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Mathematics is shit

Name: Anonymous 2012-01-09 1:19

Just look at these ugly swatches of code:
http://www.cse.unt.edu/~idl99/Proceedings/ahn/img50.gif
http://www.cse.unt.edu/~idl99/Proceedings/ahn/node8.html

That is what you get, when programming immitates math. Even Java and PHP looks beautiful compared to that. Mathematics should be banned as a harmful and obfuscated teaching.

Name: Anonymous 2012-01-09 15:17

>>80
That's rich coming from someone who believes that the interval [0, 1] is a subset of the natural numbers. Go fuck yourself you fucking retard, you're wrong and you're a fucking moron. This is basic mathematics.

Name: Anonymous 2012-01-09 15:17

faggots everywhere, get a fucking room already and spread your AIDS amongst you

Name: kodak_gallery_programmer !!kCq+A64Losi56ze 2012-01-09 15:17

>>79
Listen you piece of shit retard, [0,1] isn't a subset of the natural numbers you fucking moron,

I have several math proofs done by some well known professors at UC Berkeley that say otherwise.

do you sincerely believe that 0.5 is a natural number?

I said the subset was the real numbers. In other words, I never implied that 0.5 is a natural number. Again, you're stupid. And again, you have no possible future as a computer programmer.

Name: kodak_gallery_programmer !!kCq+A64Losi56ze 2012-01-09 15:19

>>81
It satifies the definition of a subset. I have a lot of my math proofs from grad school at UC Berkeley to support this fact. Now shut up and go scrub another toilet.

Name: Anonymous 2012-01-09 15:21

>>83
You don't even know what a subset is you fucking piece of shit retard. For a set A to be a subset of a set B every element in A has to be an element of B, 0.5 is in [0,1] but 0.5 isn't a part of the natural numbers, so you're wrong you god damned fucking retard. You don't know what you're talking about and you're a god damned retard.

I never implied that 0.5 is a natural number.
You just did by stating that [0,1] is a subset of the natural numbers you fucking retard.

Again, you don't know what you're talking about, again you're a fucking retard.

Name: Anonymous 2012-01-09 15:21

I used to think kodak was somewhat intelligent, now I don't anymore.

Name: Anonymous 2012-01-09 15:22

Silly Kodak-san, the natural numbers are a subset of the real numbers, not the other way around, silly Kodak-san.

Name: Anonymous 2012-01-09 15:23

>>84
By the way you fucking moron you have yet to make a single correct statement in this thread beyond that the natural numbers are countable, congratulations, you're at the level of a challenged high school student you piece of shit.

Name: Anonymous 2012-01-09 15:25

>>85
You don't even know what a subset is you fucking piece of shit retard. For a set A to be a subset of a set B every element in A has to be an element of B

No. What happens if the set includes the empty set?

You just did by stating that [0,1] is a subset of the natural numbers you fucking retard.

Yes, and from this subset, I can construct set of real numbers such that each value in [0,1] maps to a real number.

Name: Anonymous 2012-01-09 15:26

>>88
Huh? You have no clue what you're talking about. Again, have you ever done a formal math proof in your entire life?

Name: Anonymous 2012-01-09 15:28

>>76
What happens if I have a set of real numbers in the interval [0,1].
The set of real numbers is not countable, your counterexample is invalid.

Name: Anonymous 2012-01-09 15:30

>>91
Yes, and this set of uncountable numbers is a subset of the set of natural numbers which is countable.

Name: Anonymous 2012-01-09 15:36

The set S defined by the interval [0,1] is:
- Finite, if ∀x∈S.x∈Z.
- Countably infinite, if ∀x∈S.x∈Q.
- Uncountably infinite, if ∀x∈S.x∈R.

Name: Anonymous 2012-01-09 15:38

>>69
It depends you idiot. What happens if I have some function called f defined over a set of natural numbers that doesn't satisfy either the onto on one-to-one condition that is necessary for countability?
For your statement to make even a shred of sense it needs to be a function defined over some other set to the natural numbers, not over some subset of the natural numbers. And even then you have proved nothing, you may always construct such a function from any set to another set, consider a set X now let f(x) = 0 for each x in X, this is a function which is neither bijective or surjective while X might be countable or uncountable.

If I'm generous I'm guessing what you really mean is that if there does not exist any function f that is an isomorphism from the set to the natural numbers then the set isn't countable.

>>71
Yeah the metric you're looking for is d(n, m) = if n != m then 1 else 0, the naturals are bounded with respect to this metric.

>>75
Is correct and the proof is even in the post.

>>83
You seem a bit confused here, [0, 1] is a real interval, it is defined as the set { x in the Reals : 0 <= x <= 1}, and that is not a subset of the natural numbers.
If I'm being generous I suppose what you're trying to state is that {0, 1} is a subset of the natural numbers, which might or might not be true, most definitions of the natural numbers don't contain 0 but some do, let's be generous and assume that 0 is a member of the natural numbers.

Name: Anonymous 2012-01-09 15:39

>>92
0 is natural, rational and real; 0.5 is not natural, it is rational and real; pi is not natural nor rational, but it is real. Thus, N is a subset of Q is a subset of R.

Seriously, go back to high school.

Name: Anonymous 2012-01-09 15:40

>>92
No for example 0.5 is in [0,1] but not in the natural numbers, so [0,1] is not a subset of the natural numbers.

Name: Anonymous 2012-01-09 15:43

Taken from http://en.wikipedia.org/wiki/Countable_set
Proposition: Any subset of a countable set is countable.
Proof: The restriction of an injective function to a subset of its domain is still injective.
There you go, it's simple and a single sentence long, I suggest that you revisit some basic high school mathematics before you run your mouth again you fucking retard.

Name: Anonymous 2012-01-09 15:45

>>95
No one said anything about 0.

Name: Anonymous 2012-01-09 15:47

>>96
Take the interval [3,4]:
∀x∈{3,4}.x∈N
∀x∈{3,...,3.001,...,3.9,3.99,...,4}.x∈Q
∀x∈{3,3.1,...,3.14,...,π≈3.1415926535897932384626433,...,3.9,3.99,...,4}.x∈R

Name: Anonymous 2012-01-09 15:47

Smart people get mathematics right, unintelligent people get mathematics wrong or think mathematics is scary.

In short: check 'em.

Name: Anonymous 2012-01-09 15:49

>>94
You're wrong.

Name: Anonymous 2012-01-09 15:50

>>99
What are you trying to prove? Again, [0,1] isn't a subset of the natural numbers, if it were then every element in [0,1] would also be in the natural numbers, 0.5 isn't a member of the natural numbers so [0,1] isn't a subset of the natural numbers. It's that fucking simple you god damned piece of shit retard, go back to special ed and try to learn something this time you fucking moron.

Name: Anonymous 2012-01-09 15:52

>>101
Please show me where you think I'm wrong and I'll be happy to correct you.

Name: Anonymous 2012-01-09 15:53

>>102
So, you think you can't take an interval of natural numbers? http://en.wikipedia.org/wiki/Interval_%28mathematics%29#Integer_intervals

Name: kodak_gallery_programmer !!kCq+A64Losi56ze 2012-01-09 15:56

>>94
You seem a bit confused here, [0, 1] is a real interval, it is defined as the set { x in the Reals : 0 <= x <= 1}, and that is not a subset of the natural numbers.

No, you're confused. Let's say I have the interval [0,1] defined on the set N. I can construct a function k* such N maps to a set real numbers, R, in [0,1].

Name: Anonymous 2012-01-09 15:56

>>104
If you see >>76 he clearly states that it's the real number interval, but let us be generous, from now on, let [0,1] be the set {0,1} and please, I'd very much like you to attempt to continue with the argument in >>76 with this new piece of information, I'm sure that the result will be very humorous.

Name: Anonymous 2012-01-09 15:59

Stop talking past each other already.

Natural numbers are defined rather simply by 0 and all finite applications of the successor function. Natural numbers are provably enumerable/countably finite.

Real numbers are defined by dedekind cuts, not always accessible in a 'constructive' way, which is why one would need the "Axiom of Choice"(when talking about set theory) if they want to quantify over reals. Real numbers are provably non-enumerable/uncountably finite. The reals as a set/structure do contain the naturals, although one needs to consider what axiomatic system, model and interpretation is being used when talking about something. Some concepts don't exist or mean different things in different interpretations, such as various things being argued in this thread.

It would do well for people to specify exactly which systems they are talking about, but civilized discussion has become hard, mostly due to a certain kodak or kodak-like person who insists on insulting everyone he responds to and forcing those replying to him to insult back. I'll take my leave and try not to reply any further, at least until the discussion has returned to more civilized levels.

Name: kodak_gallery_programmer !!kCq+A64Losi56ze 2012-01-09 15:59

>>106
I said the set of real numbers *in* the interval [0,1]. That implies there function that defines a mapping from N to R. Where N is the set of natural numbers and R is the set of real numbers.

Name: Anonymous 2012-01-09 16:01

>>105
Let's say I have the interval [0,1] defined on the set N
All right so [0,1] means {0,1} now, that is fine.
I can construct a function k* such N maps to a set real numbers, R, in [0,1]
So now there is a function which maps from N to R, and it only has values in [0,1]? ([0,1] I presume means {x : 0 <= x <= 1} now, correct me if I'm wrong).

And then what? Please continue your argument.

Name: kodak_gallery_programmer !!kCq+A64Losi56ze 2012-01-09 16:05

>>109
Yes!

Name: Anonymous 2012-01-09 16:06

>>106
>>76 is:
- claiming that the real interval [0,1] is a subset of the natural numbers.
- using that claim to prove that subsets of countable sets can be uncountable.

The fallacy here is that the real interval [0,1] is not a subset of the natural numbers, because 0.5 nor pi are natural numbers, but are contained in [0,1].

Let's use the integer interval [0,1] instead, which is the set {0,1}. It has 2 elements: 0, 1. It is countable. It is a subset of the natural numbers, because 0 and 1 are natural numbers.

We proved that kodak-kun is retarded. QED

Name: Anonymous 2012-01-09 16:07

>>110
Well mister Kodak, I am not impressed, both of these still hold true
the real interval [0,1] is not a subset of the natural numbers
any subset of a countable set is also countable
Go on, prove them wrong now.

Name: Anonymous 2012-01-09 16:09

I used to think Kodak had at least some entry level knowledge of mathematics and a basic level of intelligence, both were proved wrong today.

Name: Anonymous 2012-01-09 16:12

>>111
pi isn't contained in [0,1] but the point still stands as 0.5 is.

Name: Anonymous 2012-01-09 16:14

>>113
He's just another programmer without any mathematical ability and no mathematical talent, just another code monkey. He should stick to the stuff he actually knows which is programming.

Name: kodak_gallery_programmer !!kCq+A64Losi56ze 2012-01-09 16:15

>>112
any subset of a countable set is also countable

And what happens if I have two ore more subsets that map to the same set?

the real interval [0,1] is not a subset of the natural numbers

Let me find the proof.

Name: Anonymous 2012-01-09 16:16

>>114
Oops, sorry, I was still thinking of [3,4].

Name: Anonymous 2012-01-09 16:19

>>116
And what happens if I have two ore more subsets that map to the same set?
Not anything that makes that statement false.

Let me find the proof.
No need to, 0.5 is in the interval [0,1] while 0.5 isn't in the natural numbers, therefore the interval [0,1] is not a subset of the natural numbers.

Name: >>118 2012-01-09 16:21

>>116
Actually on second thought, do find the proof, I think it will be quite humorous.

Name: kodak_gallery_programmer !!kCq+A64Losi56ze 2012-01-09 16:24

>>116
Let ri be in x for all i = 1, 2, 3,...

r1 = 0.d11d12d13..
r2 = 0.d21d22d23...
r3 = 0.d31d32d33...

I can construct a k*

0 < k* < 1

such that k* = d1d2d3...di... for all 1, 2, 3

This function, which is defined over a real interval, is a subset of the natural numbers.

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