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Existence of the Square Root Proof

  1. Jun 21, 2012 #1
    I was playing trying to work through a proof in Apostol's Calculus and can't quite understand a step noted. This is from chapter 3, theorem 1.35. Every nonnegative real number has a unique nonnegative square root. The part where you are establishing the set S as nonempty so you can use LUB it is stated that a/(1+a) is in the set S. I've seen different choices for this on other versions of this proof. When I first looked at this I figured it was in S for the reason that that would produce a square of a fraction which would produce something smaller than a. But it looks like this is then used with the binomial theorem to finish off the proof. I don't follow it. Can someone walk me through the logic in this one?
  2. jcsd
  3. Jun 21, 2012 #2
    Indeed, [itex]\frac{a}{1+a}[/itex] is in S because it is positive and because

    [tex]\frac{a^2}{(1+a)^2}\leq a[/tex]

    To see this, note that this is equivalent to

    [tex]a\leq (1+a)^2[/tex]


    [tex]a\leq 1+a^2+2a[/tex]

    And this is certainly true.
  4. Jun 21, 2012 #3
    This book never ceases to make me feel stupid. Thanks Micro for making that clearer.
  5. Jun 21, 2012 #4


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    do you believe the intermediate value theorem? If so you only need to prove that squaring is continuous. since (a+h)^2 = a^2 + 2h + h^2, it is clear that making h small will make a^2 close to (a+h)^2. qed.
  6. Jun 22, 2012 #5
    Now i'm lost again. We are trying to prove that the LUB^2 (LUB=b) cannot be any other value but a. From this point on I don't follow the proof at all. For instance to test if LUB^2>a he sets a number c=b-(b^2-a)/(2b). Where did that come from?
  7. Jun 22, 2012 #6
    I know this is binomial trickery but I just don't see it. Any ideas?
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