/sci/ - Science & Math

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Anonymous Sat May 21 19:01:01 2022 No.14504005 [Reply] [Original]

>There is a rational number between any 2 distinct irrational numbers. Therefore there are at least as many rationals as irrationals.

uh…cantorbros??

>>	Anonymous Sat May 21 19:04:07 2022 No.14504016 >>14504005 so?

>>	Anonymous Sat May 21 19:08:38 2022 No.14504025 >>14504016 If there as many rationals as irrationals, then there as many natural numbers as reals

>>	Anonymous Sat May 21 19:12:24 2022 No.14504029 >>14504005 Eh? There are more irrational numbers between any two irrational numbers than rational ones.

>>	Anonymous Sat May 21 19:13:25 2022 No.14504030 >>14504025 >If there as many rationals as irrationals, then there as many natural numbers as reals True. You still didnt prove rationals and irrationals are equinumerous.

>>	Anonymous Sat May 21 19:28:20 2022 No.14504055 there are as many numbers between 1 to 2 vs 1 to infinity

>>	Anonymous Sat May 21 20:00:13 2022 No.14504115 >>14504005 You can find infinitely many irrationals between each irrational. Therefore there are infinitely many more irrationals than rationals

>>	Anonymous Sat May 21 20:05:59 2022 No.14504132 >>14504115 rationals are also infinitely dense.

>>	Anonymous Sat May 21 20:15:05 2022 No.14504153 these observations itt are completely pointless

>>	Anonymous Sat May 21 20:18:49 2022 No.14504158 >>14504005 Theorem 1: All separable linear orders are countable

>>	Anonymous Sat May 21 20:19:19 2022 No.14504160 This thread feels like a popsci video on infinity

Anonymous Sat May 21 23:15:16 2022 No.14504492
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Proof that the reals are countable:

1. The algebraic numbers are countable.
2. Only countably many transcendental numbers are known: pi, e, sin(x) for x algebraic, etc ... and countable combinations thereof (multiples, sums)

QED

Anonymous Sat May 21 23:24:49 2022 No.14504509
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The diagonal argument for the proof of the uncountability of the reals is flawed btw. It relies on decimal representations. But as the "0.999 = 1" disaster shows, decimal representations don't need to be unique. So while there might be uncounably many decimal representations, infinitely many of them could refer to the same numbers, thus leaving the actual unique reals countable.

Anonymous Sat May 21 23:32:42 2022 No.14504522

>>14504509
I’m not sure if the argument is flawed for this reason. I’m binary, every number ends in repeating 0’s or 1’s, so there are only two representations. Sure, you could create an arrangement such that the main diagonal doesn’t work, but if you just switch the place of a few numbers, it’s possible that the new sequence will be a new number.

I simply don’t believe that it’s possible to create a new number in the first place. Or, in the spirit of >>14504492
perhaps not all decimal reapresentaríamos are real numbers. Maybe only certain types of real numbers exist. I’m not sure if they are uncountable but the diagonal argument doesn’t work because the new sequence isn’t necessarily a proper real number

>>	Anonymous Sat May 21 23:34:03 2022 No.14504526 >>14504509 >the "0.999 = 1" disaster What disaster was that? Did a plane crash because of decimal representation or something?

Anonymous Sat May 21 23:34:12 2022 No.14504527

heres a question, why does it matter if a number has a decimal point that is so many digits long? can you think of ANY application for something with a million decimal points? surely there is some length of decimal where any more no longer has any application in either math or physics making it meaningless

>>	Anonymous Sat May 21 23:45:26 2022 No.14504544 >>14504527 What would be the point of declaring a boundary on how many decimal places everyone's allowed to use? A lot of mathematics exists without practical applications.

>>	Anonymous Sat May 21 23:52:11 2022 No.14504552 >>14504005 You're supposing that there is a space between all irrational numbers such that there are consecutive irationals. Your proof is faulty from the get-go.

Anonymous Sat May 21 23:59:48 2022 No.14504565

>>14504492
you’re right. There are only a countable amount of meaningful irrational numbers. Either they are roots to algebraic equations, or they are infinite sums with a defined rule, or some function, etc. We are supposed to believe that most real numbers are completely inaccessible to us. They may as well not exist.

>>	Anonymous Sun May 22 00:11:36 2022 No.14504575 >>14504492 Not a proof, because a slave boy can't follow it.

>>	Anonymous Sun May 22 00:41:50 2022 No.14504612 >>14504005 There is also at least 1 natural number between any 2 distinct natural numbers n and 3n.

>>	Anonymous Sun May 22 00:41:57 2022 No.14504613 Predicative mathematics completely destroys the notion of uncountability, as it does away with circularity completely, thus killing these paradoxes once and for all.

Anonymous Sun May 22 00:56:12 2022 No.14504635

>the majority of modern “mathematicians” believe in the existence of numbers which can’t be defined in any way, and that almost all numbers are of this type
We will NEVER interact with these numbers in any way. It is impossible. So how can we say they exist? They can’t even be related to the numbers that we do know. The reals are countable.

>>	Anonymous Sun May 22 00:58:41 2022 No.14504638 >>14504005 There is a butthole between any 2 of your buttcheeks. Therefore you have at least as many buttholes as buttcheeks.

Anonymous Sun May 22 01:15:05 2022 No.14504662

cardinality has nothing to do with there being "as many rationals as irrationals"
the number of rationals and irrationals are both infinite, so you can't say that one has more than the other
what you can say is that the cardinality of the set of rationals is less than the cardinality of the set of irrationals, but that is unrelated to the "amount" of rationals or irrationals there are. it solely has to do with possible mappings from one set to the other.

>>	Anonymous Sun May 22 01:16:08 2022 No.14504664 >>14504527 in cryptography insane precision is very important. as well for predicting the behaviour of chaotic systems.

>>	Anonymous Sun May 22 02:24:14 2022 No.14504769 >>14504635 >The reals are countable produce a bijection between Q and R immediately fag

Anonymous Sun May 22 02:26:48 2022 No.14504775

>>14504769
Not every infinite sequence digits represents a real number. Real numbers are defined (distinguished) by sets. Every real number that is defined by a well-defined set is a countable set. So these real numbers are countable. Undefinable (non-existent) numbers are the reason for the uncountability of the “real numbers”

>>	Anonymous Sun May 22 02:28:54 2022 No.14504784 >>14504775 >So these real numbers are countable So it should be easy to produce a bijection between R and a countable set, like Q. So do it.

>>	Anonymous Sun May 22 02:28:56 2022 No.14504785 >>14504775 >the set of reals I can count is countable Well. You're not wrong but you're still an idiot.

>>	Anonymous Sun May 22 02:32:08 2022 No.14504789 >>14504785 If something can’t be counted, it doesn’t exist. This should be intuitively obvious

>>	Anonymous Sun May 22 02:38:20 2022 No.14504800 >>14504789 Stop. You're retarded.

Anonymous Sun May 22 02:48:09 2022 No.14504816

>>14504784
The definable real numbers is a countable set because the finite definitions are countable. So uncountability of reals comes from infinite definitions. The problem with an infinite definition is that there is NO WAY for us to differentiate such numbers. You could accept such nonsense if you’d like, but at the very least, you should prove that these numbers are necessary for the reals to be complete, and I haven’t seen such a proof

>>	Anonymous Sun May 22 03:06:10 2022 No.14504852 >>14504816 >The definable real numbers is a countable set because the finite definitions are countable Great. Still waiting on the bijection.

Anonymous Sun May 22 03:13:50 2022 No.14504865

>>14504852
I don’t have to give a bijection. It’s enough to show that the definitions are countable, since they use a countable alphabet, and the definitions are finite. You can’t use the diagonal argument here. There is no contradiction. Just because we don’t know an exact bijection doesn’t mean these bijections don’t exist.

>>	Anonymous Sun May 22 03:36:47 2022 No.14504912 >>14504865 >I don’t have to give a bijection Yes you do. This is the literal fucking meaning of the word "equinumerous"

>>	Anonymous Sun May 22 03:47:13 2022 No.14504926 >>14504912 The union of countably many countable sets is countable. So you can show that a set is countable without giving a bijection. You’re embarrassing yourself

>>	Anonymous Sun May 22 07:51:32 2022 No.14505318 >>14504926 >So you can show that a set is countable without giving a bijection And yet it must still be possible to produce one since there is equinumerosity. Are you stupid? Is there a specific reason you cannot do this?

>>	Anonymous Sun May 22 15:18:18 2022 No.14505905 >>14504865 The catch here is that the "set" of definable numbers is not itself definable, so you really can’t prove anything about it.

>>	Anonymous Mon May 23 03:12:25 2022 No.14507358 >>14504662 >are both infinite I request proof

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