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No exclamation mark after hello I'm afraid, I'm frustrated with the following proof.

Okay, here's the theorem first:

Every nonnegative real numberahas 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 [tex]\sqrt(a)[/tex]=0. If a>0 let S be the set of all positive x such that [tex]x^2 \leq a[/tex]. Since (1+a)^{2}>a the number (1+a) is an upper bound for S. Also, S is non-empty because the number [tex]\frac{a}{(1+a)}[/tex] is in S. In fact [tex]a^2 \leq a(1+a)^2[/tex] and hence [tex]\frac{a^2}{(1+a)^2} \leq a[/tex]. S has a least upper bound, b; [tex]b\geq \frac{a}{(1+a)}[/tex] so b>0. Therefore, the three possibilities are b^{2}>a, b^{2}<a or b^{2}=a. It then goes on to prove b^{2}=a after some further working out.

Okay, so [tex]x^2 \leq a[/tex] 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, a^{2}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 [tex]\frac{a}{(1+a)}[/tex] 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, a^{2}<=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.

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# Unique nonneg. square root for nonneg. real number

Can you offer guidance or do you also need help?

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