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Range of sqrt

  1. Mar 13, 2009 #1


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    Apologies for the trivialness of the question, but I'm not so great at this. I was wondering why the square root of a real number is positive. Why is sqrt(9) = 3, and not -3 as well, since (-3)² would give 9. Is it just a condition you set, that the function values must be positive? At least I thought it was, googling for it produces sites that tell the opposite, such as this one: http://thesaurus.maths.org/mmkb/entry.html?action=entryByConcept&id=1015 [Broken]. Are they wrong?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 13, 2009 #2
    The principal square root function of x is defined to give the positive value that when squared gives x. It is only a convention.
    Last edited: Mar 13, 2009
  4. Mar 13, 2009 #3
    every number has two square roots, however to differentiate between them, [tex]\sqrt{x}[/tex] is defined to be positive. The notation for both roots is [tex]\pm \sqrt{X}[/tex]
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