3 ms·
Silly solution: The rationals are countable. Just use any bijective mapping of the natural numbers to the rationals and then use the scheme for the natural numb
by neunhoef 9y ago
Silly solution: The rationals are countable. Just use any bijective mapping of the natural numbers to the rationals and then use the scheme for the natural numbers. (I know, a mathematician's answer! :-) )
- Veedrac 9y agoThat would imply the mapping preserved ordering, which doesn't seem possible.
- neunhoef 9y agoYou are of course right. My mistake.
- bmm6o 9y agoIt's not possible, since between any two rationals there exists another: (a+b)/2.
- deleted 9y ago[deleted]
- rocqua 9y agoWait, do the natural numbers and rational numbers have the same ordinality? It seems not because of the whole 'a number in between every 2 rational numbers'. But then what is the ordinality of the rational numbers?
- Veedrac 9y agoThey have the same cardinality, but there is no order-preserving bijection.
- bmm6o 9y agoThe standard argument for showing that they have the same cardinality is to consider the matrix (let's see if I can render this): 1/1 2/1 3/1 4/1 ... 1/2 2/2 3/2 4/2 ... 1/3 2/3 3/3 4/3 ... ... You can traverse each SW-NE diagonal, producing the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... This will eventually hit every positive rational (several representations, in fact). If you want, you can start with 0 and insert -q when you hit q.