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baldr
Tomasz Grysztar,
padic number field? I love topology. 

07 Nov 2008, 16:43 

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 

24 Jan 2009, 10:10 

sleepsleep
i think this line is somehow incorrect
Quote:


24 Jan 2009, 14:20 

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. 

24 Jan 2009, 14:41 

revolution
LocoDelAssembly wrote: It should be 

24 Jan 2009, 14:49 

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. 

24 Jan 2009, 15:14 

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
_________________ Previously known as The_Grey_Beast 

24 Jan 2009, 15:15 

revolution
sleepsleep wrote: but 

24 Jan 2009, 15:16 

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 

24 Jan 2009, 15:18 

Borsuc
That square roots have 2 roots?


24 Jan 2009, 15:26 

LocoDelAssembly
wikipedia wrote: If an equation in algebra is known to be true, the following operations may be used to produce another true equation: 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)? 

24 Jan 2009, 15:30 

revolution
Borsuc wrote: That square roots have 2 roots? LocoDelAssembly wrote: So, is it valid the multiplication by (Pi  3)? 

24 Jan 2009, 15:57 

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? 

24 Jan 2009, 16:06 

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? 

24 Jan 2009, 16:10 

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. _________________ Previously known as The_Grey_Beast 

24 Jan 2009, 16:21 

revolution
Borsuc wrote: also I searched the net, and the 'solution' is just what I mentioned above. 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? 

24 Jan 2009, 16:25 

Borsuc
No, I didn't search for the solution, I searched whether this (the square root thing) is enough
_________________ Previously known as The_Grey_Beast 

24 Jan 2009, 16:26 

revolution
Borsuc wrote: No, I didn't search for the solution, I searched whether this (the square root thing) is enough 

24 Jan 2009, 16:27 

Borsuc
EDIT: oops, seems not. Somehow I saw a 3 on the left hand side (not 6), sorry my bad
At least if pi=3, then x=3 as well 

24 Jan 2009, 16:28 

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