Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Unique nth Roots in the Reals - Rudin 1.21

  1. Oct 25, 2011 #1
    Unique nth Roots in the Reals -- Rudin 1.21

    In Principles of Mathematical Analysis 1.21, Rudin sets out to show that for every positive real x there exists a unique positive nth root y. The proof is rather long and I would like to zoom into the portion of it where it seems that Rudin takes too many steps to show a sub-conclusion. This is where Rudin chooses an element h ∈ R such that the following two conditions hold:

    (1) 0 < h < 1
    (2) h < (x-yn) / n(y+1)n-1

    He does this to say that since

    (y+h)n - yn < hn(y+h)n-1 < hn(y+1)n-1 < x - yn

    follows, we can then observe (y+h)n < x which comes from the first and last terms in the longer inequality above.

    So my question/objection is as follows: why not just merely choose "h" to satisfy

    (1a) 0 < h
    (2a) h < (x - yn) / n(y+h)n-1

    ...? This way we could get to our conclusion much faster since

    (y+h)n - yn < hn(y+h)n-1 < x - yn

    implies with fewer steps that

    (y+h)n < x.
  2. jcsd
  3. Oct 25, 2011 #2
    Re: Unique nth Roots in the Reals -- Rudin 1.21

    There is an h in the right-hand side and an h in the left-hand side. So showing the very existence of such an h will be less trivial than how Rudin does it.
  4. Oct 27, 2011 #3
    Re: Unique nth Roots in the Reals -- Rudin 1.21

    Well, here is my attempt at a proof that there exists an h with the two conditions Rudin lays out:

    (1) 0 < h < 1
    (2) 0 < h < [x – yn] / [n(y+1)n-1]

    Let ⊥ = [x – yn] / [n(y+1)n-1]

    First observe that ⊥ > 0 since it is really just the two positive quantities of (x-yn) and 1/[n(y+1)n-1] being multiplied together (which is positive by axiom).

    Next note that by trichotomy ⊥ < 1 or ⊥ = 1 or ⊥ > 1.

    If ⊥ ≤ 1, then let H = {h ∈ Q: 0 < h < ⊥}. We know that H is non-empty since we can say from Rudin 1.20(b) that there always exists a rational number in between two reals. In this case, there will be an h between 0 and ⊥ (which we know is real by the closure axiom).

    Similarly, if ⊥ > 1, then let H = {h ∈ Q: 0 < h < 1 < ⊥}. Likewise we know that H is non-empty since we know there exists a rational between the two reals of 0 and 1.

    Hence we have shown that ∃h(h ∈ Q ∧ (0 < h < 1) ∧ h < ⊥)

    Relating this proof to my original question, we could show using a similar (but again much easier!) proof that there exists an h which satisfies the following criteria:

    (1a) 0 < h
    (2a) h < (x - yn) / n(y+h)n-1

    …which would, with fewer steps, lead to the conclusion that (y+h)n < x, which is what Rudin is going after to begin with. Is there anything I’m missing? Can anyone think of why Rudin would introduce the extra criteria for h which leads to extra steps to show his sub-conclusion?
  5. Oct 31, 2011 #4
    Re: Unique nth Roots in the Reals -- Rudin 1.21

    Kind of off topic but another way to show the exsistence of nth roots is consider the function f(x) = x^n. It is increasing on [0, 00), and since the function is one to one, you can take its inverse and then with the intermediate value theorem to get the result.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook