Edgard Neuman 11 Posted April 16 (edited) Hi, Ok, I was looking for a way to apply the cantor argument to rational number (because I'm masochist) and I think I found one. I'm sure somebody will be happy to disprove my idea : I write rational numbers in a special way. Let's say you have a/b I decompose a and b into sums of 1s.. a= 1 + 1 + .... + 1+1+1 b= 1+ 1+ .. +1+1.. first I simplify, then I use it to create a pseudo number like this : when there's a 1 in the first line, and 1 in the second : it's 3. When 0 and 0 => 0, when 1 and 0 => 2 and when 0 and 1 => 1 So for instance 4 / 5 = > 33331000000.... 12/6 = 2/1 => 3200000... I then use it to fill a cantor-like array now I use the diagonal to create a new row If it's a 0, i put a 1 in the new row If it's a 1, i put a 2 in the new row If it's a 2, i put a 3 in the new row If it's a 3, i put a 0 in the new row Now we are sure that are new number is either +1 or -1 or - 1 -1 on a or b from each previous row.. So it's a new row. (ok there may be a special case when 1+1 / 1+1 = 1/1 but we can obviously exclude 1 from rows from the beginning) Did I just prove that Card (ℚ)> Card (ℕ) ? Edited April 16 by Edgard Neuman 0 Share this post Link to post Share on other sites

Edgard Neuman 11 Posted April 17 (edited) ok i see the problem now : once you rearrange the new number, it could be already in the rows.. But I found a fix : Instead of writing the number a and b in sums of 1, you write it in product of primes For instance 275/12 275 =2^0 * 3^0 * 5 ^0 * 7^0 * 11 ^ 1 12 = 2^2 * 3^1 * 5 ^0 * 7^0 * 11 ^ 0 so you get in the row 275/12 = > (0;2) (0;1) (0;0) (0;0) (1;0) ... (0;0) and then to get a new row, you just add 1 to the bigger number (up or down) in each column now you are sure the new number wasn't in any row that a least would prove that there is more rational numbers than prime numbers Edited April 17 by Edgard Neuman 0 Share this post Link to post Share on other sites

MotleyNoumenon 2 Posted July 7 I the first post you construct a map from the rationals to the integers (since we dont have infinite integers i propose the small change that if we have 0 over 0 then we adjoin nothing to the sequence, this way it is well defined.) In fact since any element in the image is of one of the forms 3...32...2, 2...21.....1, 3...31.....1, 3...32...2, 3...3, 2...2, 1.....1, we can produce a preimage by constructing a rational using the following formula: if a is the number of 3s, b is the number of 2s and c is the number of 1s, then the desired rational is (a+b)/(a+c). (Note that b is non zero if and only if c is zero by the construction---at least that is the way i think you intended for it to be like, as otherwise there would be problems with the order of things in the definition.) Thus we can even construct the inverse map and we see that the very construction gives that Card (ℚ)= Card (ℕ). 0 Share this post Link to post Share on other sites