Para-consistent logic

[RBerenguel]

Integer division is closed already ( i.e. defined on Z X Z0 ). What you close is some (proto-
) division defined on a specific subset of Z X Z0 ( consisting of the (proto-)dividable tuples ). a is protodivable by b if there is a number c such that bc==a. Thus you generate the rationals. That is what they taught me in my first year course "number theory" some forty+ years ago. Somewhere I must still have Walter Rudin's "Principles of Analysis", a very nice, be it dry book. It seems you mix up integer division with this proto division. But both are stricly no divisions on Z.
So what about SmoothOper? Strangely enough Cantor's proof has nothing to do with division or rationals so this whole discussion is futile.

I don't exactly remember the details (a shame given that you almost do after 40 years and I don't with 10,) but we defined divisibility properties (seems similar to what you write as proto-division) via ideals and ideal intersections and other properties, so it could be extended to any ring.

And yes, the proof has nothing to do with division, so I'm not sure what the point is either.

[SmoothOper]

Integer division is closed already ( i.e. defined on Z X Z0 ). What you close is some (proto-
) division defined on a specific subset of Z X Z0 ( consisting of the (proto-)dividable tuples ). a is protodivable by b if there is a number c such that bc==a. Thus you generate the rationals. That is what they taught me in my first year course "number theory" some forty+ years ago. Somewhere I must still have Walter Rudin's "Principles of Analysis", a very nice, be it dry book. It seems you mix up integer division with this proto division. But both are stricly no divisions on Z.
So what about SmoothOper? Strangely enough Cantor's proof has nothing to do with division or rationals so this whole discussion is futile.

I don't exactly remember the details (a shame given that you almost do after 40 years and I don't with 10,) but we defined divisibility properties (seems similar to what you write as proto-division) via ideals and ideal intersections and other properties, so it could be extended to any ring.

And yes, the proof has nothing to do with division, so I'm not sure what the point is either.

The point is there are more rationals than integers.

[cyclops]

The point is there are more rationals than integers.

Well there are some rationals that are not an integer but there are no integers that are not rational. In that sense you are correct. In the same sense there are more integers than even integers.
But,
... -4, -2, 0, 2, 4, 6 ......
... -2, -1, 0, 1, 2, 3 ......
the upper row is as long as the lower row so there are not more integers than even integers.
So "there are more" is not an usefull concept unless you define it better. So please give me a definition of "there are more" that doesn't run in this contradiction and then proof your statement. You will be famous thereafter, a SaintOper.

[lightvector]

And yes, the proof has nothing to do with division, so I'm not sure what the point is either.

The point is there are more rationals than integers.

The real answer is that it depends on your definition of "more". And I mean this in a completely serious way.

If by "more" you mean that the rationals have greater *cardinality*, that there exists an injective function but no bijective function from the integers to the rationals, then that's false, there are actually not more rationals than integers.

If by "more" you mean that the rationals are a *strict superset of* the integers (when the integers are identified or viewed as a subset of the rationals in the natural way), then that's true, there are more.

If by "more" you mean that the rationals have strictly greater *area*, as in Lebesgue measure, as a subset of the reals than do the integers, then that's false, both have measure 0.

If by "more" you mean that the rationals form a *topologically-dense subset* of the real numbers whereas the integers do not (when both are viewed as a subset of the reals in the natural way), then that's true, there are more.


A important part of mathematics is being precise in communicating what you mean. It turns out that people can easily disagree on colloquial concepts like "more", because they actually mean different things by the word. A lot of base-level mathematics is simply about providing a way to separate out and talk about the different intuitions people might have about concepts like "more" in different contexts so that the relationships between them can be studied. Being precise about what you mean is critical when doing this.

Unfortunately, it can be easy for people to forget this when explaining or popularizing mathematics. Particularly because it's easy to forget that an audience may have much less mathematical background. Whereas someone with more background can often pick up enough of the context to infer what was meant, allowing one to get away with being imprecise, with a more popular audience, this may just produce confusion and misinformation.

[Bonobo]

Does there exist a last post in this thread?

Sorry, no.

Well, that’s wrong. There always existed and exists a last post in this thread since was started.

[lemmata]

And yes, the proof has nothing to do with division, so I'm not sure what the point is either.

The point is there are more rationals than integers.

The real answer is that it depends on your definition of "more". And I mean this in a completely serious way.

If by "more" you mean that the rationals have greater *cardinality*, that there exists an injective function but no bijective function from the integers to the rationals, then that's false, there are actually not more rationals than integers.

If by "more" you mean that the rationals are a *strict superset of* the integers (when the integers are identified or viewed as a subset of the rationals in the natural way), then that's true, there are more.

If by "more" you mean that the rationals have strictly greater *area*, as in Lebesgue measure, as a subset of the reals than do the integers, then that's false, both have measure 0.

If by "more" you mean that the rationals form a *topologically-dense subset* of the real numbers whereas the integers do not (when both are viewed as a subset of the reals in the natural way), then that's true, there are more.


A important part of mathematics is being precise in communicating what you mean. It turns out that people can easily disagree on colloquial concepts like "more", because they actually mean different things by the word. A lot of base-level mathematics is simply about providing a way to separate out and talk about the different intuitions people might have about concepts like "more" in different contexts so that the relationships between them can be studied. Being precise about what you mean is critical when doing this.

Unfortunately, it can be easy for people to forget this when explaining or popularizing mathematics. Particularly because it's easy to forget that an audience may have much less mathematical background. Whereas someone with more background can often pick up enough of the context to infer what was meant, allowing one to get away with being imprecise, with a more popular audience, this may just produce confusion and misinformation.

This came up earlier in the thread, but unfortunately, the person you are responding to has written that axioms are things to be believed or disbelieved. Any hope of rational discussion ends at that point. It is not a stretch to conjecture that he also thinks that definitions (which axioms often imply) are a matter of belief, too.

This thread has taught me to appreciate Robert Jasiek. His threads are often tiring to go through, but he exhibits an internally consistent view that he presents in coherent ways and disagreements with him are always a matter of vocabulary or opinion. We often disagree with him, but we know where our differences stand. He never loses composure when he is attacked. Most importantly, when people talk with RJ, both parties are always sharing the same conversation (or eventually do through a series of vocabulary clarifications from RJ). Compare and contrast with this thread.

[Solomon]

SmoothOper has been banned for violating the rules a third time after a week-long suspension per guidelines. Thank you.