Baby Rudin Proof

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Main Question or Discussion Point

I think I've found a proof where Rudin is actually too wordy! For your welcomed inspection, I will type a part of said proof, then comment.

Theorem: For every x>0 and every integer n>0, there is a unique real y such that [itex] y^{n} = x [/itex]
Now the uniqueness portion is easy, since the reals are an ordered field.

Proof: Let E be the set of all positive real numbers t such that [itex] t^{n} < x [/itex]. If t=x/(1+x), then [itex]t^{n} \leq t < x [/itex] and t is not empty.
If t=1+x, then [itex]t^{n} \geq t > x [/itex], so 1+x is an upper bound for E.
This implies the existence of a y=supE.
At this point the strategy is to show that the cases where [itex]y^{n} <x [/itex], and [itex] y^{n} > x[/itex], are contradictory. To save time, I want to get to the point where I think Rudin is too wordy, that is the case where [itex] y^{n} > x [/itex].

Oh here's an identity he uses in the proof,

The identity [itex]b^{n} - a^{n} = (b-a)(b^{n-1} + b^{n-2}a + ... + a^{n-1}) [/itex] yields the inequality [itex] b^{n} - a^{n} < (b-a)nb^{n-1} [/itex], when 0<a<b.
Which isn't an identity for n=1. Did I just mess up this proof?


Back to the proof...

Assume [itex] y^{n} > x [/itex]. Put [itex] k = \frac {y^{n} - x}{ny^{n-1}} [/itex]. Then 0<k<y. If [itex]t \geq (y-k)[/itex], we conclude that [itex]y^{n} - t^{n} \leq y^{n} - (y-k)^{n} < kny^{n-1} = y^{n} - x [/itex]. Thus [itex]t^n > x [/itex] so [itex] t \notin E [/itex]. It follows that y-k is an upperbound of E. But (y-k)<y, which contradicts the fact that y is the least upper bound of E.
Besides the identity that I'm hung up on (which I didn't catch until I started to type this proof). I don't think it's necessary to point out that [itex] t \geq (y-k) [/itex], since, assuming I am mistaken about the said identity, I can write with less words that...

since 0<y-k<y, we have, [itex]y^{n} - (y-k)^{n} < kny^{n-1} = y^{n} - x [/itex]. This implies that (y-k)>x, and so (y-k) is an upper bound that is less than the Least Upper Bound y, which is a contradiction.

Any thoughts?
 

Answers and Replies

  • #2
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Which isn't an identity for n=1. Did I just mess up this proof?
Thought about this, and I realized that this doesn't matter. If n=1, then

[itex] y^{1}=y = x [/itex].

I'm glad I caught this. Still, any thoughts on the wordiness? Is it necessary to state that [itex] t \geq (y-k) [/itex] in the case where [itex] y^{n} > x [/itex] is considered?

Considering another train of thought, the whole point of this theorem is to show that there are many irrationals in R, right?
 
  • #3
WWGD
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The point is to show every positive real x has an unique n-th root, for n=1,2,3,..... Notice the third line in your 1st post.
 
  • #4
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The point is to show every positive real x has an unique n-th root, for n=1,2,3,..... Notice the third line in your 1st post.
Yes, this is a good/useful point. What do you think about the former question? Was it necessary to write [itex] t \geq (y-k) [/itex]?
 
  • #5
WWGD
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I see, so (y-k)^n >x , so y-x is not in E, so y-k is an LUB <y . Yes, that seems like a correct point to me.
 
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