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02022004, 03:42 PM

#1

Member
Registered: Oct 2003
Distribution: fedora cor 5 x86_64
Posts: 639
Rep:

diagonalization proof...
i have a question for a cs class that is impossible... the technique to solve it is called diagonalization. i have to prove that all positive rational numbers are denumerable (meaning have a 1 to 1 correspondence with all positive integers). here is how your supposed to do it...
... ... ... ... ...
3/0 3/1 3/2 3/3 3/4 ...
2/0 2/1 2/2 2/3 2/4 ...
1/0 1/1 1/2 1/3 1/4 ...
0/0 0/1 0/2 0/3 0/4 ...
then to show the correspondence you say that
f(0/0) = 0
f(1/0) = 1
f(0/1) = 2
f(2/0) = 3
f(1/1) = 4
f(0/2) = 5
and so on... where the term diagonalization comes from... if you cant see it, you start at 0/0 then move one up and go down the diaginal, and so on... well, i thought that that is enough to prove that there is a 11 correspondence, but i have to find a function that will show that f(x) = y ... the only function that i have came up with was this
f(x/y) = (x+y)(x+y+1)/2 + x
does anyone know if there is a function with just 1 variable?? thank



02022004, 10:26 PM

#2

Senior Member
Registered: Oct 2003
Location: hopefully not here
Distribution: Gentoo
Posts: 2,038
Rep:

i dono if i read whatever equation u sadi right, but i got that whatever you used is impossible, i x=0 y=0 and ur euation says the answer is 0.5 when it should be 0 right?,
neways thats a nice little trick, it moves out in 3 directions, across y,x and between x and y, only there looks as its a small inconsitiency at the start when it sets up, making it look like a L a bit, as there are 2 distint lines that show b4 3 and then start a solid pattern up



02022004, 11:44 PM

#3

Member
Registered: Oct 2003
Distribution: fedora cor 5 x86_64
Posts: 639
Original Poster
Rep:

((0 + 0)(0 + 0 + 1)/2) + 0 = 0
i made the same mistake you did, you have to realize that your multiplying the numerator by 0



02032004, 09:59 AM

#4

Member
Registered: Mar 2003
Location: PA
Distribution: Ubuntu (x2)
Posts: 158
Rep:

Ah. Cantor's diagonalization theroem.
To prove the cardinality of the naturals and rational is equal you need to find a bijection between the two.
Set up a cartesian plain.
now assume the x coordinate is the denominator, and the y the numerator (or vice versa)
imagine a diagonization snaking thorugh all the numbers.
you appear to have control of the part above, i was just clarifing,
the numbering i was taught was not as formulaic as you seem to require but its a valid function.
f(x) = the xth unique rational on the diagonalization (discarding 0 denominators and unsimplfied fractions)
Google it for better results or possibility a more acceptable expression of the function, my math is rusty.



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