this post was submitted on 19 Dec 2025
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[–] procrastitron@lemmy.world 16 points 4 days ago* (last edited 4 days ago) (1 children)

The problem is that rational numbers can be mapped (1 to 1) to the integers (e.g. just encode each rational number as an integer), so there are not more rational numbers than integers.

[–] BarbedDentalFloss@lemmy.dbzer0.com -5 points 4 days ago* (last edited 4 days ago) (1 children)

No that's not true. There are rational numbers in between the integers and all integers are rational. Therefore the mapping from integers to rational numbers is injective and thus there are more rational numbers than integers.

[–] berber@feddit.org 8 points 4 days ago (1 children)

"the" mapping? there is no "the" mapping.

you are talking about the canonical inclusion mapping 1 in N to 1 in Z (restriction of the canonical inclusion of rings of integers Z into any other ring, Z is an initial object), which can be seen as a non-generic canonical mapping of semigroups.

but as sets, there is no inherent structure, there are injection, surjections, and of course bijections in both directions.

the only way one can call one set "bigger" is in the very strict sense of sets, N being a true subset of Q. however, this assumes N to be an actual subset of Q, which is a matter of definition and construction. so we say there is some embedding included, which is the same as (re)defining N as that embedded subset, so we are at your canonical inclusion of semigroups again. if you view this as inherent to N and Q, then there are "more" elements in Q as in N, but not in terms of cardinality.

transfinite hocuspocus bullshit is what it is