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sleepsleep

Joined: 05 Oct 2006
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sleepsleep
emm, i would like to use my first helping tools. i need some hint.
is the question you posted is err on the fundamental logic or err on the way/equation that was presented/written up there?
24 Jan 2009, 19:22
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17273
revolution
sleepsleep wrote:
emm, i would like to use my first helping tools. i need some hint.
is the question you posted is err on the fundamental logic or err on the way/equation that was presented/written up there?
The proof is written wrongly (intentionally of course). You must find a way to correct it so that mathematically it is either correct, or at the least, comes to a proper conclusion without some weird non-sensible result.
24 Jan 2009, 19:28
Borsuc

Joined: 29 Dec 2005
Posts: 2466
Location: Bucharest, Romania
Borsuc
Why non-sensible? pi has nothing to do with it. You can put there y and you arrive at the conclusion that x=y if you take the positive root (pi just a random variable here...)

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Previously known as The_Grey_Beast
24 Jan 2009, 20:52
sleepsleep

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sleepsleep
well, borsuc.
if 2x = y + 3 is ==> y = 2x -3
then how come y = 3 in the end.
24 Jan 2009, 21:11
Borsuc

Joined: 29 Dec 2005
Posts: 2466
Location: Bucharest, Romania
Borsuc
2x = y + 3

x = 3
y = 3

==> 2*3 = 3 + 3

6 = 6

x=y (y, no matter what it is! there's nothing special here about 'pi')
25 Jan 2009, 01:09
sleepsleep

Joined: 05 Oct 2006
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sleepsleep
if you change x=4 and y = 5
so 2(4) = 5+3

it will count correct until this line.
3 - x = y - x
25 Jan 2009, 01:22
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17273
revolution
Borsuc wrote:
(pi just a random variable here...)
Okay, so you discovered my slight distraction about making it pi. Yes, the "pi" variable is just a distraction to attempt to throw the unwary off the scent for a short while. Hehe, seems to have worked as intended. But, of course, that is not the answer. I'm still waiting ...
25 Jan 2009, 08:31
nop

Joined: 01 Sep 2008
Posts: 165
Location: right here left there
nop
Code:
```If          x = (a + n) / 2       ;x is the average of 2 constants a, n
then       2x = a + n
and 2x(a - n) = a^2 - n^2     ;multiply by the difference of a and n
so a^2 - 2x(a - n) - n^2 = 0
=> a = n                   ;is only valid if a = n i.e. a - n = 0

e.g.          x = (p + 3) / 2
.....
.....
=> p = 3 = x

x = (q + 4) / 2
.....
.....
=> q = 4 = x    ```
25 Jan 2009, 09:57
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17273
revolution
nop: Can you formulate that into an answer? I'm not 100% clear about what you imply.
25 Jan 2009, 10:03
nop

Joined: 01 Sep 2008
Posts: 165
Location: right here left there
nop
revolution wrote:
nop: Can you formulate that into an answer? I'm not 100% clear about what you imply.
that is my answer, just gimme the star dammit i need that star its all i've ever wanted pleeze just one little star <beg> <beg> <sob> <sob>
25 Jan 2009, 10:11
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17273
revolution
nop wrote:
that is my answer, just gimme the star dammit i need that star its all i've ever wanted pleeze just one little star <beg> <beg> <sob> <sob>
Haha. LOL.

I would be very happy to give you the star, but you must work hard for it (they are expensive and thus not given freely). Your "answer" above is not correct, sorry no star just yet!
25 Jan 2009, 10:14
nop

Joined: 01 Sep 2008
Posts: 165
Location: right here left there
nop
geez rev your too tough i worked relly hard finding that stuff on the net
25 Jan 2009, 10:44
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17273
revolution
nop wrote:
geez rev your too tough i worked relly hard finding that stuff on the net
Hehe, these days it seems there is no need to think anymore, what with the 'net available it has all the answers to the questions anyone could ever ask.
25 Jan 2009, 10:55
sleepsleep

Joined: 05 Oct 2006
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sleepsleep
emm..
i think this line weird.
9 - 6x = pi^2 - 2pix
9 - 6x + x^2 = pi^2 - 2pix + x^2

it must be
(9 - 6x) + x^2 = (pi^2 - 2pix) + x^2

so, this
(3 - x)^2 = (pi - x)^2 is invalid. because (9 - 6x) and (pi^2 - 2pix) must be solve first before integrate with x^2.

well, if it is still incorrect, i wait for ur answer then.
i like maths, but i am not doing so well with it.
25 Jan 2009, 12:17
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17273
revolution
sleepsleep: Keep trying.
25 Jan 2009, 12:21
Borsuc

Joined: 29 Dec 2005
Posts: 2466
Location: Bucharest, Romania
Borsuc
revolution wrote:
Hehe, these days it seems there is no need to think anymore, what with the 'net available it has all the answers to the questions anyone could ever ask.
Heh that's until you NEED to make up your own algorithms to fit your purpose (as a guy who programmed 3D software, I do know how math should be 'visualized', not by learning formulas ).

Is it because of pi^2, that taking the square root means the absolute value of it? (not original number)

I mean, when 'expanding' (pi + 3)(pi - 3) -- it only works the other way around? (otherwise it is absolute value)

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Previously known as The_Grey_Beast
25 Jan 2009, 16:22
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17273
revolution
The attempts so far seem to be sort of on the right track, but so far no one has yet been able to bring all the ideas together into a full coherent answer.

There are bits of an answer here and parts of an answer there etc.
25 Jan 2009, 16:41
sleepsleep

Joined: 05 Oct 2006
Posts: 8897
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sleepsleep
15 Feb 2009, 04:15
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17273
revolution
sleepsleep wrote:
Ah, such impatience. There are currently three open puzzles not solved in this thread. To which are you referring?

http://board.flatassembler.net/topic.php?p=83742#83742 (half solved)
http://board.flatassembler.net/topic.php?p=88366#88366 (half solved)
15 Feb 2009, 05:03
windwakr

Joined: 30 Jun 2004
Posts: 827
Location: Michigan, USA
windwakr
Ok, I'm probably wrong and looking stupid here but.....

If Pi does indeed equal 3, then on line 4 you would be dividing by zero to remove the (pi-3), because (3-3)=0. Thus making the formula FALSE, right?
Code:
```x = (pi + 3) / 2
x=(3+3)/2=3

2x = pi + 3
6=6

2x(pi - 3) = (pi + 3)(pi - 3)
0=6(0)

2pix - 6x = pi^2 - 9
YOU DIVIDED BY ZERO TO REMOVE (pi-3), the rest is just filler to make it look real

9 - 6x = pi^2 - 2pix

9 - 6x + x^2 = pi^2 - 2pix + x^2

(3 - x)^2 = (pi - x)^2

3 - x = pi - x

pi = 3    ```

I WANT ANOTHER STAR lol

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12 Aug 2009, 05:25
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