Nikita: the latest news

Nikita expands his internet presence by attacking the Russian functional programming community: http://ru-declarative.livejournal.com/108182.html

Cute!

>>189
LLLLLEEEEEEEEEEEEEEEEEEEEEEEEEEELLLLLLLLLLLLLLLLLLLLLLLLLL
Dude, it's one person.

>>190
Yes, we've become popular enough to attract the single insufferable faggot that is yourself. Congrats man.

>>184

That comment makes me think I have the wrong idea of what we are talking about here, but here is my proof of what I gathered from >>180-182

Suppose that f is a function on the reals, satisfying the following:

P1: for all a,b in R, f(a + b) = f(a) + f(b)
P2: for all a,c in R, f(ac) = f(a)f(c)

Part 1: f(0) = 0.

let a,b = 0 from P1.
f(0 + 0) = f(0) + f(0)
f(0) = f(0) + f(0)
f(0) - f(0) = (f(0) + f(0)) - f(0)
0 = f(0) + 0
0 = f(0) XD

Part 2: f(1) = 1, or f(1) = 0

f(1) might be zero. In the case that f(1) is not zero, we have the following:

let a,c = 1 from P2.
f(1*1) = f(1)f(1)
f(1) = f(1)f(1)
f(1)(1/f(1)) = (f(1)f(1))(1/f(1))
1 = f(1)1
1 = f(1) XD

Part 3: f(1/x) = 1/f(x)
let x be a real, with x not 0

1 = f(1) = f(x(1/x)) = f(x)f(1/x)
because f(x)f(1/x) = 1, f(x) is the multiplicative inverse of f(1/x).
Hence, 1/f(x) = f(1/x) XD

Part 4: f(-x) = -f(x)

0 = f(0) = f(x + -x) = f(x) + f(-x)
Because 0 = f(x) + f(-x), f(x) is the additive inverse of f(-x).
Hence, f(-x) = -f(x). XD

Part 5: f is the zero function or the identity.
to be continued.

Sorry, I can't do any more right now, not that any of what I've written so far is hard or original. The fries might be cold.

Nikita, why are you so interested in US internal politics?

I know what QED stands for, I don't know about XD though.

>>194
It's the greek analogous of LEL.

>>193
I'm interested mostly in Jewish infiltration of US.

>>171
implying kikes don't oppose the Reals

>>197
Please refrain from using ``imageboard memes'' such as ``implying'' in the textboards.

egin

please refrain from spamming.

>>199,200
Please refrain from having such awesome, yet unchecked dubs.

>>198
Please refrain from using ``/prog'' when you are a JEW.

Part 5: If n is a natural number, then we have that either ∀n f(n) = 0, or we have that ∀n f(n) = n.

It was shown in part 2 that if f(1) is not zero, then f(1) = 1.

Suppose that f(1) = 0, and n is a natural number. Then we have that:

f(n)
= f(1 + 1 + ... (n times) ... + 1)
= f(1) + f(1) + ... (n times) ... + f(1)
= 0 + 0 + ... (n times) ... + 0
= 0 XD

Now suppose it is the case that f(1) = 1. Then:

f(n)
= f(1 + 1 + ... (n times) ... + 1)
= f(1) + f(1) + ... (n times) ... + f(1)
= 1 + 1 + ... (n times) ... + 1
= n XDD

Part 6: If n is an integer, then either ∀n f(n) = 0, or ∀n f(n) = n.

In part 4 it was shown that f(-x) = -f(x). If f is zero on the naturals, we have, for all negative integers m:

f(m) = -f(-m) = -0 = 0 XD

If f is id on the naturals, we have for all such m:

f(m) = -f(-m) = -(-m) = m XDD

>>203
Do you have Tourette's? What is XD and how is it relevant to the proof?

>>203
stop it with the XD you fucking retard

>>204
OMG he doesn't get it!!!!! XD

>>206
Take your pills, retard-kun.

>>207
Ahahahaha-kun!  Touhou-kun, yoohoo chum!  Ahahahahaha, XD!  XOXOXO XD!!!

>>208
You can't say ``Touhou-kun'', retard-kun.

You can't stop yourself from responding, can you kun?  Kunny kun kun-kun?

>>210
Not really, retard-kun.

Your calm retard kunning is shaking my faith in my shitposting day.
I am going to sit here for a minute and think hard about my life.

K, I'm back!  Why won't you love me, retardo-san?

>>212
Kun is Jewish.
http://www.surnamedb.com/Surname/Kun

>>213
Oh wow!

Part 3 assumed that f(1) = 1, and ignored the possibility of f(1) = 0. A mistake! A mistake! Terrible! Terrible! Ignore it in the case that f(1) = 0 ! Ignore it! Ignore it!

Part 6: for all rationals r, we either have that ∀r f(r) = 0 or we have that ∀r f(r) = r.

Suppose we are in the case that f(1) = 0. Then it was shown that for all integers n, f(n) = 0. So for a rational, r = p/q, we have that:

f(r)
= f(p/q)
= f(p*(1/q))
= f(p)*f(1/q)
= 0*f(1/q)
= 0 XD

Now suppose that we are in the case that f(1) = 1. Then it was shown that for all integers n, f(n) = n. Also part 3 is valid. We have that:

f(r)
= f(p/q)
= f(p*(1/q))
= f(p)f(1/q)
= f(p)*(1/f(q))
= p*(1/q)
= p/q
= r XDD

Ok, I'm just going to assume that f(1) = 1.

because if f(1) = 0, then we have that, for all reals, r.

f(r)
= f(1*r)
= f(1)*f(r)
= 0*f(r)
= 0

so f has to be the zero function on all of R if f(1) = 0. So now that case is tied up. I'm going to assume that f(1) = 1 now.

>>216
Is f a linear function ?

>>217
I'm trying to show that f is the identify function in one case, but my only assumptions are given in >>192 as P1 and P2.

Sorry >>190-san. While your bot was most disruptive, it was also creative in its own right.

Try proving it isn't the identity function, i.e. assume f(n) ≠ n, then continue the proof by induction as you were.
No idea how it would go, but it's worth a try?

I'm just wondering: have you guys ever craved cock so badly that you found yourself running around outside, howling at the moon for it? Literally ROARING at the top of your lungs, wanting nothing less than a dick's head churning against your glottal stop?

Tell me I'm not alone.

>>219
dont remember euclid needing your identity function...

>>221
don't remember patrick bateman checking my TRIPS

>>222
Yeah, well, he did. And he was amazed!

Pretty interesting, >>223

>>219

Well, so far, we have that:

1. If f(1) = 0, then f is the zero function.
2. If f(1) is not 0, then f is the identity function on the rationals.

If f has to be continuous, then (2) implies that f is the identity on the reals as well, but without continuity I can't connect the rationals to non-rationals so easily.

Part 7: if f(1) is not zero, then f is id on all real nth roots of all rationals.

Let n be a power and r a non-negative rational.
Let b be the unique real nth root of r, b = root(r,n)

r
= f(r)
= f(b*b* ... (n times) ... *b)
= f(b)*f(b)*f(b)* ... (n times) ... *f(b)

Since f(b) multiplied by itself n times is equal to r, we have that f(b) is an nth root of r. Because real roots are unique, this implies that f(b) = b. XD

>>225
this implies that f(b) = b. XD
That's not even funny, fagshit.

dont remember euclid needing your identity function.

wut?

NEETkita Le Hikikomori

Well, it can be shown by induction of each of the parts that any number that can be formed by a finite number of additions, subtractions, multiplications, divisions, and square roots, applying the said operations in any order, then f fixes this number. So that covers quite a few numbers, but I really doubt that every real number can be expressed in this way. I could prove that by showing the set of all finite length algebraic expressions using sums, differences, multiplications, quotients, and roots, to be countable. I guess that is obvious, since the set of such algebraic expressions would correspond to a subset of the language of all strings containing '( ) + * - / ^ 1' which is a countable set.

So no, I don't have the reals yet. I think continuity is needed. I'll try to produce a counter example.

>>230
Prove that "induction" is a provable way prove anything.

>>231
Prove that russians aren't subhuman scum.

>>231
Give me an expression using n applications of the said operations, and I'll write you a proof that uses n lines, using the parts already shown at each step.

but you are right. I can't prove that induction works unless you are willing to assume something else that is just about as controversial.