Existence of the Square Root Proof

Ronnin

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?

micromass

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]

or

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

And this is certainly true.

Ronnin

This book never ceases to make me feel stupid. Thanks Micro for making that clearer.

mathwonk

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.

Ronnin

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?

Ronnin

I know this is binomial trickery but I just don't see it. Any ideas?