RE: If 0.999 (etc.) = 1, does 1 - 0.999 = 0?
March 4, 2012 at 2:57 pm
(This post was last modified: March 4, 2012 at 3:04 pm by Child of Stardust.)
(March 4, 2012 at 2:52 pm)LastPoet Wrote: Yes. 1-0.9999... =0 because 1 and 0.9999 are the same number and x + (-x) = 0, although the decimal representations are different as I outlined in the other thread. This is because the decimal (base 10) system we use, although sufficient to represent a real number, the same number can have two different decimal representations.
Thanks very much for the answer!
Aw, well that's kind of sad. This is universally agreed on, then? Do you know anywhere I can look at a "layman's terms" explanation of why? (If one exists...I'm probably the only person on earth who both failed high school math and actually cares about the question!
)EDIT:
(March 4, 2012 at 2:56 pm)Categories+Sheaves Wrote: There are a lot of practical (and aesthetic) reasons to not permit the existence of infinitesimals. Which is why you don't see infinitesimals given much attention.
The most popular modern formulation of analysis with infinitesimals (Abraham Robinson's Hyperreal numbers) involves the construction of an ultrafilter on the natural numbers, and the uniqueness of the Hyperreals as a system of numbers is equivalent to the continuum hypothesis. It's all well-defined and whatnot, but it's a bit further... out there.
Sorry...didn't see your response! Unfortunately, I don't know the meanings of most of those terms yet. My interest in math is very lay at this point. Sorry in advance for any confusion or slowness...it has more to do with my lack of knowledge than actual denseness.
