/sci/ - Science & Math

[Anonymous No.7181112]

Lets say I found a way to make a bijection between every number of N and R in a Cantor's diagonal argument sort of way.
Is this anything special? Would this mean R is countably infinite? Should I Arxiv it?

[Anonymous No.7181115]

>>7181112
You would have come up with a flawed proof.
If you really think you're right, show it to a professor to check it over, before you ruin your reputation.

[Anonymous No.7181126]

>>7181115
Thanks for the honest answer.

My proof is assuming that 0.10 != 0.1 , is this already bad?
If 0.1 == 0.10 it wouldnt be a 1:1 relation anymore but still work.

[Anonymous No.7181129]

>>7181112
Let's see it OP, I need a good laugh.

[Anonymous No.7181131]

>>7181126
Quit math.

[Anonymous No.7181133]

>>7181126
>My proof is assuming that 0.10 != 0.1

[Anonymous No.7181134]

>>7181126
>>7181129
Did not disappoint OP, did not disappoint.

[Anonymous No.7181139]

>>7181126
decimal representations of numbers are not numbers themselves. look at how R is constructed for the reason for this. decimal representation is only unique if you are careful with infinities like

0.99999...=1

and disallow adding zeroes.

[Anonymous No.7181140]

>>7181112
post it bro you probably broke math

[Anonymous No.7181141]

>>7181133
>>7181134
>>7181131
guys wait

Im enumerating binary numbers and for the part after the comma i would need to so something like 0.1, 0.10 0.100 etc. but when I do 1->0.1 , 2->0.10 , 3->0.100 one could argue that they are the same numbers so its not bijective anymore, - but seen as strings of symbols they are different, so from this point of view it would be a bijection.

[Anonymous No.7181144]

>>7181141
You're only listing numbers with finite decimal expansion. Real numbers have infinite decimal expansion.

[Anonymous No.7181146]

>>7181141
You know where your proof fails? You can't do that because the set of strings over a finite alphabet is countable. You're just playing around with the representation.

[Anonymous No.7181157]

>>7181146
>You can't do that because the set of strings over a finite alphabet is countable.
I dont understand what you mean.

[Anonymous No.7181502]

>>7181126
I think it's already bad, yes, but let's see the rest of your proof.

[Anonymous No.7181947]

>>7181502
ok here is the proof it will take a few posts

Step 1 is to write the binary expansion after the comma in a 2d plane like pic related.
With this we will get every number after the comma.

[Anonymous No.7181957]

>>7181947
Step 2 is to find a way to write every number in front of the comma,
For this we copy the first plane and shift it behind the first plane in 3d, then we change the 0 to 1
we do the same thing again (copy the new 2d plane behind itself) and replace the 1 with 10. like pic related. repeat this.

Now we have a way to display all real number in a 3d space.

[Anonymous No.7181962]

>>7181957
Now we take a 3d hilbert curve and go through every point in our 3d space.

The curve gives us a 1:1 relation between N and R.

q-q.e.d.

[Anonymous No.7181964]

>>7181947
here's the flaw: your columns are countable sets but your rows aren't. the " ... " in the rows are misleading: no countable sequence can contain what you mean by " ... ". Define them formally and you'll see.

[Anonymous No.7181968]

>>7181964
adding to this: see cantor's proof for the uncountability of reals. the proof's simple, and it implies yours is wrong quickly.

[Anonymous No.7181979]

>>7181964
The rows are just N written in binary, 1,10,11,100,101,..., n
The colums are just adding a zero in front of them.

[Anonymous No.7181985]

>>7181964
Not OP but I don't understand why. Can't a bike toon be made with the natural numbers after a countable number of elements that need to be tagged on at the beginning

[Anonymous No.7181989]

>>7181968
>see cantor's proof for the uncountability of reals
I looked at it and decided that its shit

[Anonymous No.7181990]

R is countable. It is easy to see that one can count a group of numbers. Simply start a counter and count each new number that you see. By this procedure, R is countable.

[Anonymous No.7181991]

>>7181985
*bijection
...
I guess my phone doesn't know bijection...

[Anonymous No.7181993]

>>7181989
it's not shit. it's valid. but i'm looking at what you said in >>7181979. I know this is wrong because [0,1) is uncountable too. but let me figure out why

[Anonymous No.7181998]

Suppose R is uncountable. Then the set R is not well-defined, since we can't figure out what is inside of it. Thus, R is countable.

[Anonymous No.7182001]

Im hoping this is making it more clear

[Anonymous No.7182004]

You couldn't do such a thing if the words bijection, N and R mean what they actually mean.

[Anonymous No.7182006]

>>7181993
there's no way for me to explain this without using cantor's argument again. so i'll try to put this intuitively

every row is uncountable, because your row isn't actually 0,1 and then N in binary, your row is actually "the set of all infinite strings in {0,1} that start with 0,1". this set is much bigger than N. N can easily biject the finite ones, and any arbitrarily large string. but N doesn't have an "infinitely long string" like R does: trascendental numbers, for one will not be in your representation. i think irrationals in general won't be there but i'll be careful and leave it at that.

[Anonymous No.7182009]

>>7182001
you don't need to make it any clearer, the mistake is definitely in step 1 because [0,1) is uncountable. from there it's easy to biject it to R, you don't need to go through those hoops you did.

[Anonymous No.7182010]

Ok the problem is that you can't just tag on that string of natural numbers represented in binary after having infinitely many 0's. It would take 'too long' for you to reach there

[Anonymous No.7182011]

>>7182010
he doesn't have inifinitely many 0s: he has an arbitrarily large number of 0s. and that's fine.

[Anonymous No.7182015]

>>7182006
>"the set of all infinite strings in {0,1} that start with 0,1". this set is much bigger than N
But in Cantors proof of the countability of Q he does pic related.
If we use the 2d plane I made and do what Cantor did then my projection must also be 1:1 with N.

[Anonymous No.7182021]

>>7182011
>inifinitely many
>arbitrarily large
Whats the difference

[Anonymous No.7182023]

>>7181947

see
>>7181144

You haven't even proven that the rationals are countable, much less the reals.

[Anonymous No.7182026]

>>7182011
Hmm I don't think so. Because how a bijection works is that a formula needs to be produced and not some hand wavy line drawn through the numbers. From what's been given, give me a formula which bisects [0,1) to N. It won't be possible for the exact reason you've given, it has arbitrarily many 0's at the front of the expression. I could be wrong but I don't think so

[Anonymous No.7182045]

op: your sequence does not contain the number 0.010101010101...

[Anonymous No.7182047]

>>7182045
Yes it does

[Anonymous No.7182048]

>>7182021
let's do a metaphor: you prove something on a set by induction. so you prove the base case, 1 element is ok. Then you prove that if n elements are ok, n+1 are. You have proved it for any number of elements in the set. Have you proved it for an infinite set? No! There are several counterexamples for this, but i hope you get the idea.

>>7182015
cantor proves that you can write any number in Q inside of the 2d grid. this is the first step. you haven't proved that you can write any real there: as I said, the trascendentals aren't in your grid, any infinitely long number isn't there.

>>7182023
his proof actually proves the rationals are countable. it's interesting. step 1 works for Q in [0,1)

>>7182026
arbitrarily large isn't the same as infinite. arbitrarily large is fine. the "rest" of the sequence, to fill it with inifinite numbers, is what the rows are about and where it fails

[Anonymous No.7182049]

>>7182045
of curse you can find it in the first row, it's equivalent to the number 10101010101.... in N just with a 0, written in front of it

[Anonymous No.7182051]

>>7182049
that "number" isn't in N

[Anonymous No.7182052]

>>7182048
let me correct something i said here.

>>7182023
the proof doesn't work for rationals either. sorry about that.

[Anonymous No.7182056]

>>7182047
i mean oscillating infinitely, your sequence contains the numbers
0.01
0.0101
0.010101
etc but not
0.0101010101...
(after the first 0, every even digit is 1, every odd digit is 0)

[Anonymous No.7182058]

>>7182048
>you haven't proved that you can write any real there
Its only a part of the proof m8 Im drawing the reals in step 3 with a 3d hilbert curve

>any infinitely long number isn't there.
Thats like saying N itself isnt infinitly long, or like "the number 10^x {x eof N, assuming N has inf many numbers} doesnt exist".

[Anonymous No.7182060]

>>7182058
the issue is a misunderstanding of infinities.

N is infinite, but no infinitely long number is in N. see >>7182056

[Anonymous No.7182068]

>>7182060
>N is infinite, but no infinitely long number is in N.
Ok so I guess an infinitly long number would be 1/3 = 0.3...
But 1/3 is just an operation.

f(x)0 = 3
f(x)n+1 = f(x)n*10+3

Is also just an operation that gives 3333....
Every recursive iteration will give a number in N but the operation itself isnt in N?

[Anonymous No.7182083]

Cantor's proof works because any rational can be reached in a finite number of steps- for instance, I can reach the number 1/9 in 37 steps using his diagonalization method. By contrast, 1/9 in base 2 is 0.000111000111..., and so your argument will NEVER reach it in ANY number of steps. You haven't even defined the rationals by your system, much less the reals.

[Anonymous No.7182089]

>>7182068
every infinite long number isn't a countable operation. there's a name for those numbers: algebraic. those are actually countable.

the numbers that can't be expressed as root of a polynomial are called trasendental

[Anonymous No.7182094]

>>7182089
actually trascendentals are also products of countable operations, i meant finite here.

[Anonymous No.7182136]

wait so are these comma notations indicating a termination after a bunch of zeroes?

[Anonymous No.7182138]

>>7182136
>wait so are these comma notations indicating a termination after a bunch of zeroes?
yes

[Anonymous No.7182142]

>>7182138
how would this represent the reals? wouldn't that only show values that have rational representations? that in itself is already countable using the squiggly line approach.

[Anonymous No.7182147]

>>7182142
>>7182142
also, how are the bijections even taking into account rationals or reals that aren't just repeating zeroes? this is all confusing.

[Anonymous No.7182154]

>>7182142
We can systematically build a decimal expansion for each element in R. (http://en.wikipedia.org/wiki/Construction_of_the_real_numbers#Stevin.27s_construction (at the bottom)) So the idea was that if we can plot each possible decimal representation on a line we also we get every possible real number.

[Anonymous No.7182156]

>>7181947
>>7181957
The numbers you list are rationals, and not even all rationals. With your method you will never reach a periodic number or an irrational number.

[Anonymous No.7182157]

>>7182154
but doesn't that only work for CONVERGING series?? i don't see how such a sequence could show, say, 2.235492509482..., but perhaps 2.22222222.., since you could simplify it as 2,30 as you've shown. i mean, is this trying to suggest that you could take finite expansions to form infinite ones? is this something that's done?

[Anonymous No.7182159]

>>7182157
disregard my statement about 2.2222... i am an idiot.

[Anonymous No.7182165]

>>7182157
>but doesn't that only work for CONVERGING series??
why should it ? I dont get it

[Anonymous No.7183733]

>>7181126
0.10 can mean <span class="math">0.10 \pm 0.005[/spoiler] and 0.1 can mean <span class="math">0.1 \pm 0.05[/spoiler]. But if you view the numbers in a purely mathematical sense, then 0.10 = 0.1