
    You just have to understand the meaning of the word "definition".

    An example: By definition, the word "knife" refers to a metallic or plastic object used for cutting stuff. You can use it to stab someone or you can use it when you eat. That is how "knife" is defined. IT IS COMPLETELY ARBITRARY. You might as well define "knife" to mean the same as "monkey" or whatever. Go ahead, invent your own language.

    The same thing goes for the symbol "sqrt". Is has a certain definition, a certain meaning.
     
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