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Analysis - problem on completeness of the reals

  1. May 26, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    Show that for all x[itex]\geq 0[/itex] in R, there is a real y with y²=x.


    3. The attempt at a solution

    Let U={[itex]u\in R:u^2\leq x[/itex]}. U is bounded above so let y=sup(U). I'm trying to show that y²=x by showing that the two other alternatives lead to contradictions. First, suppose y²>x. My idea is that I should be able to find a u in U close to y such that x<u²<y². Because y is the supremum of U, for any 0<epsilon<y, there is a u in U with u>y-epsilon <==> u²>(y-epsilon)². Thus, if I find an epsilon<y such that (y-epsilon)²>x, then I will have won. We have (y-epsilon)²>x <==> epsilon² - 2y*epsilon + (y²-x) > 0. If I solve this quadratic inequality for epsilon<y, it gives 0<epsilon<y-[itex]\sqrt{x}[/itex]. So epsilon=(y-[itex]\sqrt{x}[/itex])/2 would do.

    But isn't that solution redudant? Because it involves [itex]\sqrt{x}[/itex], a number that when squared, gives x!
     
  2. jcsd
  3. May 30, 2007 #2

    morphism

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    Well, you can apply the intermediate value theorem to [itex]\sqrt{x}[/itex] to get a quick proof. If you want to stick to first principles, then I don't think you should be using [itex]\sqrt{x}[/itex] at all.

    Let's look at your argument. We want v=y-e with e>0, so v<y, and we want v^2>x. Since we're already assuming that y^2>x, maybe it'd be a good idea to set e=y^2-x. If we try this, we'll find that the y^2 term is annoying. How about we try e=(y^2-x)/y, then? If you try that, you'll see that halving e will be a good idea, i.e. set e=(y^2-x)/(2y).

    v^2 = (y-e)^2 = y^2 - (y^2-x) + [(y^2-x)/(2y)]^2 > x

    i.e. v is an upperbound for U that is less than y, a contradiction.

    (Of course in order to divide by y I had to assume that y>0, i.e. in essence that x>0. But this isn't a problem, because if x=0, then we can deal with this trivial case separately. That is, 0^2=0!)
     
    Last edited: May 30, 2007
  4. May 31, 2007 #3

    NateTG

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    You can't use [itex]\sqrt{x}[/itex], but perhaps you can come up with an alternative upper bound for [itex]\epsilon[/itex] that doesn't involve [itex]\sqrt{x}[/itex]?
     
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