# 0 = 100... money is nothing

Discussion in 'BlackHat Lounge' started by Gogol, Jul 29, 2013.

1. ### GogolElite Member

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I originally posted it here
Code:
`http://www.blackhatworld.com/blackhat-seo/blackhat-lounge/591138-wheres-missing-dollar.html`
but still, I am making a new thread because I think I am hijacking sn0rt's thread which I do not want to do. So guys, this is another riddle for you... can you tell me where I am wrong?

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2. ### aspe_heatPower Member

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So it's something like :

Code:
```x = y.
Then x2 = xy.
Subtract the same thing from both sides:
x2 - y2 = xy - y2.
Dividing by (x-y), obtain
x + y = y.
Since x = y, we see that
2 y = y.
Thus 2 = 1, since we started with y nonzero.
Subtracting 1 from both sides,
1 = 0.```

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3. ### GogolElite Member

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Wow I think I am drunk now :O Looks perfectly valid to me lol

4. ### Goal Line TechnologySenior Member

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I didn't think that you could have a square root of an imaginary number

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nothing

6. ### back2formJr. VIPJr. VIP

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"Absolut 100" If you had this one, then you can prove any number = 100

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7. ### ZuuuuRegular Member

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haha I like your post back2form )))))))

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8. ### GogolElite Member

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You are close

but where exactly did I do the mistake?

Let everyone have a go at it and I will post the answer if no one is able to answer it within tomorrow

9. ### Junkfood00Elite Member

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Goddamit, how many times are we going to have to say this? You can't divide by zero, x-y=0.

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10. ### Junkfood00Elite Member

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Dude, you killed the minus right there which is required to form the imaginary number. -1 to +1.

You should've square rooted the square before multiplying (-1) by itself.

11. ### Olivier7t1Newbie

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Wrong wrong wrong. Was still fun knowing you spent your time trying to figure it out.

12. ### JFouldsPower Member

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i^2 = (âˆš-1)^2
(âˆš-1)^2 = (âˆš-1)*(âˆš-1)
(âˆš-1)*(âˆš-1) != (âˆš(-1)(-1))

It's been a while since I've done complex mathematics, but pretty sure that's your mistake, I'll leave the mystery in why...

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13. ### davids355Jr. VIPJr. VIPPremium Member

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The problem is, 0 doesn't = 100.

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14. ### GogolElite Member

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You have the correct answer man.

You can not make whole root when any one of the rooted numbers are imaginary.

For example
(âˆš2)*(âˆš2) = (âˆš(2 * 2))

but,
(âˆš-1)*(âˆš-1) != (âˆš(-1) (-1))

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Last edited: Jul 30, 2013
15. ### umerjutt00Jr. VIPJr. VIP

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Your first step is wrong. THe square root will be cancelled by the power of 2. So only -1 remains... Stop playing games man....

16. ### GogolElite Member

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Thanks for the big fonts but the first step is perfectly valid lol. Check JFloud's post, and my last reply before this.

17. ### GogolElite Member

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I wish it was. In that way, I owed you \$100 in paypal

Last edited: Jul 30, 2013
18. ### jazzcModeratorStaff MemberModeratorJr. VIP

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As JFoulds said, sqrt(x) * sqrt(y) does not necessarily equals sqrt(x*y) in the complex plane

The reason is hidden deep in the fact that when you deal with the complex plane using 1-dimensional functions (i.e. f(x)), you need to cut the complex plane into special pieces, do your stuff there and glue the results together. This means shit 's got discontinuities - which are to math what Matt Cutts is to seo.

For a result to be true, it has to be the same, no matter how you cut those pieces.When you want to find the principal of an sqrt and you 've got multiplication in there, the possible ways of cutting and gluing do not yield the same result, so the equality can't generally stand.

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19. ### umerjutt00Jr. VIPJr. VIP

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(âˆš-1)^2

power of square root is 1/2

= (-1)^1/2 x 2

= -1

= i^2

It all ends up like this......

(âˆš-1)^2 is not equal to âˆš-1 x âˆš-1.

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Last edited by a moderator: Jul 30, 2013

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