Exactly. Err....well, its an italicized .

Therefore, i equals the square root of -1.

:)

That's not true, actually.

If you substitute -3 for x, you get $$\sqrt{3} = i\sqrt{-3}$$, that is, $$\sqrt{3} = -\sqrt{3}$$.

>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:

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:
$$e^{i\pi} + 1 = 0$$
though it's also fun to toy around with
$$e^{\frac{i\pi}{2}} = i$$

>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?

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