>>2
Congrats and wow, I didn't think any of you retards would get it. You are all monotonous drones who don't truly understand the nature of the equation you utilise. Nice thinking.
Your reward is 1 free 4chan. Enjoy!
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Anonymous2006-01-07 12:56
actually you got it wrong
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Anonymous2006-01-07 12:56
bullshit! you can't just derive like that!
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Anonymous2006-01-07 13:24 (sage)
Yes you can, just fill it in:
(1/ln(a))^(1/(1/ln(a)))=a
(1/ln(a))^(ln(a))=a
ln(a)^(-ln(a))=a
Which is obviously true.
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Anonymous2006-01-07 14:51
taking a=1 instantly disproves the x=1/log(a) hypothesis.
even if I do a proof you won't understand it because you haven't mastered ZF, FOL or elementary analysis.
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Anonymous2006-01-24 15:45
>>37
If you can't make a proof after spending 3 years studying maths in college or whatever it means you are stupid and inferior.
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Anonymous2006-01-25 8:50
This equation doesn't have a definite solution without YOF or RV
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Anonymous2006-01-25 12:34
>>38
I said that you're too dumb to understand it, not that I can't do it.
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Anonymous2006-01-25 13:25
>>40
You are even too stupid to realise the burden of proof is on you.
I didn't do maths in college, I have an A in A-level human biology, a B in chemistry and a B in English literature. I don't know the basic of maths, but this isn't because I'm stupid, but because I'm not omnipotent and haven't been taught maths past what I need to know for A-level chemistry. Now provide a proof.
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Anonymous2006-01-25 13:26
>>40
I would also like to metnion you are inferior for not realising this.
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Anonymous2006-01-25 17:13
>>41 >>42
Relax buddy. All that stress ain't good for yer heart.
Proof by contradiction:
blah blah blah
QED.
Since you don't know enough maths, then it's all the same to you anyway.
Assume f(x)=x^(1/x). We are looking for it's inverse g(x)=f^-1(x) that can be expressed with a finite combination of the field operations (addition, multiplication, multiplicative inverse, addative inverse). We restrict ourselves to the rational numbers; since the rational numbers are closed under these operations it means that for each value of g(x), the argument of f(x) should be rational. However, taking g(2) we get 2=x^(1/x), for which no rational solution exists. We have a contradiction, which completes the proof.
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Anonymous2006-01-27 2:00 (sage)
>>46
*clap clap clap* good sir, you have sucessfully ended this thread.
The root of 2 is irrational (2^(1/2)). It's usually the first proof you learn when using contradiction and is accepted without having to state the proof over...unless of course you're a noob.
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Anonymous2006-02-04 16:45
>>46
Wrong. Chaos theory implies otherwise. Let's take an ordinary equation.
Y = X^2
According to chaos theory you can find X in terms of Y. According to your theory this cannot be done.
X = Y^0.5
Now take a look at this equation, if you can't understand it you are an idiot and should gb2 high school as this thread is getting too complex for you.
>>54
The thing is they're not trolling, they just like to sit and jerk off at how smrt they are, when not everyone goes to college and does a course in Riemann geometry. It is sort of like taking 2 people, whisperring to one that sin(118^7)*65 in radians is 0.58407, then asking them the question and then going AHAHAHAH U R DUMB LOL when the one who wasn't told the answer says he can't do such an equation using mental arithmetic.
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Anonymous2006-02-07 14:51
sin((118^7)*65)
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zeppy2006-02-07 16:04 (sage)
z=xy
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Anonymous2006-02-08 2:18
>>55 you must be either in grade-school or stuck in the 18th century, because mental arithmetic is pretty fucking useless at higher maths these days
no it wasn't. you just couldn't think of a better example because you're fucking clueless.
if you think being a math nerd is somehow wrong then why are you even here? go back to your cave, turn on the tv and glorify all the beautiful celebrities or whatever it is you do