# are there more multiples of 3 or 7? page 2

31 posts

 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 The square root of negative 1 doesn’t help you divide by 0. It does when you define z/0 to equal something. 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. Oh jeez. You’re dividing by zero…fiugiuj#%&&&@@dqwxf…AAHHHH!0111001011000011110001010111100010011101001001 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.