are there more multiples of 3 or 7? page 2

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The reals are a subset of the complex numbers (with a 0 argument). I don’t know how wolfram evaluates it, but I’m guessing that they treat it as a complex number so that they can give a well defined answer.

Or it’s possible that I’m full of shit :P

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The square root of negative 1 doesn’t help you divide by 0.

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It does when you define z/0 to equal something.

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Originally posted by BigJM:

It is unable to be determined; therefore, it is indeterminate.

Heh, I guess I wasn’t as sure about translation of mathematical terms as I thoughts. To me, 1/0 would be “undefined” and 0/0 would be “not well-defined”. I guess I’m just handicapped from learning math in my native tongue.

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Oh jeez. You’re dividing by zero…fiugiuj#%&&&@@dqwxf…AAHHHH!0111001011000011110001010111100010011101001001

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lim(x→4) ((6x-x^2-8)/(4-x))
What is the answer?
let x = 4…
0/0 = …? 1 (n/n=1)? 0 (0/n=1)? undefined (n/0=undefined)?

lim(x→4) ((6x-x^2-8)/(4-x)) = lim(x→4) ((x-2)(4-x)/(4-x)) = lim(x→4) (x-2)
now, 4-2 = 2 o.O

So /0 does not necessarily lead to black holes. At least not all the time.
Hopefully I didn’t screw anything up though.