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Posts: 523 Joined: 25 Sep 2006 Last Visit: 24 Mar 2018
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The Maths Topic 
Posted: Sun 19 Nov, 2006 


So, have you ever wondered why the constant pi, with the symbol π, why does it have to be so complicated? Why do you always need to rely on the π button on your calculator that doesn't show all the decimals anyway but only about up to 3.1415926535? Why can't π just be 3? Well, today I cooked up a proof to show you that indeed π is 3!
Whatever π is, I can always base it of another value right? A value we call x? So now I have this little starter equation:
x = (π + 3) / 2
Shouldn't be too hard for any of you to understand. Now, I multiply both sides with 2.
2x = (π + 3)
You're following me right? Now, I will multiply both sides with (π  3). This causes:
2x(π  3) = (π + 3)(π  3)
Now, through basic algebra I simplify the terms into this.
2xπ  6x = π^2  9
The little ^ symbol means "powered by" which in this case "powered by two" or "squared" so basicly I can write 3^2 = 3 × 3 = 9 which I did with the nine up there.
Now, we rearrange the terms with some simple algebra.
9  6x = π^2  2xπ
Just for the fun of it, I'm going to add x^2 to each side (x squared).
x^2  6x + 9 = x^2  2xπ + π^2
If you might notice, each side can now be written as a squared parenthesis, using the quadration law backwards. This might help you understand:
(x  3)^2 = (x  π)^2
Now let's just drag the square root out of both sides.
x  3 = x  π
Take away x from both sides and multiply both sides with 1.
3 = π
3 = π
Ah, but look! I have successfully proved that 3 = pi! Our worries are over! Or...?
The one that can spot the mistake in this proof gets a free cookie.



 
 
   





Posts: 135 Joined: 21 Feb 2006 Last Visit: 02 Jul 2008
 


Posted: Sun 19 Nov, 2006 


It seems valid.
Except:
Quote: 
(x  3)^2 = (x  π)^2
Now let's just drag the square root out of both sides. 
You can't do that without checking.
To prove that:
(3)^2 = (3)^2
We get
9 = 9
Now, remove the square root:
3 = 3
Oops, wrong.



 
 
   





Posts: 523 Joined: 25 Sep 2006 Last Visit: 24 Mar 2018
LD count: Thrice a week
 


Posted: Sun 19 Nov, 2006 


Woho! Someone actually discovers the false lead! Goodo work Here is your free cookie!



 
 
   




26 
Posts: 219 Joined: 06 Oct 2006 Last Visit: 21 Nov 2009
Location: Las Vegas,NV  


Posted: Sun 19 Nov, 2006 


Oh man i wish i had that cookie
*KingOmar takes a cookie



 
 
   





Posts: 135 Joined: 21 Feb 2006 Last Visit: 02 Jul 2008
 


Posted: Sun 19 Nov, 2006 


Cool. Thanks.
* Piculum slaps Omar, grabs the cookie out of his hands and eats it...



 
 
   




26 
Posts: 219 Joined: 06 Oct 2006 Last Visit: 21 Nov 2009
Location: Las Vegas,NV  


Posted: Mon 20 Nov, 2006 


*KingOmar sadly cries for cookie.



 
 
   





Posts: 5954 Joined: 03 Dec 2005 Last Visit: 29 Mar 2018
LD count: a bunch.
Location: fleeting.  


Posted: Mon 20 Nov, 2006 


x = (π + 3) / 2
↓ × 2
2x = (π + 3)
↓ × (π  3)
2πx  6x = π²  9
↓ THEREFORE
9  6x = π²  2πx
↓ + x²
x²  6x + 9 = π²  2πx + x²
↓ THEREFORE
(x  3)² = (x  π)²
↓ THEREFORE
(x  3)²  (x  π)² = 0
Now given the aspect of that (the difference between two roots of multiplicity two outputting zero), this expression is not a function, and moreover it has no points anywhere in { x ϵ ℝ  x ≠ (3 + π)∕2 } whilst for { x ϵ ℝ  x = (3 + π)∕2 }, y = ] ∞, +∞ [ so that is makes true the first statement: x = (π + 3)/2
<edit>Now, whoever solves the following equation gets a biscuit:
x = ³√8
Tip: it's not as simple as you think it is. </edit>



 
 
   





Posts: 523 Joined: 25 Sep 2006 Last Visit: 24 Mar 2018
LD count: Thrice a week
 


Posted: Mon 20 Nov, 2006 


x = 2 There are no negative roots because sqrt(8) = 2 and not 2.
Now, I require a big cookie. Now.



 
 
   





Posts: 5954 Joined: 03 Dec 2005 Last Visit: 29 Mar 2018
LD count: a bunch.
Location: fleeting.  


Posted: Mon 20 Nov, 2006 


As I said, it wouldn't be as simple as you would think. I want all solutions to that expression.



 
 
   




32

Posts: 2778 Joined: 30 Jul 2005 Last Visit: 21 Nov 2018
LD count: 48
Location: Sweden  


Posted: Mon 20 Nov, 2006 


Is this right?
x=2(cos 0 + i sin 0)
x=2(cos 2π/3 + i sin 2π/3)
x= 2(cos 4π/3 + i sin 4π/3 )
/Magnus
Current LD goal(s): Play chess on the moon :), transform into a guinea pig



 
 
   





Posts: 5954 Joined: 03 Dec 2005 Last Visit: 29 Mar 2018
LD count: a bunch.
Location: fleeting.  


Posted: Mon 20 Nov, 2006 


Magnus wrote: 
Is this right?
x=2(cos 0 + i sin 0)
x=2(cos 2π/3 + i sin 2π/3)
x= 2(cos 4π/3 + i sin 4π/3 )
/Magnus 
Yes!
*Bruno gives Magnus a biscuit
x = ³√8
↓ THEREFORE
x³ = 8 → 2³ = 8 so 2 is a root
↓ BY RUFFINI'S RULE
x³  8 = 0
x³  8 ≡ (x  2)(1x² + 2x + 4)
↓ BY BHASKARA'S EQUATION
1x² + 2x + 4 = 0 → x = (2 ± √(2²  4 × 1 × 4))∕(2 × 1)
x = (2 ± √(4  16))∕2
x = (2 ± √(12))∕2
x = (2 ± 2√(3) i)∕2
x = 1 ± √(3) i
↓ THEREFORE
x = 1 + √(3) i
AND
x = 1  √(3) i
↓ THEREFORE
x³  8 ≡ (x  2)(x +1  √(3) i)(x +1 + √(3) i)
S = { 2, 1 + √(3) i, 1  √(3) i }
Magnus' method is actually easier: you know 2 is a root. Since 2 ϵ ℕ, you can also say 2 = 2(cos 0 + i sin 0). From that you need the other tree roots in the Complex space, so you just keep the argument of the number and change it's angle in 2π/3 (120°):
2(cos 0 + i sin 0)
2(cos 2π/3 + i sin 2π/3)
2(cos 4π/3 + i sin 4π/3)
Last edited by Bruno on Mon 20 Nov, 2006; edited 1 time in total



 
 
   





Posts: 523 Joined: 25 Sep 2006 Last Visit: 24 Mar 2018
LD count: Thrice a week
 


Posted: Mon 20 Nov, 2006 


It's true, but I think it's a bit overcomplicated. Although yes, you can involv real + imaginary numbers too
Here is a nice proof to show that 0 / 0 is NOT 1.
Let a = 1, b = 1. Then;
a^2  b^2 = a^2  ab
(a+b)(ab) = a(ab)
a+b = a
2 = 1
Another "division by zero" issue



 
 
   




34 
Posts: 1678 Joined: 16 Mar 2005 Last Visit: 02 Dec 2012
Location: the real world  


Posted: Mon 20 Nov, 2006 


Argh, you are all... all... NERDS!
(I'm just jealous because maths hurts my brain. And puts me to sleep, and, oh wait  zzzzzzzzzzzzz.)



 
 
   





Posts: 5954 Joined: 03 Dec 2005 Last Visit: 29 Mar 2018
LD count: a bunch.
Location: fleeting.  


Posted: Tue 21 Nov, 2006 


Stormthunder wrote: 
Argh, you are all... all... NERDS!
(I'm just jealous because maths hurts my brain. And puts me to sleep, and, oh wait  zzzzzzzzzzzzz.) 
I KNOW! Until a couple of months ago, I used to hate Maths with all my heart and it was perhaps the utmost sedative I knew! Good times. That was until I found out Maths weighted the triple relative to other subjects in the application test for Economics university. So I decided to give the whole thing a second chance and fell in love with the subject! It's actually awesome if you give it a chance! I had to learn twelve years worth of Maths in a couple of months, but it was worth it.
(I'll forget it all in a couple of months, if I pass the exam. )



 
 
   




32 
Posts: 210 Joined: 23 Dec 2005 Last Visit: 24 Aug 2010
Location: Norway  


Posted: Tue 21 Nov, 2006 


I love math
Seriously though, I've always had a knack for numbers , ever since I was a kid. I can't see how anyone can avoid liking math. Especially these kinds of funny puzzles.
Who can spot the flaw in this one?
1$ = 100c
= (10c)^2
= (0.1$)^2
= 0.01$
= 1c
A dollar is actually equal to a cent? So much for the US economy.



 
 
   
 
 