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.
And Sandra is right too: b² = 0.01$ [/b] and there are even three different ways of proving that! :shock:
/me geeft Sandraatje een biscuit 
en een knuffel 
Now, Kenneth just made another puzzle and this one is slightly more difficult to prove:
b² = 0.01 $²[/b]
But, as Sandra proved,
b² = 0.01$ [/b]
Therefore,
0.01$ = 0.01 $²
1$ = 1$²
But oh look, Sandra’s law develops as follows:
b² = 0.01$[/b]
b² = 1 c[/b]
100 c² = 1 c
1 c² = 0.01 c
The Number Theory freak who manages to tell me why 1 $² = 1 $ whilst 1 c² = 0.01 c gets two biscuits.
Oh, by the way, I found another way to answer the π problem and I think it’s actually the easiest available so far. I’ll try to post it later because right now I have to go. 
You seem almost a bit more nerdy than me! 
That problem is simple.
1$ ^2 = 1$ * 1$ = 1$
1c ^2 = 0.01$ ^2 = 0.0001$ = 0.01c
Eh, these kind of tricks are as old as the algebra itself.
I still remember how you can prove 2+2=5, had a little fun in middle school.
Engeneering is good enough, but real use of advanced math is programming. And that IS something you can really have fun with. Coding games, for example. Writing your own shaders and then admiring how shiny the armor on that trooper is…
Good, but that’s demonstrating, and I asked you to explain. 
Gnarhl looks at clock let me prove instead for you that 2 = 1.
Let a = 1, and b = 1. Then it is true that we have:
a^2 - b^2 = a^2 - ab
(a+b)(a-b) = a(a-b)
a+b = a
2 = 1
End. I can also prove that sin x / n = 6.
sin x / n = sin x / n = si x = six = 6 
Alright, I’ll explain it myself. I was expecting someone to say it’s because $ = 1 and c = 0.01, which is wrong.
Lets schedule a couple of those linear equations regarding the relation between $ and c:
1$ - 100c = 0
0.01$ - 1c = 0
↓ THEREFORE
[code] – –
| 1 -100 | 0 |
| 0.01 -1 | 0 |
L2 => (-L1/100 + L2)
| 1 -100 | 0 |
| 0 0 | 0 |
0$ + 0c = 0
0 = 0[/code]
“Sweet, Bruno, what does this mean?”
That you can’t define $ if you don’t have c, and you can’t define c if you don’t know $. That means $ isn’t necessarily 1: all we know is, it’s worth 100 c. So if 1 c is worth 5 in pure numbers, 1 $ equals 500.
“Alright, and what’s the difference?”
The difference is, since $ and c can’t be defined in terms of pure numbers, they’re said to be a number system. As a number system, it doesn’t matter how much $ is worth, what matters is $ is the basic unit of its system, so it’s worth 1 of whatever we’re measuring with it. Although $ doesn’t have to equal 1, $ α 100% (α indicates proportionality) of it’s system, whilst c α 1%.
Thus, the issue can be demonstrated as follows:
1 $ α 100%
1 c α 1%
↓ TO THE POWER OF 2
1² $² α (100%)²
1² c² α (1%)²
↓ THEREFORE
1 $² α 100%
1 c² α 0.01%
↓ THEREFORE
1 $² = 1 $
1 c² = 0.01 c
There’s a lot of nice psyched metaphysical stuff one can find out from that, a real lot, but I really don’t have the time for the crazy what the metaphysics.
Now, as promised:
x = ³√8
x³ -8 = 0 → 2³ -8 = 0
2 = 2∙(cos 0 + i sen 0)
By Euler’s fomula,
2 = 2∙e^(i 0)
Since the other two solutions must be spread around the complex field with argument 2 and equal angular distances, we divide 2π/3 and get the three solutions:
[b]α = 2∙e^(i ⅓∙2π)
β = 2∙e^(i ⅔∙2π)
γ = 2∙e^(i 2π)[/b]
↓ THEREFORE
S = { 2∙e^(i ⅓∙2π), 2∙e^(i ⅔∙2π), 2∙e^(i 2π) }
Wow that was fast!
Someone else propose a problem!
a) (easy) 1 + 1 + 1 + 1… + 1 (x times) = x
take derivatives:
0 + 0 + 0 + 0 + … + 0 = 1
0 = 1
b) (moderate) sqrt(1) = 1; but 1 = -1*-1;
sqrt([-1]*[-1]) = 1;
sqrt[-1]sqrt[-1] = 1;
ii = 1;
-1 = 1.
OR: -1 = sqrt([-1]^2) = sqrt(1) = 1.
c) (hard but involves a tiddy bit of physics) a ladder is leaning with one end of it on the wall, and the other end on the ground. It touches the ground at a distance x from the wall, and touches the wall at a distance y from the ground. If L is how long the ladder is, it’s obvious then that: y = sqrt(L^2 - x^2).
so now we grab the end of the ladder that is touching the ground and we pull it away from the wall at a constant speed vx. vx is obviously dx/dt. the other tip of the ladder slides down the wall towards the ground. let’s see how fast the other end of the ladder moves towards the ground:
vy = dy/dt = - x* vx / sqrt(L^2 - x^2). (no need to check this calculation, it’s correct).
ah but notice, the ladder hits the ground when x = L. but in the above equation, setting x to L gives us: vy = infinity. So the other tip of the ladder hits the ground at an infinite speed?