Hi KatolaZ,
Thanks to your reading list, I downloaded "The C programming language"
by (Kernighan & Ritchie) and I am reading it.
P.S.
That proof is a temptation to pass some time with when I can. I will
attempt it. :)
Edward
On 20/06/2016, KatolaZ <kato...@freaknet.org> wrote:
> On Mon, Jun 20, 2016 at 09:35:17AM +0200, Edward Bartolo wrote:
>
> [cut]
>
>>
>> Consider Set I = {...., -3, -2, -1, 0, 1, 2 , 3, ....}, the set of
>> Integers that is infinite in size having neither a lower bound nor an
>> upper bound.
>>
>> Now, consider Set M = {...., -9, -6, -3, 0, 3, 6, 9, ....}, the set of
>> multiples of 3 that also has neither a lower bound nor an upper bound.
>>
>> BOTH sets are infinite, yet, set I has 3 elements for EVERY element in
>> set M! This gives the impression infinity is graded. But does it makes
>> sense to claim a graded infinity? If it is graded, is it still
>> infinite?
>>
>
> Despite your question might be a bit off-topic in this list, I am
> sorry but there is no paradox here. The two sets belong to the class
> of numerable infinity and have the same cardinality (i.e., the same
> number of elements), however strange this might seem at a first
> sight.
>
> The proof consists into showing that both have the same size of the
> set of natural integer numbers N=0,1,2,... and proceeds by assigning
> the "0" in each set to the number "0", positive elements to odd
> integers and negative elements to even integers.
>
> QED
>
> KatolaZ
>
> --
> [ ~.,_ Enzo Nicosia aka KatolaZ - GLUGCT -- Freaknet Medialab ]
> [ "+. katolaz [at] freaknet.org --- katolaz [at] yahoo.it ]
> [ @) http://kalos.mine.nu --- Devuan GNU + Linux User ]
> [ @@) http://maths.qmul.ac.uk/~vnicosia -- GPG: 0B5F062F ]
> [ (@@@) Twitter: @KatolaZ - skype: katolaz -- github: KatolaZ ]
>
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