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Abs(x) = sqrt(x^2) Proof

  1. Dec 31, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that |x| = sqrt(x^2)

    3. The attempt at a solution
    I've written two proofs but I don't know if they can be justified as real proofs or whether they are valid or not.
    Proof 1:
    [itex] \surd x^{2} = \surd \vert x \vert ^{2} = \vert x \vert [/itex]

    Proof 2:
    First Case ) Suppose [itex] x \geq 0 [/itex] then [itex] \surd x^{2} = x = \vert x \vert [/itex]
    Second Case ) Suppose [itex] x < 0 [/itex] then [itex] \surd x^{2} = -x [/itex] where [itex] -x > 0 [/itex] therefore [itex]-x = \vert x \vert [/itex]
     
  2. jcsd
  3. Dec 31, 2011 #2

    gb7nash

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    Let's look at the definition of the square root:

    If a2 = b and a ≥ 0, then a = √b. Now look at your problem. What two things do we have to prove?
     
  4. Dec 31, 2011 #3
    That [itex] x^{2} \geq 0 [/itex] and [itex] \vert x \vert \geq 0 [/itex] ?
     
  5. Dec 31, 2011 #4

    gb7nash

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    The two bolded things are what you want to prove. Once you have those, then the conclusion follows. Before you do anything, what is a in your problem? What is b? Once you have a and b, what is the first thing you need to prove?
     
  6. Dec 31, 2011 #5
    [itex] a = \vert x \vert [/itex] and [itex] b = x^{2} [/itex]

    [itex] a^{2} = \vert x \vert ^{2} = x ^ {2} = b [/itex]
    [itex] a = \vert x \vert [/itex] which is nonnegative. correct?
     
  7. Dec 31, 2011 #6

    gb7nash

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    Correct.
     
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