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Homework Help: Proof square root bijective;

  1. Jul 20, 2012 #1
    In what branches of mathematics is this proven.. I have never seen a proof, so I wonder if anyone can give me the basics of what is done to proove it or got a link to a proof..

    Edit: By square root I mean the positive square root.
  2. jcsd
  3. Jul 20, 2012 #2


    User Avatar

    to prove square root function to be bijective you need to specify the domain and codomain of the function.
    if codomain is set of real numbers, then its not bijective.
  4. Jul 20, 2012 #3


    Staff: Mentor

    Is this a homework assignment? You should know by know that we won't do your assignments for you, but we'll guide you while you do the work.
  5. Jul 22, 2012 #4
    This is definately not a homework assignment. I'm on holidays and this speculation came to me. I study physics and thus my courses do not focus on how to get to the mathematical results rigorously but rather their applications. Therefore I have to do this work on my own and I have no idea whatsoever what tools will be needed for the above proof. The domain of the positive square root are all positive integers including zero. But to go from here to show that the square root takes one number into another number uniquely I don't know how to show. Thus I ask you.
  6. Jul 22, 2012 #5


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    Science Advisor

    Then why post it in "Homework and Classwork"? In any case "the square root takes one number into another number uniquely" is NOT "bijective". That is "injective". You can't say "bijective" without, as pcm said, specifying the domain and codomain.

    If you intend the domain and codomain as "the non-negative real numbers" then, yes, the square root function is bijective. To show that you show it is "injective" ("one to one"): if [itex]\sqrt{x}= \sqrt{y}[/itex] then x= y. That's easy to show. And "surjective" ("onto") is equally easy- given any non-negative number, y, show that there exist a non-negative number x such that [itex]\sqrt{x}= y[/itex].
  7. Jul 22, 2012 #6


    Staff: Mentor

    HallsOfIvy alluded to this, but to make it more explicit - the domain of this function is all nonegative real numbers, not just the nonnegative integers.
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