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gwsinger
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Unique nth Roots in the Reals -- Rudin 1.21
In Principles of Mathematical Analysis 1.21, Rudin sets out to show that for every positive real x there exists a unique positive nth root y. The proof is rather long and I would like to zoom into the portion of it where it seems that Rudin takes too many steps to show a sub-conclusion. This is where Rudin chooses an element h ∈ R such that the following two conditions hold:
(1) 0 < h < 1
(2) h < (x-yn) / n(y+1)n-1
He does this to say that since
(y+h)n - yn < hn(y+h)n-1 < hn(y+1)n-1 < x - yn
follows, we can then observe (y+h)n < x which comes from the first and last terms in the longer inequality above.
So my question/objection is as follows: why not just merely choose "h" to satisfy
(1a) 0 < h
(2a) h < (x - yn) / n(y+h)n-1
...? This way we could get to our conclusion much faster since
(y+h)n - yn < hn(y+h)n-1 < x - yn
implies with fewer steps that
(y+h)n < x.
In Principles of Mathematical Analysis 1.21, Rudin sets out to show that for every positive real x there exists a unique positive nth root y. The proof is rather long and I would like to zoom into the portion of it where it seems that Rudin takes too many steps to show a sub-conclusion. This is where Rudin chooses an element h ∈ R such that the following two conditions hold:
(1) 0 < h < 1
(2) h < (x-yn) / n(y+1)n-1
He does this to say that since
(y+h)n - yn < hn(y+h)n-1 < hn(y+1)n-1 < x - yn
follows, we can then observe (y+h)n < x which comes from the first and last terms in the longer inequality above.
So my question/objection is as follows: why not just merely choose "h" to satisfy
(1a) 0 < h
(2a) h < (x - yn) / n(y+h)n-1
...? This way we could get to our conclusion much faster since
(y+h)n - yn < hn(y+h)n-1 < x - yn
implies with fewer steps that
(y+h)n < x.