# Unique nonneg. square root for nonneg. real number

1. Feb 26, 2009

### nobahar

Hello,
No exclamation mark after hello I'm afraid, I'm frustrated with the following proof.
Okay, here's the theorem first:
Every nonnegative real number a has a unique nonnegative square root.
Here's the start of the proof, pretty much as they have done it but slightly condensed, all equations are included: a=0 then $$\sqrt(a)$$=0. If a>0 let S be the set of all positive x such that $$x^2 \leq a$$. Since (1+a)2>a the number (1+a) is an upper bound for S. Also, S is non-empty because the number $$\frac{a}{(1+a)}$$ is in S. In fact $$a^2 \leq a(1+a)^2$$ and hence $$\frac{a^2}{(1+a)^2} \leq a$$. S has a least upper bound, b; $$b\geq \frac{a}{(1+a)}$$ so b>0. Therefore, the three possibilities are b2>a, b2<a or b2=a. It then goes on to prove b2=a after some further working out.
Okay, so $$x^2 \leq a$$ indicates the range that the elements x in set S can fall into. The next bit, (1+a)2>a, means (1+a) is the upper bound, a2 can be equal to a and in that case the element x=a=1, so this is can't be an upper bound. So (1+a) forms the upper bound, but surely (a+tiny amount)2 would be an upper bound as I move closer to a?
Next, S is not a void set because $$\frac{a}{(1+a)}$$ is in S. I figured since it has to be <a but >0, but it doesn't imediately seem an obvious 'move', if that makes sense. The next bit throws me, a2<=a(1+a)2, and since I don't know what's going on here I can't proceed with the proof. It also may come across from what I have written, but I'm not confident with the first 'bits'. Anyone want to attempt to explain this for me? Not all the proof, preferably upto what I have included here if you don't mind.
Thanks.