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[ap-calculus] Re: An interesting continuity function
Posted:
Mar 27, 2006 11:10 AM
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Doug Shaw wrote:
http://mathforum.org/kb/thread.jspa?threadID=1354259
> f(x) = 0, for x irrational
> f(x) = 1/q for x = p/q, where p and q are integers,
> and the fraction is in lowest terms.
>
> Have them look at the continuity of f(x). It turns
> out that f is continuous where x is irrational,
> and discontinuous where x is rational.
Nit-picks: f is not defined at x=0 (or at least,
f isn't single-valued at x=0). Also, if you define
f(0) = 0, then f is continuous at the rational
number 0. In addition, you want to specify the
sign of either p or q (most people use p > 0)
so that f will be a function. Note that
(-2)/(3) = (2)/(-3) --- the left side gives
us f(- 2/3) = 1/3 and the right side gives us
f(- 2/3) = -1/3. I seem to recall that Richard
G. Brown and David P. Robbins's "Advanced Mathematics:
A Precalculus Course" overlooks one or more of
these issue when this function is defined somewhere
in their chapter on limits (in one of the exercise
sections).
This function is often called the "ruler function".
(Look at the lenghts of the marks on an English
foot or yard ruler.) If anyone is interested,
I posted 20 journal references that deal with
the *differentiability* of the ruler function
in sci.math on November 9, 2005. I gave brief
summaries for the first 13 references, and then
had to stop because I was running out of time.
http://groups.google.com/group/sci.math/msg/cee496c760dd15be
The following google searches will also prove useful:
http://www.google.com/search?q=p%2Fq+irrational+continuous
http://www.google.com/search?q=%22ruler+function%22+irrational
http://books.google.com/books?as_q=rational&as_epq=ruler+function
http://groups.google.com/groups?q=%2Bp%2Fq+irrational+continuous&filter=0
http://groups.google.com/groups?q=%22ruler+function%22+irrational&filter=0
Dave L. Renfro
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