🚀 go-pugleaf

In article <E97o4r.I6v.0.-s@inmet.camb.inmet.com>,
Don Porges <porges@sunfire.camb.inmet.com> wrote:
!
!In article <5jamlt$7ng@samba.rahul.net>, Ken Smith  <kensmith@rahul.net> wrote:
!>
!>Proof:
!>
!>Any rational number can be written in the form of N/D.  If each rational 
!>number can be mapped to a integer there is a countable number of them.  
!>For any rational number an integer can be written using the following 
!>procedure:
!>
!>Write down the value in the form N/D.  If one of the numbers has less 
!>digits than the other add leading zeros to that number to make the number 
!>of digits equal.
!>
!>Starting at the left end of the number write down the left most digit of 
!>N and then the left most digit of D.  Next write beside those digits the 
!>next digit of N and the next digit of D.  Continue this process of taking 
!>one digit from each number until all of the digits have been used.
!>
!>Example:
!>
!>           12/345  ...>   012/345   ....>  304152
!>
!>Each rational number can thus be mapped to an integer and hence the 
!>number of rationals must equal the number of integers.
!>
!True, but isn't this unnecessarily complicated?  Why not just map
!12/345 into 12345?  As long as each rational maps onto exactly one
!integer, your proof stands.  (Therefore you have to specify that
!all rationals are to be reduced as far as possible before applying 
!the transformation.)

because this wouldn't distinguish between 1/2345, 12/345, 123/45, and 1/2345.

the algorith we learned was to start at (0,1), and move through diagonals in
the first quadrant, successively, with each being the numerator and denominator
as such: (0,1),(0,2),(1,1),(0,3),(1,2),(2,1),(0,4),(1,3),(2,2),(3,1),(0,5),etc.
this will progress through the rationals.

to iclude negative rationals, simply alternate between the above method and
the equivalent w/ negative numerators (second quadrant).

thus, we can enumerate the members of the set, and therefore prove that the
set of rational numbers is countable.  and we _don't_ have to reduce the
rationals at all; this algorithm determines that both 1/2 and 2/4 are in the
set of rationals.

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