Unique nth Roots in the Reals - Rudin 1.21

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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.
 

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  • #2
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why not just merely choose "h" to satisfy

(1a) 0 < h
(2a) h < (x - yn) / n(y+h)n-1
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.
 
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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.
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?
 
  • #4
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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.
 

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