Re: Rational Numbers/Irrational Numbers

"Dave Seaman" <dseaman@xxxxxxxxxxxx> wrote in message
news:epfl09$7cg$2@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Sat, 27 Jan 2007 01:28:41 -0500, David T. Ashley wrote:
<davidmarcus@xxxxxxxxxxxx> wrote in message
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On Jan 26, 11:36 pm, "David T. Ashley" <d...@xxxxxxxx> wrote:
"Leo" <newsdon...@xxxxxxxxxxx> wrote in
messagenews:1169780763.086460.114690@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Which set has more numbers, the set of rational numbers or the set of
irrational numbers?
Well, the set of irrational numbers has at least twice as many elements
as
the set of rational numbers.

Think about the following functions:

f(x) = PI + x
g(x) = PI + PI + x

Every rational number x can be paired with at least two irrationals.

So, I'm going to go with "irrational" as being bigger.

Right answer. Wrong reason. The rationals are countable. The
irrationals are uncountable. The rationals have Lebesgue measure zero.
The irrationals in [0,1] have Lebesgue measure 1.

I was just clowning around ... but OK, let's explore your logic.

In order for me to have stated the "wrong" reason, there has to be a
counterexample where my test fits but where there are not "more".

Please provide a counterexample.

Consider one of the standard proofs that the rationals are countable. We
define a mapping f: N -> Q+ by listing the positive rationals in the
following order:

1/1, 1/2, 2/1, 3/1, 2/2, 1/3, 1/4, 2/3, ...

I'm not seeing the algorithm behind the sequence above. Could you state it
explicitly?

The concept I understand (the mapping), just I don't see how the sequence
above is generated.

I might be having a slow day.

Thanks.

--
David T. Ashley (dta@xxxxxxxx)
http://www.e3ft.com (Consulting Home Page)
http://www.dtashley.com (Personal Home Page)
http://gpl.e3ft.com (GPL Publications and Projects)


.

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