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Unique nonneg. square root for nonneg. real number

  1. Feb 26, 2009 #1
    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 [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 b2>a, b2<a or b2=a. It then goes on to prove b2=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, 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 [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, 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.
     
  2. jcsd
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