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Author
baldr

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

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

Joined: 24 Aug 2004
Posts: 17270
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.
24 Jan 2009, 10:10
sleepsleep

Joined: 05 Oct 2006
Posts: 8885
Location: ˛　　　　　　　　　　　　　　　　　　　　　　　　　　　　　⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣Posts: 334455
sleepsleep
i think this line is somehow incorrect
Quote:

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

Joined: 24 Aug 2004
Posts: 17270
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.
24 Jan 2009, 14:37
LocoDelAssembly

Joined: 06 May 2005
Posts: 4633
Location: Argentina
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
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17270
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.
24 Jan 2009, 14:49
sleepsleep

Joined: 05 Oct 2006
Posts: 8885
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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.
24 Jan 2009, 15:14
Borsuc

Joined: 29 Dec 2005
Posts: 2466
Location: Bucharest, Romania
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
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17270
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.
24 Jan 2009, 15:16
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17270
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
Okay, so what is the answer?
24 Jan 2009, 15:18
Borsuc

Joined: 29 Dec 2005
Posts: 2466
Location: Bucharest, Romania
Borsuc
That square roots have 2 roots?
24 Jan 2009, 15:26
LocoDelAssembly

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)?
24 Jan 2009, 15:30
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17270
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?
24 Jan 2009, 15:57
LocoDelAssembly

Joined: 06 May 2005
Posts: 4633
Location: Argentina
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
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17270
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?
Warmer.
24 Jan 2009, 16:10
Borsuc

Joined: 29 Dec 2005
Posts: 2466
Location: Bucharest, Romania
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
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17270
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?
24 Jan 2009, 16:25
Borsuc

Joined: 29 Dec 2005
Posts: 2466
Location: Bucharest, Romania
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
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17270
revolution
Borsuc wrote:
No, I didn't search for the solution, I searched whether this (the square root thing) is enough
Alright then. I forgive you.
24 Jan 2009, 16:27
Borsuc

Joined: 29 Dec 2005
Posts: 2466
Location: Bucharest, Romania
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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