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Is there a way of proving that all positive numbers have a real square root?

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data

    My prof showed us the proof that sqrt2 is not a rational number. She said, however, that we haven't proved that it is irrational, because we haven't proved that sqrt2 exists. How would we go about proving this?

    2. Relevant equations


    3. The attempt at a solution

    This isn't really a homework question, it's just curiosity. I think she said something about showing that sqrt2 is a solution to x^2 - 2 = 0, and that when you graph this function, it does indeed intersect the x-axis sqrt2. But that seemed a bit fishy (or incomplete, or something...) to me.
  2. jcsd
  3. Sep 27, 2009 #2
    Nothing fishy about the proof that Regina (I see that you`re in MAT157 too) gave. It just wasn't completely rigorous because we haven’t proved the intermediate value theorem yet, which is what she used. When you think of any polynomial function, it’s continuous, right? Even without defining continuity, you have some intuitive idea what a continuous function should look like. Drawing the graph of a polynomial and seeing that the graph of it “never breaks”, then intuitively you can see that it’s continuous. Regina showed that for some x, the function f(x) = x^2 – 2 was greater than 0, and for some other x it was less than 0. Remembering that f is continuous, this must mean that f has to pass through the x-axis, meaning there is a solution to the equation x^2 -2 = 0, which is what we were trying to prove.
    You can do this for any real number c. Just write f(x) = x^2 – c. It’s easy to see that it’s possible to take x large enough so that x^2 > c. This implies x^2 – c > 0. And if you take x = 0, f(0) = 0^2 – c = -c < 0. So f must have a root and thus there exists a solution to our equation.
  4. Sep 27, 2009 #3
    Thanks very much! I forgot about IVT. What are you doing up so late/early? Working on your problem set?
  5. Sep 27, 2009 #4
    Nah, I just woke up. Went to bed early last night.
  6. Sep 27, 2009 #5
    By the way, Regina never said that the square root of 2 is not rational, but we cannot say it is irrational because we haven’t proved it exists. She essentially said that under the assumption that it exists, it is irrational.
    Last edited: Sep 27, 2009
  7. Sep 27, 2009 #6


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    Homework Helper

    The existance of square roots is a basic property of the real numbers. One defines the real numbers by giving them some properties and then showing a unique system has those properties.
    common examples
    3)nested intervals
    4)least upper bound

    if we know the reals have a least upper bound property for example sup S exist for any subset so we set sqrt(a)=sup {x rational| x^2-a<0}

    The idea is the rationals have holes, and it would be good to hav a system that fills in all those holes. The reals are such a system.
    Last edited: Sep 27, 2009
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