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1 = 2 
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We all know that 0 => 1.


Tue Jan 13, 2009 8:22 am
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megamanzer0 wrote:
h4tred wrote:
Deathlike2 wrote:
funkyass wrote:
Franky wrote:
I do not suck at mathematics and logic.


this entire thread is proof to the contrary.
am I doin it rite?


humility is more impressive than lame excuses Franky.

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Tue Jan 13, 2009 9:29 am
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Substituting one x with one answer while the other one is still expecting two is a bad, bad idea :P.
Grinvader, I hope that's the correct way to express the answer, I did the steps with x² - x = 0 and it's far more obvious to see the error.

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Tue Jan 13, 2009 10:17 am
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fun^10 x int^40 = Ir2

the end:

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Tue Jan 13, 2009 7:40 pm
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MisterJones wrote:
grinvader wrote:
Now we set x != 0
x=-1-1/x


This step doesn't look right.

Why ? I just divide left and right by x, which is valid since it's not 0.

juliobbv covered the main issue.
x²+x+1 = 0 has 2 roots, none of them being 1. [ they are (-1±i*sqrt(3))/2, for those who suck ]
Since the steps are legal, these roots are also answers to x=-1-1/x.

Subtituting partially and forgetting where we come from is the fault. To be correct, you should now have a system of equations with:
x²-1/x=0 AND x=-1-1/x

The new equation, alone, is not equivalent to the old one, and introduces the third root [1], but still keeps the previous roots. So saying x³=1 <=> x=1 is also flawed in itself.

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Tue Jan 13, 2009 9:09 pm
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grinvader wrote:
I found my favourite one. It is much more subtle. No dividing by zero bullshit. No sir.
It's actually interesting to find where it fucks up. :)

x²+x+1 = 0
x² = -x-1

Now we set x != 0
x=-1-1/x

Substitute that in the first equation...
x² - 1 - 1/x +1 = 0
x² - 1/x = 0
x² = 1/x
x³ = 1
x = 1 != 0, ok...

Substitute that in the first equation again...
1² + 1 + 1 = 0
3 = 0

Have fun suckers


Well, I was tired yesterday night when I tried to find the error (hence the stupid mistakes I made).
I looked at it again for about 30 seconds and it's obvious now:
"x²+x+1 = 0" (your first equation) is false for any value of X, thus invalidating all of your other equations aswell.

Even finding the roots (you FUCKER, grinvader, you made me use imaginary numbers) did not help. The root are...
x = -1 + (((10j) ^ 0.5) / 2) and
x = -1 - (((10j) ^ 0.5) / 2)
I believe. Am I correct?

EDIT:
!! Yes, that's it. You are using imaginary numbers. That's the error.


Tue Jan 13, 2009 11:45 pm
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Franky wrote:
grinvader wrote:
I found my favourite one. It is much more subtle. No dividing by zero bullshit. No sir.
It's actually interesting to find where it fucks up. :)

x²+x+1 = 0
x² = -x-1

Now we set x != 0
x=-1-1/x

Substitute that in the first equation...
x² - 1 - 1/x +1 = 0
x² - 1/x = 0
x² = 1/x
x³ = 1
x = 1 != 0, ok...

Substitute that in the first equation again...
1² + 1 + 1 = 0
3 = 0

Have fun suckers


Well, I was tired yesterday night when I tried to find the error (hence the stupid mistakes I made).
I looked at it again for about 30 seconds and it's obvious now:
"x²+x+1 = 0" (your first equation) is false for any value of X, thus invalidating all of your other equations aswell.

Even finding the roots (you FUCKER, grinvader, you made me use imaginary numbers) did not help. The root are...
x = -1 + (((10j) ^ 0.5) / 2) and
x = -1 - (((10j) ^ 0.5) / 2)
I believe. Am I correct?

EDIT:
!! Yes, that's it. You are using imaginary numbers. That's the error.


Firstly, you posted your "solution" nearly 3 hours after he posted. Even longer if you count that other guy's post.

Secondly, you pretty much missed the point of grinvader's solution.

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Wed Jan 14, 2009 12:57 am
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Imaginary numbers are both fun and useful. I use them daily in my work.

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Wed Jan 14, 2009 5:08 am
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The roots are actually complex, Franky.


Wed Jan 14, 2009 6:02 am
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DOLLS (J) [!] wrote:
The roots are actually complex, Franky.

Hey, complex numbers, immaginary numbers. Well, maybe I did get the terminology slightly wrong, but you know what I meant.


Wed Jan 14, 2009 11:46 am
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?SYNTAX ERROR
READY.
Image

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Last edited by creaothceann on Wed Jan 14, 2009 1:49 pm, edited 2 times in total.



Wed Jan 14, 2009 1:24 pm
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Franky wrote:
I looked at it again for about 30 seconds and it's obvious now:
"x²+x+1 = 0" (your first equation) is false for any value of X, thus invalidating all of your other equations aswell.

Wrong. As any other quadratic polynomial, it has 2 roots.

Quote:
Even finding the roots (you FUCKER, grinvader, you made me use imaginary numbers) did not help. The root are...
x = -1 + (((10j) ^ 0.5) / 2) and
x = -1 - (((10j) ^ 0.5) / 2)
I believe. Am I correct?

Wrong again. I already wrote what they are in my previous post. You could use some work on that stuff.

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!! Yes, that's it. You are using imaginary numbers. That's the error.

Nope. That's irrelevant.

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Wed Jan 14, 2009 11:47 pm
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Franky wrote:
DOLLS (J) [!] wrote:
The roots are actually complex, Franky.

Hey, complex numbers, immaginary numbers. Well, maybe I did get the terminology slightly wrong, but you know what I meant.

It's not just the terminonolgy, imaginary numbers have the form ai, complex numbers can be of the form ai + b, complex numbers have an imaginary as well as a real part, hence, both real and imaginary are a subset of complex numbers, it's definitely not a syntactical issue.

You lack quite a bit of humility, dude.


Wed Jan 14, 2009 11:50 pm
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DOLLS (J) [!] wrote:
Franky wrote:
DOLLS (J) [!] wrote:
The roots are actually complex, Franky.

Hey, complex numbers, immaginary numbers. Well, maybe I did get the terminology slightly wrong, but you know what I meant.

It's not just the terminonolgy, imaginary numbers have the form ai, complex numbers can be of the form ai + b, complex numbers have an imaginary as well as a real part, hence, both real and imaginary are a subset of complex numbers, it's definitely not a syntactical issue.

You lack quite a bit of humility, dude.


There's nothing bad about admitting wrong... but it's plenty fun to poke at it if you believe otherwise.

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Thu Jan 15, 2009 6:28 am
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Sun Feb 01, 2009 4:37 am
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Sun Feb 01, 2009 11:55 am
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selmo2000 wrote:
THE MATRIX HAS YOU

Yeah, gonna skip that one.

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Sun Feb 01, 2009 10:05 pm
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