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baldr



Joined: 19 Mar 2008
Posts: 1651
baldr
Tomasz Grysztar,

p-adic number field? I love topology.
Post 07 Nov 2008, 16:43
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revolution
When all else fails, read the source


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revolution
As you will see, the end result of the following equation can not possibly be correct. You must tell me where the error in the working has occurred and how it can be made correct. NOTE: the "proof" below shows that pi (the ratio of any circle's circumference to its diameter) = 3
Code:
           x = (pi + 3) / 2

          2x = pi + 3

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

   2pix - 6x = pi^2 - 9

      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    
As usual there will be a small prize of one star to the first person the give the correct answer.
Post 24 Jan 2009, 10:10
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sleepsleep



Joined: 05 Oct 2006
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sleepsleep
i think this line is somehow incorrect
Quote:

3 - x = pi - x
Post 24 Jan 2009, 14:20
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revolution
When all else fails, read the source


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revolution
sleepsleep wrote:
i think this line is somehow incorrect
Quote:

3 - x = pi - x
What is wrong with it?

Note: I'm not saying you are right or wrong, I just want you to say specifically what is the problem that you think it may be.
Post 24 Jan 2009, 14:37
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LocoDelAssembly
Your code has a bug


Joined: 06 May 2005
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LocoDelAssembly
It should be
Code:
       |3 - x| = |pi - x|    

But I think it is still wrong since I don't see how that could be satisfied with -3 != Pi != 3.
Post 24 Jan 2009, 14:41
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revolution
When all else fails, read the source


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revolution
LocoDelAssembly wrote:
It should be
Code:
       |3 - x| = |pi - x|    

But I think it is still wrong since I don't see how that could be satisfied with -3 != Pi != 3.
Getting warm.
Post 24 Jan 2009, 14:49
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sleepsleep



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sleepsleep
but
2x = pi + 3 (on ur line 2)

if sort it now, pi = 2x - 3 already. and pi can be any value depend on the value of x, since this pi is not the π, it just a variable that using the name, pi.
Post 24 Jan 2009, 15:14
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Borsuc



Joined: 29 Dec 2005
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Borsuc
Hmm you do realize you take a square root somewhere (line before the last line)? You can't just take the positive root like that, you have to analyze them both. The negative root brings you to the original equation Razz

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Post 24 Jan 2009, 15:15
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revolution
When all else fails, read the source


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revolution
sleepsleep wrote:
but
2x = pi + 3 (on ur line 2)

if sort it now, pi = 2x - 3 already. and pi can be any value depend on the value of x, since this pi is not the π, it just a variable that using the name, pi.
Well sure, but why does the equation come out to pi=3? That is the problem you must solve.
Post 24 Jan 2009, 15:16
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revolution
When all else fails, read the source


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revolution
Borsuc wrote:
Hmm you do realize you take a square root somewhere (line before the last line)? You can't just take the positive root like that, you have to analyze them both. The negative root brings you to the original equation Razz
Okay, so what is the answer?
Post 24 Jan 2009, 15:18
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Borsuc



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Borsuc
That square roots have 2 roots?
Post 24 Jan 2009, 15:26
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LocoDelAssembly
Your code has a bug


Joined: 06 May 2005
Posts: 4633
Location: Argentina
LocoDelAssembly
wikipedia wrote:
If an equation in algebra is known to be true, the following operations may be used to produce another true equation:

1. Any quantity can be added to both sides.
2. Any quantity can be subtracted from both sides.
3. Any quantity can be multiplied to both sides.
4. Any nonzero quantity can divide both sides.
5. Generally, any function can be applied to both sides. (However, caution must be exercised to ensure that one does not encounter extraneous solutions.)


Interesting...
Let's see:
Code:
x^2 - x       = 0    ; Solutions x = {0,1}
(x^2 - x) * 0 = 0 * 0 ; Used property 3. Solutions x = {R}    


So, is it valid the multiplication by (Pi - 3)?
Post 24 Jan 2009, 15:30
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revolution
When all else fails, read the source


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revolution
Borsuc wrote:
That square roots have 2 roots?
They sure do. But that is not the answer required. You need an extra step.
LocoDelAssembly wrote:
So, is it valid the multiplication by (Pi - 3)?
Is that an answer posed as a question?
Post 24 Jan 2009, 15:57
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LocoDelAssembly
Your code has a bug


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LocoDelAssembly
A question, since Pi value is unknown at that point then, how to be sure I'm not changing the solution set by doing such multiplication?

I changed from warm to very cold, right? Razz
Post 24 Jan 2009, 16:06
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revolution
When all else fails, read the source


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revolution
LocoDelAssembly wrote:
A question, since Pi value is unknown at that point then, how to be sure I'm not changing the solution set by doing such multiplication?

I changed from warm to very cold, right? Razz
Warmer.
Post 24 Jan 2009, 16:10
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Borsuc



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Borsuc
If pi is 3, (pi - 3) is 0, so what do you get?
0 = 0?

not sure.
also I searched the net, and the 'solution' is just what I mentioned above. However, since maybe you want more details, I'm going to think this up a bit.

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Post 24 Jan 2009, 16:21
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revolution
When all else fails, read the source


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revolution
Borsuc wrote:
also I searched the net, and the 'solution' is just what I mentioned above.
Folks, we have a cheater in our midst!

But you see how even searching the 'net didn't help you to answer my problem. I didn't want to make it that simple now did I. Where is the fun in simply searching for someone else's answer?
Post 24 Jan 2009, 16:25
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Borsuc



Joined: 29 Dec 2005
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Borsuc
No, I didn't search for the solution, I searched whether this (the square root thing) is enough Wink

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Post 24 Jan 2009, 16:26
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revolution
When all else fails, read the source


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Location: In your JS exploiting you and your system
revolution
Borsuc wrote:
No, I didn't search for the solution, I searched whether this (the square root thing) is enough Wink
Alright then. I forgive you. Razz
Post 24 Jan 2009, 16:27
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Borsuc



Joined: 29 Dec 2005
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Borsuc
EDIT: oops, seems not. Somehow I saw a 3 on the left hand side (not 6), sorry my bad Embarassed

At least if pi=3, then x=3 as well Confused
Post 24 Jan 2009, 16:28
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