Trolling in [field of study].

[Redundant]

@Jordus

The dividing by x is perfectly valid, as x is nonzero. It's not standard for dealing with quadratics, but this does tell you that all solutions are of the form x = -1-1/x.

[Jordus]

@Jordus

The dividing by x is perfectly valid, as x is nonzero. It's not standard for dealing with quadratics, but this does tell you that all solutions are of the form x = -1-1/x.

This is why I never claimed to be a mathmatician

[flOvermind]

Hi all - it's my first post; sadly I don't know that much about Go (although I hope to improve at that, and might eventually post about some Go-related stuff), but as I've spent the past 10 years studying Maths (and now have a PhD - not that that is needed to debunk this particular argument), it's fairly simple to say what the flaw is:

The original quadratic is fine, and has 2 roots, which are both of the complex cube roots of 1 (e^2pi*i/3 and e^-2pi*i/3). Both are also roots of x^3=1, of course, but that equation has a third root, 1. You can get from the original equation x^2+x+1=0 to x^3=1 just by multiplying it by x-1 on both sides. The fatal flaw is in going from x^3=1 to x=1 - this only works if you assume x has to be real, which isn't the case.

You can't just multiply with (x-1) on both sides to get x^3=1, because then x=1 is a solution, making the previous step a multiplication by zero. That introduces more solutions.

As for the last step: I don't see a problem there. The only flaw it introduces is that it ignores the other solutions. If you interpret the original question as "find some x such that ..." instead of "find all x such that ...", that's a perfectly valid step.
Just assume the last step would have been from x^3=1 to x=(-1/2 + i sqrt(3)/2). That would be a correct solution to the original equation. Not all solutions, but a correct solution.
The actual error must be at an earlier point, where the equation changes from two solutions to three solutions.

EDIT:
D'oh. It's so easy

The error must obviously be in the transition from the fourth line to the fifth. That's where the third solution (x=1) appears.

In that line, *one* x is substituted by another formula that contains another x. That in itself is not a problem. But now, you suddenly have two different xs in the equation, the one from the substitution and one "leftover" x. In the remaining steps, these different xs are mixed up. You have to substitute *both* the x and the x^2, then it all works out correctly

[Stable]

What is trolling?


I don't have any good pictures myself, but I like this site a lot: http://fakescience.tumblr.com/

[Sverre]

The error must obviously be in the transition from the fourth line to the fifth. That's where the third solution (x=1) appears.

In that line, *one* x is substituted by another formula that contains another x. That in itself is not a problem. But now, you suddenly have two different xs in the equation, the one from the substitution and one "leftover" x. In the remaining steps, these different xs are mixed up. You have to substitute *both* the x and the x^2, then it all works out correctly

The preceding lines imply line 5, and line 5 is true. The error is that one-way implication is not equivalence.

Can you give me a precise definition of when substitution is safe and when it is unsafe? You claim that partial substitution is risky, can you prove that "full" substitution never introduces extraneous solutions?

[Gresil]

Very well, I will do my best.


It is essential for the good and healthy functioning of an internet forum to have a steady and balanced influx of 4chan memes. Anyone who thinks otherwise is an idiot.

[Monadology]

As it is not possible to demonstrate that any choice or action is not predetermined, it must therefore be true.

This is just trolling anyone who understands how basic reasoning works, not just philosophy.

[Chew Terr]

Good thing nobody's started calling you on this one, or else you would be feeling the heat, no?

...I think I'm funny, sometimes.

[topazg]

As it is not possible to demonstrate that any choice or action is not predetermined, it must therefore be true.

This is just trolling anyone who understands how basic reasoning works, not just philosophy.

I know, I'm so disappointed no-one rose to this

(Even if just to nitpick the subject area )

Good thing nobody's started calling you on this one, or else you would be feeling the heat, no?

...I think I'm funny, sometimes.

*Groan*

[hyperpape]

Continuing on Araban's example

Thanks guys, I didn't see that there's no problem until you actually take a step (x^3 = 1 => x = 1) that depends on the solution being real.

[Solomon]

As it is not possible to demonstrate that any choice or action is not predetermined, it must therefore be true.

This is just trolling anyone who understands how basic reasoning works, not just philosophy.

I know, I'm so disappointed no-one rose to this

(Even if just to nitpick the subject area )

Philosophy is a minefield; I could end up trolling even if my intention wasn't to.

[emeraldemon]

woo equations:

just to make things concrete:
x^2+x+1 = 0 (eqn1)
implies
x = -1/2 + i*sqrt(3)/2 OR
x = -1/2 - i*sqrt(3)/2

Meanwhile
x^3 = 1 (eqn2)
implies
x = -1/2 + i*sqrt(3)/2 OR
x = -1/2 - i*sqrt(3)/2 OR
x = 1

Once you know this, it should be easy to see that eqn1 implies eqn2: any solution of eqn1 is also a solution of eqn2. But There is a solution of eqn2 that isn't a solution of eqn1: x=1. The mistake (which several have already pointed out) is assuming that all solutions of eqn2 are solutions of eqn1, and inserting x=1 into eqn1 to get the wrong answer.

The way to avoid falling into these traps: when solving for x, don't multiply both sides by x. It adds additional solutions you probably didn't intend! Similarly dividing by x can remove solutions you wanted.

Most of y'all have probably seen this one, but it's one of my favorites:
Three guests decide to stay the night at a lodge whose rate they are initially told is $30 per night. However, after the guests have each paid $10 and gone to their room, the proprietor discovers that the correct rate should actually be $25. As a result, he gives the bellboy the $5 that was overpaid, together with instructions to return it to the guests. Upon consideration of the fact that $5 will be problematic to split three ways, the bellboy decides to pocket $2 and return $1 each, or a total of $3, to the guests. Upon doing so, the guests have now each paid a total of $9 for the room, for a total of $27, and the bellboy has retained $2. So where has the remaining $1 from the initial $30 paid by the guests gone?!

[palapiku]

Go Seigen: The go board has 6 sides

$$B
$$ ------------
$$ | . . 3 . . |
$$ | . . . . . |
$$ | 4 . . . 5 |
$$ | . . . . . |
$$ | . . 6 . . |
$$ ------------

Click Here To Show Diagram Code

[go]$$B
$$ ------------
$$ | . . 3 . . |
$$ | . . . . . |
$$ | 4 . . . 5 |
$$ | . . . . . |
$$ | . . 6 . . |
$$ ------------[/go]

[daniel_the_smith]

Most of y'all have probably seen this one, but it's one of my favorites:
Three guests decide to stay the night at a lodge whose rate they are initially told is $30 per night. However, after the guests have each paid $10 and gone to their room, the proprietor discovers that the correct rate should actually be $25. As a result, he gives the bellboy the $5 that was overpaid, together with instructions to return it to the guests. Upon consideration of the fact that $5 will be problematic to split three ways, the bellboy decides to pocket $2 and return $1 each, or a total of $3, to the guests. Upon doing so, the guests have now each paid a total of $9 for the room, for a total of $27, and the bellboy has retained $2. So where has the remaining $1 from the initial $30 paid by the guests gone?!

27 - 2 = 25 ... don't add! (yes, I've seen it before)

[Liisa]

Hegel trolled astronomy by giving an a priori proof that there were seven planets. And I just trolled philosophy by mentioning it.

I do not know how Hegel proved it, but there really is seven planets. Mercury, Venus, Mars, Jupiter, Saturn, Uranus and seventh Neptune. Earth-Moon system is a double planet, so it should not be counted as a planet in singular.