/sci/ - Science & Math » Thread #11391175

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Alright Nerds

Anonymous Sun 16 Feb 22:46:39 2020 No.11391175 View Reply Original Report

Quoted By: >>11391177 >>11391179 >>11391387 >>11391764 >>11391907

Ok nerds, answer me this. If Q is dense in R then how can Q and R have different cardinalities? Pro tip: you can't.

Anonymous

Anonymous Sun 16 Feb 2020 22:47:45 No.11391177 Report

Quoted By: >>11391184 >>11391187 >>11391387

>>11391175
Sqrt2 is in R but not in Q therefore R has strictly higher cardinality. The end.

Anonymous

Anonymous Sun 16 Feb 2020 22:48:19 No.11391179 Report

Quoted By:

>>11391175
Nigga what

Anonymous

Anonymous Sun 16 Feb 2020 22:52:43 No.11391184 Report

Quoted By: >>11391191

>>11391177
So you're saying that Q and I have the same cardinality? I'm pretty sure that's what you're saying.

Anonymous

Anonymous Sun 16 Feb 2020 22:55:05 No.11391187 Report

Quoted By: >>11391191

>>11391177
Anon that's not how cardinalities work...

Anonymous

Anonymous Sun 16 Feb 2020 22:55:45 No.11391191 Report

Quoted By: >>11391198

>>11391184
>>11391187
troled

Anonymous

Anonymous Sun 16 Feb 2020 22:57:31 No.11391198 Report

Quoted By:

>>11391191
lole

Anonymous

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Anonymous Sun 16 Feb 2020 23:15:27 No.11391237 Report

Quoted By: >>11391240

Because Q is still equal to R retard

>pic is you

Anonymous

Anonymous Sun 16 Feb 2020 23:15:50 No.11391238 Report

Quoted By:

itt: retarded pseuds. OP is pretty clear

Anonymous

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Anonymous Sun 16 Feb 2020 23:17:06 No.11391240 Report

Quoted By:

>>11391237
Read the question moron

Anonymous

Anonymous Mon 17 Feb 2020 00:01:58 No.11391355 Report

Quoted By:

What's the matter, nerds? I knew you couldn't explain this. Math confirmed fake and gay.

Anonymous

Anonymous Mon 17 Feb 2020 00:13:04 No.11391387 Report

Quoted By: >>11391422 >>11391427

>>11391177
Insanely stupid argument
$(0,1)$ and $[0,1]$ have exactly the same Cardinality.

>>11391175
They havr different cardinalities because you can't construct a function from Q to R thats 1-1 and onto.
If R are pigeons and Q is wholes then you will always have more than 1 pigeon in at least 1 hole.
Look at Cantor's diagonalization argument.

Anonymous

Anonymous Mon 17 Feb 2020 00:25:52 No.11391422 Report

Quoted By: >>11391448

>>11391387
Yeah I know Cantor's diagonalization argument but how does the density of Q in R not contradict it? Between any two points in R you can find a point in Q, so how can they not have the same cardinality? It doesn't make any sense.

Anonymous

Anonymous Mon 17 Feb 2020 00:28:32 No.11391427 Report

Quoted By:

>>11391387
Wait, there's 1 in [0,1] but it's not in (0,1), therefore [0,1] must have more elements right? Since (0,1) is clearly inside [0,1]. So basically I have exhibited an element of [0,1]-(0,1), which is therefore non-empty. What's more, the cardinality of {1} is 1 which is >0, so card([0,1])=card(0,1)+1

Anonymous

Anonymous Mon 17 Feb 2020 00:36:08 No.11391448 Report

Quoted By: >>11391503

>>11391422
Regardless of how you construct the reals both things hold and don't contradict each other.
Q is dense in R and Q is countable but R isnt.
Look up the definition of dense in chapter 2 of rudin and read the first chapter of abbotts understanding analysis for more info on cardinality.
Math doesnt have to be intuitive.

Anonymous

Anonymous Mon 17 Feb 2020 00:42:07 No.11391461 Report

Quoted By: >>11391503

goddamn why can't /sci/ make a reasonable answer for this?

Q = {r/s | r is a natural number and s is a natural number}
therefore since r has cardinality aleph_0 and s has cardinality aleph_0 then Q has cardinality aleph_0 x aleph_0 which is also aleph_0

this can be proved by showing that {natural numbers}x{natural numbers} has the same cardinality as {natural numbers}, which is a sophomore undergrad thing [e.g. 0->(0,0), 1-> (1,0), 2->(0,1), 3->(2,0), 4->(1,1), 5->(0,2), 6->(3,0), ....)

R on the other hand has the cardinality 2^(aleph_0) if you believe cantor. and you can prove this using the cantor diagonal slash.

how is this not undergrad level shit? or is /sci/ all highschoolers nowadays?

Anonymous

Anonymous Mon 17 Feb 2020 00:59:34 No.11391503 Report

Quoted By: >>11391507 >>11391642

>>11391448
R is uncountable
Between any two points in R I can find a point in Q
Between any two uncountably many points in R I can find a point in Q
But Q is somehow countable.

No sir, I don't buy it.

>>11391461
what I'm saying is that the density of Q in R contradicts Cantor's diagonalization argument. Let a, b be any two points in R. Then I can find a point x in Q such that a<x<b. So if I can map every point a in R to an x in Q between a and any other point b in R, then I have a one-to-one mapping from R to Q. So they have the same cardinality.

Anonymous

Anonymous Mon 17 Feb 2020 01:01:38 No.11391507 Report

Quoted By: >>11391513

>>11391503
>So if I can map every point a in R to an x in Q between a and any other point b in R, then I have a one-to-one mapping from R to Q. So they have the same cardinality.
but you can’t. try it, construct such a map without some hand waving like you alluded to. hand waving does not amount to a rigorous bijection

Anonymous

Anonymous Mon 17 Feb 2020 01:03:20 No.11391513 Report

Quoted By: >>11391523

>>11391507
For any a in R there is an x in Q between a and any other point a<b in R. How is that not a one-to-one mapping?

Anonymous

Anonymous Mon 17 Feb 2020 01:06:31 No.11391523 Report

Quoted By: >>11391745

>>11391513
because you are saying that there is a Q that approximates R. it does not map 1-1 with the element of R. "approximations" don't jive well if your task is to create a bijection in a set-theoretic sense

Anonymous

Anonymous Mon 17 Feb 2020 01:55:38 No.11391642 Report

Quoted By: >>11391651

>>11391503
>No sir I don't buy it

This isn't about "buying it" or not, were taking math here and this are proven facts. You certainly can't object to them.
Find a flaw in Cantor's proof, find a flaw in the proof that Q is dense in R or find how one contradicts the other.
Pro tip, you can't because theres no problem.
Ill repeat myself, math doesn't have to be intuitive.

Anonymous

Anonymous Mon 17 Feb 2020 02:02:01 No.11391651 Report

Quoted By:

>>11391642
it’s not unintuitive anon. bijections and mapping elements of sets to one another is as basic as it gets. playing the “math is too hard for normies” in this case is a bad argument.

the proof is simple enough to explain to a 8 year old so don’t get on any high horses, that only feeds the trolls

Anonymous

Anonymous Mon 17 Feb 2020 03:01:06 No.11391745 Report

Quoted By: >>11392045

>>11391523
If I have an ordered row of dogs, and between any two dogs I can find a cat, how can I have a countable number of cats and an uncountable number of dogs?

Anonymous

Anonymous Mon 17 Feb 2020 03:08:27 No.11391764 Report

Quoted By: >>11391866

>>11391175
Consider the sets {a,b} and {c}. Order them as a < c < b.

Between any two elements of {a,b}, there exists an element of {c}.

Therefore |{a,b}| = |{c}}.

Anonymous

Anonymous Mon 17 Feb 2020 04:05:21 No.11391866 Report

Quoted By: >>11391893 >>11391902

>>11391764
That's not what I'm saying. What I'm saying is, if I can place an element of Q between any 2 elements of R, then how can Q be countable and R not countable?

Anonymous

Anonymous Mon 17 Feb 2020 04:24:21 No.11391893 Report

Quoted By:

>>11391866
And what I'm saying is, if you can put one foot in front of the other indefinitely, how come you only have 2 feet?

There's no connection between the two things you're talking about.

Anonymous

Anonymous Mon 17 Feb 2020 04:30:39 No.11391902 Report

Quoted By: >>11391915

>>11391866
Maybe you're thinking that if you can map pairs of real numbers to rational numbers *injectively*, or uniquely, then we should have that |R| <= |Q|. This is true. But there is no such *injective* mapping. Between any two pairs of real numbers, there is a rational number. But it is (provably) not the case that there is a distinct rational number between each pair of real numbers.

Anonymous

Anonymous Mon 17 Feb 2020 04:33:10 No.11391907 Report

Quoted By:

>>11391175
If two sets have equal cardinalities, then there exists a bijective function between them. It was proven that no such function exists between Q and R, therefore they must have different cardinalities. And since Q is trivially a subset of R, that means R is "larger" than Q.

Anonymous

Anonymous Mon 17 Feb 2020 04:37:37 No.11391915 Report

Quoted By: >>11391928 >>11391969 >>11392023

>>11391902
Finally, someone actually gets what I'm trying to say. But why is it not true that there is a distinct rational number x between each pair a<b of real numbers such that a<x<b? You say it is provable, can you link me to a proof?

Anonymous

Anonymous Mon 17 Feb 2020 04:42:38 No.11391928 Report

Quoted By: >>11391969

>>11391915
The proof is that Q and R have different cardinalities. I can replicate that proof for you here.

Suppose |Q| = |R|. We know that |Q| = |N|, so there exists a bijection f : N --> R. We can think of this bijection as indexing the real numbers: so f(1) is the 1st real number, f(2) is the 2nd real number, and so on. We may now define a real number x between 0 and 1, whose nth decimal place is *distinct* from that of f(n). So if f(3) = 0.3214..., then the third decimal place is not 1. We can do this schematically, for example by making the decimal place 1 greater (if it is not 9) or 1 lesser if it is 9. Then x is a well defined real number, and it is not equal to f(n) for any n by construction. This contradicts f being a surjection.

Anonymous

Anonymous Mon 17 Feb 2020 04:59:51 No.11391969 Report

Quoted By: >>11391978 >>11391993

>>11391915
>>11391928
Maybe I should finish the argument. This proves R and Q have the same cardinality. But why does that mean there's no unique map from pairs of real numbers to rationals?

First, |R| = |R x R|, where R x R is the set of pairs of real numbers. Can you show this?

There is an injective map from the rationals into pairs of reals, namely r --> (r, 0).

Now let's say we have a function which took a pair of real numbers (a,b) and picked a rational r in between them. If it were injective, then the Cantor-Schroeder-Bernstein theorem implies that |R x R| = |Q|. But we showed |R| != |Q|.

Anonymous

Anonymous Mon 17 Feb 2020 05:01:53 No.11391978 Report

Quoted By:

>>11391969
>This proves R and Q have the same cardinality.
I meant different cardinality.

Anonymous

Anonymous Mon 17 Feb 2020 05:13:36 No.11391993 Report

Quoted By:

>>11391969
There is a subtlety here. Without the Cantor-Schroeder-Bernstein theorem, all we know is that any map f : R x R --> Q is not *both* surjective and injective. We could, a priori, have an injective map which is not surjective. But the CSB theorem says if you have an injection f : A --> B, and an injection g : B --> A, then there is a bijection h : A --> B. This theorem is very much not obvious. But it does imply that there is no injective-but-not-surjective map f : R x R --> Q.

Anonymous

Anonymous Mon 17 Feb 2020 05:33:32 No.11392023 Report

Quoted By: >>11392059

>>11391915
>But why is it not true that there is a distinct rational number x between each pair a<b of real numbers such that a<x<b?

That is true, but it isn't a mapping. By thinking of it as a mapping your are forcing both sets to have the same cardinality.

Anonymous

Anonymous Mon 17 Feb 2020 05:53:29 No.11392045 Report

Quoted By:

>>11391745
Do you understand the concept of a continuum?

Anonymous

Anonymous Mon 17 Feb 2020 06:06:24 No.11392059 Report

Quoted By:

>>11392023
There is no way to pick a unique rational number for each pair of real numbers.

Anonymous

Anonymous Mon 17 Feb 2020 15:06:55 No.11392922 Report

Quoted By:

List out all rationals. create a number where the ith digit is different than the digit in the ith digit of the ith number in the list of rationals. Congratulations you have created a new number, but wait, it differs from any other number by at least one digit. OHHH NOEEESS. this new number is not a rationals because we already listed all the rationals. Congrats, you proved that there is not a objective mapping from rationals to reals....retard

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