# The Maths Topic

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.

It seems valid.

Except:

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.

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

Oh man i wish i had that cookie
/me takes a cookie

Cool. Thanks.

• Piculum slaps Omar, grabs the cookie out of his hands and eats it…

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

[color=#333366]Now, whoever solves the following equation gets a biscuit:

x = ³√8

Tip: it’s not as simple as you think it is. [/color]

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

Now, I require a big cookie. Now.

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

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!

/me gives Magnus a biscuit

x = ³√8

[color=#333333]↓ T[size=100]HEREFORE[/size][/color]

x³ = 8 [color=#3366ff]→ 2³ = 8 so 2 is a root [/color]

[color=#333333]↓ B[size=100]Y [/size]R[size=100]UFFINI’S RULE[/size][/color]

x³ - 8 = 0

```2 1 0 0 -8 1 2 4 0```

x³ - 8 ≡ (x - 2)([color=#cc3333]1[/color]x² + [color=#cc9933]2[/color]x + [color=#339966]4[/color])

[color=#333333]↓ B[size=100]Y [/size]B[size=100]HASKARA’S EQUATION[/size][/color]

[color=#cc3333]1[/color]x² + [color=#cc9933]2[/color]x + [color=#339966]4[/color] = 0 → x = (-[color=#cc9933]2[/color] ± √([color=#cc9933]2[/color]² - 4 × [color=#cc3333]1[/color] × [color=#339966]4[/color]))∕(2 × [color=#cc3333]1[/color])

x = (-2 ± √(4 - 16))∕2

x = (-2 ± √(-12))∕2

x = (-2 ± 2√(3) i)∕2

x = -1 ± √(3) i

[color=#333333]↓ T[size=100]HEREFORE[/size][/color]

x = -1 + √(3) i

[size=100]AND[/size]

x = -1 - √(3) i

[color=#333333]↓ T[size=100]HEREFORE[/size][/color]

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)

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

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 [size=100]KNOW[/size]! 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. )

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.

I don’t it might have somting with that teacher hwo told me: “If you can’t figure that out you will never get algebra!”.

The flaw is in—
1\$ = (10c)²
—the actually valid line would be:
1\$ = 10²c
Which develops as folows:
1\$ = 10²c

1\$ = 10∙10c

1\$ = 10∙0.1\$

1\$ = 1\$

And the US economy is saved once again only—this time not by the superman, but rather SuperBruno. pokes Sandra

[color=#333366]Another biscuit to whoever manages to tell me how much (10c)² is worth in US Dollars and prove it. [/color]

SuperBruno (I started to smile again)

We already know that
10c = 0.1\$

So logically we can conclude that

b² = (0.1\$)²
(0.1\$)² = 0.1\$ ∙ 0.1\$ [/b]
b² = 0.01\$ [/b]

Why do I have the feeling I made a flaw somewhere

I don’t hate math, but I don’t love it either. Math is one of by best subjects and I even have the hardest math possible on my school. Sometimes I still make flaws in easy things and I do it right with harder things =P I dislike math on the computer, writing it makes it easier.

Because it’s totally abstract and meaningless? Sure you can rearrange Pi and make 100 cents equal 1 cent, but that achieves absolutely nothing. I can’t think of a single thing on this planet more boring.

Now, maths with a practical purpose I can understand. Engineering, for example. If you can’t figure out the correct dimensions for a bridge, it’ll fall over and people will most likely die. But even then, I’d hate to be the person doing the equations and figuring out the dimensions. It’s just… incredibly dull. I honestly don’t know how to explain it. I simply have no interest in maths whatsoever.

Bruno - I took several pure maths units at uni (unfortunately it’s compulsory for a business degree), and they were utterly PAINFUL, let me tell you. All those a’s and b’s and c’s and x,y,z’s… ARGH! They didn’t even tell you what the numbers were! How am I supposed to get any kind of meaning out of something like 2f(x)’ = x^2/xyz * zy^52 + 3yx/z? It means absolutely nothing! You can’t build a bridge with x and z - you build bridges with 5000 tons and 60000 steel poles. I somehow managed to pass the units (don’t ask me how), but, oh god, they were the most boring lectures I’ve ever sat through in my life.

I have the utmost respect for people who understand maths, though, because as I see it, they’re the brave people who are willing to take on challenges that I’d run screaming from. I think the world needs mathematicians - just not me, oh no!

Sure you can. If the function describes, for example, one type of bridge, then you might be able to enter for instance any length x, and get back y, which might be the number of tons of concrete needed. Thus, the function would describe not only one bridge, but ANY bridge with the same specifications with regard to lenght and usage of concrete. That’s the beauty of functions, once you have one that fits you just enter a new value any time and immedeately get the solution

I can see how people think that math is pointless sometimes, though. I don’t really know why I like math, I just have ever since I was little. I just think numbers are fun, even if you’re not really using them for anything. For example, I think it is fun to do untraditional math puzzles just because the numbers don’t behave as you’d expect, the answer is strange and exciting. Also, I’ve always liked trying to find quicker ways to do calculations in my head, so that I can calculate, for example, 108^2 = 11664 in my head in a second or two, though I guess that’s not exactly useless (I hope ).

And SuperBruno() was correct about the puzzle. It messes up the dimensions, so you get
(10c)^2
= (0.1\$)^2
= 0.01(\$^2)
The answer returned is thus 0.01 square dollars, not 1 cent.