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Rudin PMA Theorem 1.21 Existence of nth roots of positive reals

  1. Nov 5, 2014 #1
    1. The problem statement, all variables and given/known data
    For every real x>0 and every n>0 there is one and only one positive real y s.t. yn=x

    2. Relevant equations
    0<y1<y2 ⇒ y1n<y2n
    E is the set consisting of all positive real numbers t s.t. tn<x
    t=[x/(x+1)]⇒ 0≤t<1. Therefore tn≤t<x. Thus t∈E and E is non-empty.
    t>1+x ⇒ tn≥t>x, s.t. t∉E, and 1+x is an upper bound of E.
    By the existence of an ordered field ℝ which has the least-upper-bound property we see y=SupE.

    The identity bn-an=(b-a)(bn-1+abn-2+...+an-2b+an-1) yields the inequality bn-an<(b-a)nbn-1 when 0<a<b.


    3. The attempt at a solution
    Assume yn<x. Choose h so that 0<h<1 and
    h<(x-yn)/(n(y+1)n-1).

    Put a=y, b=y+h. Then (y+h)n-yn<hn(y+h)n-1<hn(y+1)n-1<x-yn ⇒ (y+h)n<x and y+h∈E but y+h>y in contradiction to y=SupE.


    *Now herein lies my problem understanding this proof.*

    Assume yn>x.

    *Put k=(yn-x)/(nyn-1) Then 0<k<y. If t≥y-k, we may conclude
    yn-tn≤yn-(y-k)n<knyn-1=yn-x.
    Thus tn>x, and t∉E.
    It follows that y-k is an upper bound of E. However, y-k<y in contradiction to y=SupE.
    Hence yn=x.

    I have tried to go through this proof repeatedly in order to understand why Rudin has chosen an equality and the particular 'k' for this case.
    If anyone could help elucidate this problem, i.e. the seemingly random selection of the k-equality, I would greatly appreciate the light shed upon this dark spot.

    Thank you for your effort.
     
  2. jcsd
  3. Nov 10, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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