Rudin 1.21 Problem understanding proof of unique positive root to the

  • Thread starter EdMel
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  • #1
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Main Question or Discussion Point

Firstly, the set E is defined:
"Let E be the set of all positive real numbers t such that tn<x."

Later on the proof goes:
"Assume yn>x. Put k=yn-x / nyn-1. Then 0 < k <y. If ty - k, we conclude that
yn-tn ≤ yn-(y-k)n < knyn-1 = yn-x.
Thus tn>x, and t is not a member of E. It follows that y - k is an upper bound of E."

Why does it follow? Is it because the possibility that t = y - k combined with the fact that tn>x mean that y-k always has to be an upper bound of E? Or is there some other reasoning?

Thanks in advance.
 

Answers and Replies

  • #2
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Hi All,

I feel I understand this now.

There were two things I did not have right in my head.

1) To prove the assumption yn > x is false all you need to do is find one instance when it is false. Having t ≥ y - k provides one instance. I was not recognizing this.

I was getting confused about what would happen if you set t = y - k, or t > y - k, or t < y - k; it seems that using the either of the first two would suffice in finding one instance to disprove the assumption, but you can treat them together as is done in the proof. Using t < y - k doesn't get you anywhere.

2) It is obvious that y - k is an upper bound of E if you picture E on the number line and y-k somewhere to the right of it, and the set of t ≥ y - k pointing from here out to the right.

Peace out.
 
  • #3
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Hi All,

I feel I understand this now.

There were two things I did not have right in my head.

1) To prove the assumption yn > x is false all you need to do is find one instance when it is false. Having t ≥ y - k provides one instance. I was not recognizing this.

I was getting confused about what would happen if you set t = y - k, or t > y - k, or t < y - k; it seems that using the either of the first two would suffice in finding one instance to disprove the assumption, but you can treat them together as is done in the proof. Using t < y - k doesn't get you anywhere.

2) It is obvious that y - k is an upper bound of E if you picture E on the number line and y-k somewhere to the right of it, and the set of t ≥ y - k pointing from here out to the right.

Peace out.
My question is why didn't Rudin simply define t as [tex]t=y-k[/tex] instead of [tex]t\ge y-k[/tex]?
 
  • #4
GMB
1
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I know this thread is quite old, but nevertheless I've been asking myself the same question as bhagwad for some days now. I don't even see the necessity of defining t. When showing the contradiction of assuming yn<x, Rudin does it directly after defining h. On the contrary, when looking for the contradiction of assuming yn>x, he does introduce t. He could have simply said (since 0<k<y, and hence 0<y-k<y) that yn-(y-k)n<knyn-1=yn-x. We deduce from this that (y-k)n>x and thus y-k is an upper bound of E.
Now, if this were an average textbook I would have simply thought "great! I thought of a simpler proof!". But this is Rudin, who frankly deserves reverence for his rigorous yet streamlined proofs (as is the case with the rest of this proof: most introductory analysis books prove in the same or more space the existence of square roots only, or even just the existence of √2, and don't even dare to tackle n-th roots!), so I've been thinking that maybe I'm missing something here. Any ideas?
 

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