# Baby Rudin Proof

1. Aug 28, 2014

### 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.

Now the uniqueness portion is easy, since the reals are an ordered field.

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,

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

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 $t \geq (y-k)$, 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?

2. Aug 28, 2014

### Mogarrr

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

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?

3. Aug 28, 2014

### 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.

4. Aug 28, 2014

### Mogarrr

Yes, this is a good/useful point. What do you think about the former question? Was it necessary to write $t \geq (y-k)$?

5. Aug 29, 2014

### 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.