Exactly. Err....well, its an italicized .
[img]http://upload.wikimedia.org/math/0/e/c/0ec6fcb6da4903ce355359d219ef3f5c.png[/img]Therefore, i equals the square root of -1.
:)
That's not true, actually.
If you substitute -3 for x, you get [tex]\sqrt{3} = i\sqrt{-3}[/tex], that is, [tex]\sqrt{3} = -\sqrt{3}[/tex].
>After "Da Vinci Code" who wouldn't?
I have no idea, I've never read it. ;)
(Obviously there's an error somewhere in this formula, but that's the fun of it :mrgreen:
Your error was here:
5. 2a^2 - 2ab = a^2 - ab
Since ab= a^2, then your expression above is the same as below:
2a^2 - 2a^2 = 0
Muses at how much of a math nerd I am. ;)
________________________________________
That's not true, actually.
If you substitute -3 for x, you get \sqrt{3} = i\sqrt{-3}, that is, \sqrt{3} = -\sqrt{3}.
Actualllly....;)
if x=-3 then sqrt{-(-3)} = isqrt{-3}
Then, we know that sqrt{a} x sqrt{b} = sqrt{a x b}
Therefore, we have:
sqrt{3} = sqrt{-3 x -1}
Which equals:
sqrt{3} = sqrt{3}
:)
I tend to favor odd numbers... but not odd in the sense of 2n+1, but rather numbers like pi, phi, e, sqrt(2), 1/7 (don't ask why), and i... the last of which is really fun to italicize in bbcode... :p
and of course, one of my favorite formulas would have to be:
[tex]e^{i\pi} + 1 = 0[/tex]
though it's also fun to toy around with
[tex]e^{\frac{i\pi}{2}} = i[/tex]
>After "Da Vinci Code" who wouldn't?
I have no idea, I've never read it. ;)
I thought it was a movie. :cheesy:
>I thought it was a movie.
I guess it was only a matter of time, what with the popularity of the book. Now when are they going to finally make the Dragonriders of Pern?
We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.
