Rationals and Irrationals
Maury Barbato
consider the well-known function f:(0,1)->R defined by
f(x) = 1/n if x = m/n, with m and n relatively prime,
f(x) = 0 if x is irrational.
Does there exist the right (left) limit of the
difference quotient for x irrational?
I think the answer is no, but a proof is not so
trivial, beacuse it implies to study the set
S={ m - n*x | m<n are positive integers relatively prime}
Some idea?
Thank you veru much for your attention.
My Best Regards,
Maury Barbato
Arturo Magidin
> Hello,
> consider the well-known function f:(0,1)->R defined by
>
> f(x) = 1/n if x = m/n, with m and n relatively prime,
>
> f(x) = 0 if x is irrational.
>
> Does there exist the right (left) limit of the
> difference quotient for x irrational?
> I think the answer is no, but a proof is not so
> trivial, beacuse it implies to study the set
>
> S={ m - n*x | m<n are positive integers relatively prime}
>
> Some idea?
It looks vaguely like the kind of considerations made in Hurwitz's
theorem and variants thereof. For example,
Hurwitz's Theorem. Given any irrational number x, there exist
infinitely many different rational numbers h/k (with h and k
relatively prime) such that
| x - (h/k)| < 1/[sqrt(5)k^2].
Moreover, the sqrt(5) factor is best possible.
Not sure if that helps, but perhaps a way to start looking around.
In general, the subject of "Diophantine Approximation" deals with the
problem of: given an irrational number x, to determine all solutions
of |qx - p| < psi(q), where psi is a positive, decreasing function of
real variable, and p and q are integers.
Don't know if that will help, but it looks similar to what you are
looking at.
--
Arturo Magidin
Dan Cass
For x irrational and rational r=h/k in lowest terms,
the difference quotient is
** [f(r)-f(x)]/(r-x) = [1/k - 0]/(h/k - x).
From Arturo's post, for infinitely many fractions h/k,
| x - (h/k)| < 1/[sqrt(5)k^2].
From this, 1/|h/k - x| > sqrt(5)*k^2.
So for these fractions h/k, the difference quotient
is at least sqrt(5)*k in absolute value.
So the difference quotient is unbounded (from either side).
On the other hand, if y is an irrational near the irrational x, the difference quotient is 0.
So I think this shows your function f(x) is not
differentiable from either side at irrational x.
W^3
<790314738.46370.12578...@gallium.mathforum.org>,
Maury Barbato <maurizi...@aruba.it> wrote:
Just to look at one specific irrational, consider [f(e/3) -
f(S_n/3)]/[(e/3) - (S_n/3)], where S_n = sum(k=0,n) 1/k!. S_n/3 =
integer/(3n!), so f(S_n/3) >= 1/(3n!). So in absolute value this
difference quotient is >=
[1/(3n!)]/[e/3 - S_n/3]
= (1/n!)/(e - S_n) (1).
Now e - S_n = sum(k=n+1,oo) 1/k! < 1/(n+1)![1 + 1/(n+2) + 1/(n+2)^2 +
...]. In the brackets we have a geometric series that sums to
(n+2)/(n+1), which shows (1) is on the order of n as n -> oo.
Maury Barbato
Good, this surely works!
Thank you very very much, Dan.
Maury Barbato
Numbers are a wonderful wood, where I often get
lost, but when I always get out of it, taking some
hidden treasure!
Thank you, very much, Arturo (this name sounds
quite familiar to me: I'm Italian).
Friendly Regards,
Maury Barbato
William Hughes
Avoiding Hurwitz's Theorem.
All we need is | x - (h/k)| < 1/k
which is immediate.
Let y_1, y_2, y_3 be a sequence of irrationals
with limit x. Then the difference quotient is 0.
For each n, choose an integer k(n) so that
k(n)/n is smaller that x and x-k(n)/n is a minimum.
Clearly k(n)/n has limit x.
Now consider (f(k(n)/n) - f(x))/ (k(n)/n - x).
The denominator has is absolute value less than 1/n.
The numerator is greater than or equal to 1/n.
Thus the difference
quotient has absolute value greater than 1.
- William Hughes