this post was submitted on 23 Oct 2024
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[–] [email protected] 13 points 1 month ago (4 children)

It's called countable and uncountable infinity. the idea here is that there are uncountably many numbers between 1 and 2, while there are only countably infinite natural numbers. it actually makes sense when you think about it. let's assume for a moment that the numbers between 1 and 2 are the same "size" of infinity as the natural numbers. If that were true, you'd be able to map every number between 1 and 2 to a natural number. but here's the thing, say you map some number "a" to 22 and another number "b" to 23. Now take the average of these two numbers, (a + b)/2 = c the number "c" is still between 1 and 2, but it hasn’t been mapped to any natural number. this means that there are more numbers between 1 and 2 than there are natural numbers proving that the infinity of real numbers is a different, larger kind of infinity than the infinity of the natural numbers

[–] [email protected] 5 points 1 month ago

Great explanation by the way.

[–] [email protected] 3 points 1 month ago (3 children)

I get that, but it's kinda the same as saying "I dare you!" ; "I dare you to infinity!" ; "nuh uh, I dare you to double infinity!"

Sure it's more theoretically, but not really functionally more.

[–] [email protected] 5 points 1 month ago* (last edited 1 month ago)

It's like when you say something is full. Double full doesn't mean anything, but there's still a difference between full of marbles and full of sand depending what you're trying to deduce. There's functional applications for this comparison. We could theoretically say there's twice as much sand than marbles in "full" if were interested in "counting".

The same way we have this idea of full, we have the idea of infinity which can affect certain mathematics. Full doesn't tell you the size of the container, it's a concept. A bucket twice as large is still full, so there are different kinds of full like we have different kinds of infinity.

[–] [email protected] 4 points 1 month ago

When talking about infinity, basically everything is theoretical

[–] [email protected] 1 points 1 month ago (1 children)

but not really functionally more.

Please show me a functional infinity

[–] [email protected] 1 points 4 weeks ago (1 children)

Right, an asymptote I guess, in use, but not a number.

[–] [email protected] 1 points 4 weeks ago

It's been quite some time since I did pre-calc, but I remember there being equations where it was relevant that one infinity was bigger than another.

[–] [email protected] 1 points 4 weeks ago

This reminds me of a one of Zeno's Paradoxes of Motion. The following is from the Stanford Encyclopaedia of Philosophy:

Suppose a very fast runner—such as mythical Atalanta—needs to run for the bus. Clearly before she reaches the bus stop she must run half-way, as Aristotle says. There’s no problem there; supposing a constant motion it will take her 1/2 the time to run half-way there and 1/2 the time to run the rest of the way. Now she must also run half-way to the half-way point—i.e., a 1/4 of the total distance—before she reaches the half-way point, but again she is left with a finite number of finite lengths to run, and plenty of time to do it. And before she reaches 1/4 of the way she must reach 1/2 of 1/4=1/8 of the way; and before that a 1/16; and so on. There is no problem at any finite point in this series, but what if the halving is carried out infinitely many times? The resulting series contains no first distance to run, for any possible first distance could be divided in half, and hence would not be first after all. However it does contain a final distance, namely 1/2 of the way; and a penultimate distance, 1/4 of the way; and a third to last distance, 1/8 of the way; and so on. Thus the series of distances that Atalanta is required to run is: …, then 1/16 of the way, then 1/8 of the way, then 1/4 of the way, and finally 1/2 of the way (for now we are not suggesting that she stops at the end of each segment and then starts running at the beginning of the next—we are thinking of her continuous run being composed of such parts). And now there is a problem, for this description of her run has her travelling an infinite number of finite distances, which, Zeno would have us conclude, must take an infinite time, which is to say it is never completed. And since the argument does not depend on the distance or who or what the mover is, it follows that no finite distance can ever be traveled, which is to say that all motion is impossible. (Note that the paradox could easily be generated in the other direction so that Atalanta must first run half way, then half the remaining way, then half of that and so on, so that she must run the following endless sequence of fractions of the total distance: 1/2, then 1/4, then 1/8, then ….)

[–] [email protected] -1 points 1 month ago* (last edited 1 week ago) (3 children)

Your explanation is wrong. There is no reason to believe that "c" has no mapping.

Edit: for instance, it could map to 29, or -7.

[–] [email protected] 2 points 1 month ago

Give me an example of a mapping system for the numbers between 1 and 2 where if you take the average of any 2 sequentially mapped numbers, the number in-between is also mapped.

[–] [email protected] 2 points 1 month ago* (last edited 1 month ago) (1 children)

Yeah, OP seems to be assuming a continuous mapping. It still works if you don't, but the standard way to prove it is the more abstract "diagonal argument".

[–] [email protected] 1 points 1 week ago* (last edited 1 week ago) (1 children)

But then a simple comeback would be, "well perhaps there is a non-continuous mapping." (There isn't one, of course.)

"It still works if you don't" -- how does red's argument work if you don't? Red is not using cantor's diagonal proof.

[–] [email protected] 1 points 1 week ago* (last edited 1 week ago)

Yeah, that was actually an awkward wording, sorry. What I meant is that given a non-continuous map from the natural numbers to the reals (or any other two sets with infinite but non-matching cardinality), there's a way to prove it's not bijective - often the diagonal argument.

For anyone reading and curious, you take advantage of the fact you can choose an independent modification to the output value of the mapping for each input value. In this case, a common choice is the nth decimal digit of the real number corresponding to the input natural number n. By choosing the unused value for each digit - that is, making a new number that's different from all the used numbers in that one place, at least - you construct a value that must be unused in the set of possible outputs, which is a contradiction (bijective means it's a one-to-one pairing between the two ends).

Actually, you can go even stronger, and do this for surjective functions. All bijective maps are surjective functions, but surjective functions are allowed to map two or more inputs to the same output as long as every input and output is still used. At that point, you literally just define "A is a smaller set than B" as meaning that you can't surject A into B. It's a definition that works for all finite quantities, so why not?

[–] [email protected] 1 points 1 month ago (1 children)

because I assumed continuous mapping the number c is between a and b it means if it has to be mapped to a natural number the natural number has to be between 22 and 23 but there is no natural number between 22 and 23 , it means c is not mapped to anything

[–] [email protected] 1 points 1 week ago* (last edited 1 week ago)

Then you did not prove that there is no discontiguous mapping which maps [1, 2] to the natural numbers. You must show that no mapping exists, continugous or otherwise.