Re: Irrational numbers questions

In article <e0e0d$45a50461$82a1e228$31456@xxxxxxxxxxxxxxxx>,
Han de Bruijn <Han.deBruijn@xxxxxxxxxxxxxx> wrote:
Jesse F. Hughes wrote:

Han de Bruijn <Han.deBruijn@xxxxxxxxxxxxxx> writes:

Quoted from this page: Let's associate with any irreducible fraction
p/q the number w(p/q) = 1/pq - its simplicity (: by Pierre Lamothe).
There is a theorem that the sum of simplicities of all fractions in any
row (with 2^n elements if n = row #) of the Stern-Brocot tree equals 1.

Obvious facts are: the simplicity of 1/2 is 1/2. The simplicity of 5/12
is 1/60. The simplicity of sqrt(2) is .. 0. In general: the simplicity
of any irrational number is 0 .

The obvious fact is that the simplicity of any irrational number is
undefined.

No. But I did some cut and paste in the wrong order. It's defined in the
sequel, where it reads:

Any real number can always be enclosed between two fractions, say m1/n1
and m2/n2 in the Stern-Brocot tree: m1/n2 <= r <= m2/n2. We cannot just
define the simplicity of a real number. What we can do, however, is to
establish an upper bound for it. The lower this upper bound, the less
"simple" the real number is, hence the more "irrational" it looks like.

Any of the three RS-definitions can be used for this purpose. Let some
irrational r be enclosed by two neighbouring fractions in the S-B tree:
m1/n1 < r < m2/n2 . Then finding another (more accurate) nesting for r
results in either L = m1/n1 < r < (m1+m2)/(n1+n2) = U
or L = (m1+m2)/(n1+n2) < r < m2/n2 = U .
The simplicities of the composed fractions (m1+m2)/(n1+n2) are always
_smaller_ than the simplicities of the constituents: m1/n1 and m2/n2 .

It is expected that the simplicity of r is somewhere in between the RS
of L and U. We play a safe game by considering the upper bound of RS(L)
and RS(U) and attaching it to r : RS(r) < maximum( RS(L) , RS(U) ) .

You've just contradicted yourself: if it's less than the maximum, it's
certainly not between RS(L) and RS(U). And you still haven't
_defined_ it for irrational numbers.

In any case, since given 0 < a < b < infty and epsilon > 0 there are
only finitely many rationals in [a,b] with simplicity > epsilon, your
inequality clearly implies RS(r) <= 0 for any irrational r. If you
require RS(r) >= 0, that means RS(r) = 0. And that says that RS is
_not_ any kind of irrationality measure, precisely because it does
not distinguish between different irrational numbers.

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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