Proof of Baby Rudin: Unique Real y for x & n

[Mogarrr]

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 $y^{n} = x$

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 $t^{n} < x$. If t=x/(1+x), then $t^{n} \leq t < x$ and t is not empty.
If t=1+x, then $t^{n} \geq t > x$, 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 $y^{n} <x$, and $y^{n} > x$, are contradictory. To save time, I want to get to the point where I think Rudin is too wordy, that is the case where $y^{n} > x$.

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

The identity $b^{n} - a^{n} = (b-a)(b^{n-1} + b^{n-2}a + ... + a^{n-1})$ yields the inequality $b^{n} - a^{n} < (b-a)nb^{n-1}$, when 0<a<b.

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


Back to the proof...

Assume $y^{n} > x$. Put $k = \frac {y^{n} - x}{ny^{n-1}}$. Then 0<k<y. If $t \geq (y-k)$, we conclude that $y^{n} - t^{n} \leq y^{n} - (y-k)^{n} < kny^{n-1} = y^{n} - x$. Thus $t^n > x$ so $t \notin E$. 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 $t \geq (y-k)$, since, assuming I am mistaken about the said identity, I can write with less words that...

since 0<y-k<y, we have, $y^{n} - (y-k)^{n} < kny^{n-1} = y^{n} - x$. 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?

 

[Mogarrr]

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

$y^{1}=y = x$.

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

Considering another train of thought, the whole point of this theorem is to show that there are many irrationals in R, right?

 

[WWGD]

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.

 

[Mogarrr]

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 $t \geq (y-k)$?

 

[WWGD]

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.