Three Equals Negative Nine
Upon taking a math test and getting bored doing so, I came upon a problem that turns out to have no answer:
(√3 + i )3
The correct answer is 8i:
(√3 + i) (√3 + i) (√3 + i)
3 + 2i×3^(1/2) -1
2 + 2i×3^(1/2)
2×√3 + 2i + 6i − 2*√3
8i
Now let’s reverse that:
(√3 + i)3 = 8i
√3 + i = 3√(8i)
√3 = -2i − i
√3 = -3i
3 = (-3i)2
3 ≠ -9
I’m not really sure what this proves, but it surely does not support euclidean mathematics. Keep in mind that all 3^(1/2)’s are really just the square root of 3.
EDIT: I figured out the square roots =] √3
4 Responses
noga
Apr. 5/2007/4:50:13 pm
haha i will never forget that day in math… i wonder what else we could prove over in the corner by the window?
Lamp
Apr. 5/2007/4:59:37 pm
Maybe how to display the square root sign in html…
Leigh
Sep. 4/2007/4:46:43 am
Not quite!
You are making a mistake with the (8i)^(1/3), this is not equal to 2i as you might expect.
To see that this is in error think back to what square, cube, etc roots actually are… does (2i)^3 = 8i ? No, so 2i cannot be the cube root of 8i.
To calculate roots of complex numbers it is best to conver to polar form (I believe it is possible to do without this and work in the cartesian form, but then we dont have Euler and De Moivre’s theorems to simpify things) using Eulers theorem. Then applying De Moivre’s theorem we get a fairly simple trig equation to satisfy that will give us the roots (note I say roots, since in general the nth root of a complex number has n values).
I wont give you the working here, but just to know (8i)^(1/3) = 3^(1/3) + i
This fixes your proof that (3)^(1/3) = (3)^(1/3) - who knew!
Have a search around Wikipedia for the the theorems I mentioned if you are still interested.
- Laters,
- Leigh
Lamp
Sep. 4/2007/9:31:31 pm
Well… Actually I have there that the cubed root of 8i equals -2i. Now, -2i cubed equals 8i, but when you try to cube root 8i, it gives you the square root of 3 plus i. So there is a mistake… sort of…
Yes, I forgot to edit this, but I did consult my professor and she explained that is is why we use De Moivre’s Theorem to solve it…
But thank you for your time in explaining!