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Every nonnegative real has an nth root proof

  1. Feb 8, 2015 #1
    I need help understanding this proof. The textbooks says,

    For every k∈{1,...,n} there exists an mk∈ℕ such that (n choose k) rn-k 1/mkk< δ/n
    His justification for this, is that for every real number x there exists an natural number n such that n≥x.
    I fail to see the connection between the two. Please help :(
     
  2. jcsd
  3. Feb 8, 2015 #2
    Some background information:

    Given a nonnegative real number a, I need to show that there exists a nonnegative real number r such that rn=a. The proof involves showing that rn is not less than a, and not greater than a.
    If rn<a, then rn<rn+δ<a for some δ.

    The author then writes the statement in the previous post. I'm lost, does anybody know how this proof works. I've looked up other proofs on the internet but they take a slightly different approach.
     
  4. Feb 9, 2015 #3

    mathwonk

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    have you tried using the intermediate value theorem? or are you trying to bash it out just from the axiom of least upper bounds, without even using continuity? your question does not seem to make sense because of not defining the number r. i.e. you have still left out too much information from the book.

    preferably, one goes through the ideas of continuity and proves the IVT, and then because the function f(x) = x^n is continuous, and since f(0) = 0 ≤ a, and either f(1) or f(a) is > a, according as whether a≤ 1 or a ≥ 1, we can invoke the IVT to get an r≥0 such that f(r) = a.

    but it sounds as if you are in the early stages of the somewhat tedious proof that x^n really is continuous. this proceeds by staring at (x+e)^n = x^n + e.g(x,e) and finding an upper bound for |g(x,e)| for all numbers ≤ e, and using that to show that e.g(x,e) -->0 as e -->0.
     
    Last edited: Feb 9, 2015
  5. Feb 9, 2015 #4
    The book uses the axiom of least upper bounds, not the intermediate value theorem.

    Let R = {x∈ℝ: x≥0 and xn≤a}. Then, 0∈R and R is bounded above by max {1, a} (verifying this is tedious, but it's true).
    Since R is bounded above, it has a supremum. Let r=Sup(R). To show rn=a, we will show that rn is not < a and rn is not > a. Suppose for a contradiction that rn < a, then there exists a δ>0 such that rn+δ<a.

    From here he writes the above statement I'm struggling with.
     
  6. Feb 9, 2015 #5

    mathwonk

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    well i hate this kind of proof, and don't think it does you any good really to bash through it but i'll try to say something. since r is larger than all number s with s^n < a, the monotonicity of nth powers says that if s^n < a then also s^n < r^n < a. i.e. r^n is squeezed in between a and all smaller numbers of form s^n. so to show that r^n is not less than a, we just have to show there is no room in there for it to fit. i.e. we have to show that s^n gets arbitrarily close to a. i.e. for all d > 0, we must show there is an s with a-d < s^n < a. that's the hard part. what can we use? do we know rational numbers are dense? he is showing you how to do something equivalent.

    i.e. you wrote:

    For every k∈{1,...,n} there exists an mk∈ℕ such that (n choose k) r^(n-k) 1/mk^k< δ/n
    His justification for this, is that for every real number x there exists an natural number n such that n≥x.


    since you are choosing mk, you just need it to be larger than, lets see, switch stuff to the other side,...so you need
    (n/delta)(n choose k) r^(n-k)< mk^k. so just take mk larger than the left side, and then so will mk^k be larger.


    so that's the connection between your two sentences i quoted.


    next presumably this will show that if you take the largest of all these mk's and add 1 over that to r, you will get, by the binomial theorem,

    that (r + 1/mk)^n < r^n + delta < a. this will contradict r being bigger than all numbers with nth power less than a, since r + 1/mk will be bigger than r,

    and yet still have nth power less than a.


    i hate these, i'm cleverer than you, proofs. they contain no ideas and teach nothing. but there it is.

    in my opinion it would be much better to define continuity and then do this same argument to prove nth power is continuous, and then this same work would give you some conceptual result. you aren't by any chance reading my "favorite" analysis book "baby rudin" are you? if so you are in for a whole book full of such un-insightful stuff.
     
    Last edited: Feb 9, 2015
  7. Feb 9, 2015 #6

    mathwonk

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    instead of reading that turgid stuff, why not try it yourself for n= 2? try to show that if m is large enough, then [r+(1/m)]^2 is as close as you like to r^2. i.e. just estimate the error. in particular then if r^2 < a, then so also will (r + (1/m))^2 be less than a. that's all he's doing.
     
  8. Feb 10, 2015 #7

    Svein

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    This is what is known as the Axiom of Archimedes. A consequence of this is: Between any two real numbers, there is a rational number.
    You have half of the proof, xn≤a. But since a is the least upper bound, [itex] a-\frac{1}{p}[/itex] is not an upper bound for R. Thus [itex] x^{n}\geq a-\frac{1}{p}[/itex] for any p∈ℕ....
     
  9. Feb 10, 2015 #8
    This was the part I was struggling with. I understand the reasoning now. Thank you!

    I'm reading Mathematical Analysis: A Concise Introduction by Schroder. This proof is exceptionally ugly, and I assume there will be some more ugly ones along the way, but overall it's a decent book.
     
  10. Feb 11, 2015 #9

    mathwonk

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    it's the "concise" part that seems to be the problem. that is not usually a virtue in introductions, only in review treatments.
     
  11. Feb 11, 2015 #10

    Svein

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    OK. Let w= (n choose k). Since [itex]\delta >0 [/itex], [itex]\frac{n\cdot w\cdot r^{n-k}}{\delta} [/itex] is a real number. Thus, there exists an integer mk such that [itex] m_{k}> \frac{n\cdot w\cdot r^{n-k}}{\delta}[/itex]. Rearrange....
     
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