# Existence of the Square Root Proof

1. Jun 21, 2012

### 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?

2. Jun 21, 2012

### micromass

Staff Emeritus
Indeed, $\frac{a}{1+a}$ is in S because it is positive and because

$$\frac{a^2}{(1+a)^2}\leq a$$

To see this, note that this is equivalent to

$$a\leq (1+a)^2$$

or

$$a\leq 1+a^2+2a$$

And this is certainly true.

3. Jun 21, 2012

### Ronnin

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

4. Jun 21, 2012

### 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.

5. Jun 22, 2012

### 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?

6. Jun 22, 2012

### Ronnin

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