On Wednesday, 23 October 2013 18:00:51 UTC+2, Michael F. Stemper wrote:
> Here is a simple mapping that turns any rational into a natural.
"Any" does not imply "all".
You can subtract any finite initial segment 1, 2, 3, ..., n from the set of natural numbers. Does that mean that you can extend that method such that you have nothing left?
Your argument is of the same quality.
By the way, a contradiction is not removed by supplying a better and more sophisticated method to prove one of the contradicting results.
Not only that, each rational is turned into a different natural, so it is an injection Q->N. 1. Given a rational number, r, eliminate all common factors from the numerator and denominator. 2. Call the resulting denominator "q". 3. If r is non-negative, call the resulting numerator "p". Calculate (2^p)*(3^q) 4. If r is negative, call the resulting numerator "-p". Calculate (2^p)*(5^q) A few examples: 1.4 = 14/10 = 7/5 yields (2^7)*(3^5) = (128)*(243) = 31104 -3/6 = -1/2 yields (2^1)*(5^2) = 50 If you don't think that this mapping does exactly what I say that it does, you have two options: 1. Give a specific rational number that does not get mapped to a natural number. 2. Give two distinct rational numbers that get mapped to the same natural number. No jabbering about bins or vases or tasks. Just a simple counter-example that shows this mapping is not an injection Q->N. -- Michael F. Stemper This sentence no verb.