# Very difficult algebra problem (real analysis)

1. Aug 28, 2010

### mynameisfunk

Goal: to show yn=x

This particular part of the proof supposes that yn>x. So we want
an h>0 such that (y-h)n>x

yn-(y-h)n<yn-x

yn-(y-h)n=(y-(y-h))(yn-1+yn-2(y-h)+...+(y-h)n-1)<hnyn-1

this yields h=(yn-x)/(nyn-1)

my question: how the heck does one derive h from this????

2. Aug 28, 2010

1. It'd be nice if you could type out the whole problem. Or at least give it in a bit more context.
2. Your logic is not sound. You have:

$$y^n-(y-h)^n \ < \ y^n-x$$
$$y^n -(y-h)^n \ < \ hny^{n-1}.$$

However, this does not necessarily imply that

$$hny^{n-1} \ < \ y^n-x$$.

3. Aug 28, 2010

### mynameisfunk

I do not see where I implied this to be true... Also, i do not see where my logic is unsound

4. Aug 28, 2010

How did you go from

$$y^n -(y-h)^n \ < \ hny^{n-1}.$$

to

$$h = \frac{y^n-x}{ny^{n-1}}?$$

I'm assuming you substituted $$hny^{n-1}$$ on the left in

$$y^n-(y-h)^n \ < \ y^n-x$$

to get

$$hny^{n-1} \ < \ y^n-x$$
$$h \ < \frac{y^n-x}{ny^{n-1}}$$

But you cannot do this, as I explained in my post above.

5. Aug 28, 2010

### mynameisfunk

Haha, so you sympathize with my problem then... I do not see how h was derived. That would be the issue at hand indeed. This proof is on http://www.scribd.com/doc/9654478/Principles-of-Mathematical-Analysis-Third-Edition-Walter-Rudin page 18 of the scribd scroller and page 10 of the actual text. If you are interested in the further details of the proof. However, getting a value for h is the issue for me. In the text, Rudin just gives it without showing how he came up with it.

6. Aug 28, 2010

The proof in the text seemed pretty straightforward to me. We've already proved y^n < x leads to contradiction. Now:

Let y^n > x. We define k as $$k = \frac{y^n - x}{ny^{n-1}}.$$ So 0 < k < y. Now we consider all t such that $$t \geq y - k.$$ For such t, we have the following:

$$y^n - t^n \leq y^n - (y - k)^n$$ by substituting in for t.

Now, since $$b^n - a^n < (b-a)nb^{n-1}$$ for 0 < a < b, and since 0 < y - k < y, we have

$$y^n - (y - k)^n < kny^{n-1}.$$

Which, by our definition of k, is equivalent to

$$y^n - (y-k)^n < y^n - x.$$

Thus, $$(y - k)^n = t^n > x.$$

Therefore, $$t \notin E$$, and is an upper bound of E. Thus, y - k is an upper bound of E, which contradicts our definition of y as the least upper bound of E. Hence, since we proved y^n < x and y^n > x both lead to contradictions, we find that y^n = x.

....I know I just repeated the whole proof....but I added a few clarification notes that I hope help because the rest should be pretty straightforward. By the way, what was your fixation with deriving h? That was for the first part. And it wasn't so much as derived as it was simply chosen because it works. That's often how math proofs like this work. The key is realizing the connection in the end.

Last edited: Aug 29, 2010
7. Aug 29, 2010

### mynameisfunk

The proof was not my problem. The problem WAS deriving h. I did not find the proof to be overly difficult but how does one go about picking an h that works?? My professor went through the process of showing us how he picked it, but this is wherein my problem lies-I did not understand it. If I were to be doing an original of this proof, how would i pick h?

8. Aug 29, 2010

Oh, you just need to know what you want to end up. Well you want to show in the end that y^n > x leads to contradiction. So we would want an h that will give us (y - h)^n > x, thus contradicting the fact that y is the least upper bound. We've already seen that the identity b^n - a^n < (b - a)nb^(n-1) is useful, so we want our inequality in that form.

So we rewrite our previous step as:
y^n - (y - h)^n < y^n - x.

Well we also know from the above identity that:
y^n - (y - h)^n < hny^(n-1).

So we set the two right sides equal:
hny^(n-1) = y^n - x.

Or h = (y^n - x)/ny^(n-1).

There's no point in asking "well how was I supposed to think of that?" It's a nice proof. It requires an insight as to what you want to end up with.

9. Aug 30, 2010

Thanks!