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.(adsbygoogle = window.adsbygoogle || []).push({});

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

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].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.

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

Which isn't an identity for n=1. Did I just mess up this 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.

Back to the proof...

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... 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.

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 theLeast Upper Boundy, which is a contradiction.

Any thoughts?

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# Baby Rudin Proof

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