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Bombax
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The Maths Topic
PostPosted: Sun 19 Nov, 2006  Reply with quote

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.


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piculum
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PostPosted: Sun 19 Nov, 2006  Reply with quote

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. wink5
To prove that:

(-3)^2 = (3)^2
We get
9 = 9

Now, remove the square root:
-3 = 3

Oops, wrong. smile


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Bombax
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PostPosted: Sun 19 Nov, 2006  Reply with quote

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

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KingOmar
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PostPosted: Sun 19 Nov, 2006  Reply with quote

Oh man i wish i had that cookie hand voor hoofd
*KingOmar takes a cookie cookiemonster


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piculum
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PostPosted: Sun 19 Nov, 2006  Reply with quote

Cool. Thanks. smile

* Piculum slaps Omar, grabs the cookie out of his hands and eats it... tounge2 wink5


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KingOmar
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PostPosted: Mon 20 Nov, 2006  Reply with quote

*KingOmar sadly cries for cookie. cry

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Bruno
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PostPosted: Mon 20 Nov, 2006  Reply with quote

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 smile

<edit>Now, whoever solves the following equation gets a biscuit:

x = ³√8

Tip: it's not as simple as you think it is. kiekeboe</edit>


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Bombax
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PostPosted: Mon 20 Nov, 2006  Reply with quote

x = 2 There are no negative roots because sqrt(-8) = -2 and not 2.

Now, I require a big cookie. Now.


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Bruno
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PostPosted: Mon 20 Nov, 2006  Reply with quote

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

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Magnus
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PostPosted: Mon 20 Nov, 2006  Reply with quote

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



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Bruno
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PostPosted: Mon 20 Nov, 2006  Reply with quote

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 cookiemonster

x = ³√8

↓ THEREFORE

x³ = 8 → 2³ = 8 so 2 is a root siiw

↓ BY RUFFINI'S RULE

x³ - 8 = 0

Code:
2 1 0 0 -8
  1 2 4  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)

smile




Last edited by Bruno on Mon 20 Nov, 2006; edited 1 time in total
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Bombax
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PostPosted: Mon 20 Nov, 2006  Reply with quote

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)(a-b) = a(a-b)
a+b = a
2 = 1

Another "division by zero" issue wink5


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Stormthunder
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PostPosted: Mon 20 Nov, 2006  Reply with quote

Argh, you are all... all... NERDS! tounge2

(I'm just jealous because maths hurts my brain. And puts me to sleep, and, oh wait - zzzzzzzzzzzzz.)


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Bruno
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PostPosted: Tue 21 Nov, 2006  Reply with quote

Stormthunder wrote:
Argh, you are all... all... NERDS! tounge2

(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! yes 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! 8D 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. siiw

(I'll forget it all in a couple of months, if I pass the exam. kiekeboe)


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Kenneth
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PostPosted: Tue 21 Nov, 2006  Reply with quote

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. eh 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.


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