comp_sci_kid Wrote: ------------------------------------------------------- > bchadwick Wrote: > -------------------------------------------------- > ----- > > So comp-sci… prove that > > > > (1 - 0.99999repeat) != 0 for me. > > Easy. What is a definition of 0? > > 0*a = 0. > > now, lets say we have 9*10^-n. Let’s multiple this > by a = 0^n we get 9*10^(n-n) = 9. > > 9 != 0 I don’t see the proof here. a = 0^n = 0 (except when n=0, in which case it’s undefined) [9*10^(-n)] * a = [9*10^(-n)] * 0 = 0 != 9 And in any case, even if you had proven that 9*10^(-n) != 0, you still haven’t made the connection that would show (1 - 0.9999repeat) = 9*10^(-n) Therefore the fact that 9*10^(-n) is <> 0 doesn’t imply that (1 - 0.9999rept) != 0. I still maintain that (1 - 0.9999repeat) = 0, based on the algebra in my earlier post.
strikethree Wrote: ------------------------------------------------------- > It’s definitely been well debated: > http://en.wikipedia.org/wiki/0.999… > > Like the fractional equation, the algebraic > equation is still a manipulation of an equation. > It’s not necessarily an explanation since the > fractional equation alone could be regarded as a > proof. > > In calculus, we go into explaining this phenomenon > through the use of limits and note that the > repeating number is a process. In order to > understand that process, one needs to understand > the concept of infinity. > > Unlike the originally posted PEMDAS equation, > there are still mathematicians and critics out > there who challenge this concept of the repeating > 9’s. For logic would tell us that a repeating > decimal would not equal a real number; we are > taught that point the repeating number is that it > should never lead to a fixed number – again, it > goes against logic. (even if our calculations tell > us otherwise) It isn’t the same same as x/x = 1 > because that is easily understood. Ask people to > put .9999 repeating on a number line compared to 1 > and how many would say .9999 repeating < 1? > > As a fifth grader, this problem blew my mind. Interesting Wiki. It seems to suggest that in the standard real number system, 0.9999rept = 1, but that there are number systems in which case they could be different. A bit like non-euclidean geometry and the parallel postulate. In the real numbers world that we operate in most of the time, 0.999…=1, but it is conceivable that there can be a world in which 0.9999… != 1. It is weird to think that 0.99999… might equal 1, and I remember struggling with this in 5th or 6th grade. In my mind, I resolve the wierdness by realizing that 1 can have more than one representation, like 5/5, or 10^0, or -e^(i*pi). So, 0.99999… is another representation of 1, just as 0.33333… is another representation of 1/3. The wiki had another proof which I hadn’t seen, which was that, if we accept that 1/3 = 0.33333… then 3 * (0.3333…) = 0.99999… = 3 * (1/3) = 1 I thought that was an interesting way around that.
Ha - I’m going 488 and 1. 3/3 = 1 aaaaaaalllllll day.
bchadwick Wrote: ------------------------------------------------------- > comp_sci_kid Wrote: > -------------------------------------------------- > ----- > > bchadwick Wrote: > > > -------------------------------------------------- > > > ----- > > > So comp-sci… prove that > > > > > > (1 - 0.99999repeat) != 0 for me. > > > > Easy. What is a definition of 0? > > > > 0*a = 0. > > > > now, lets say we have 9*10^-n. Let’s multiple > this > > by a = 0^n we get 9*10^(n-n) = 9. > > > > 9 != 0 > > > > I don’t see the proof here. > > a = 0^n = 0 > (except when n=0, in which case it’s undefined) > > [9*10^(-n)] * a = [9*10^(-n)] * 0 = 0 != 9 > > > > And in any case, even if you had proven that > 9*10^(-n) != 0, you still haven’t made the > connection that would show > > (1 - 0.9999repeat) = 9*10^(-n) > > Therefore the fact that 9*10^(-n) is <> 0 doesn’t > imply that (1 - 0.9999rept) != 0. > > > I still maintain that (1 - 0.9999repeat) = 0, > based on the algebra in my earlier post. I see where my confusion was, i was not properly using http://en.wikipedia.org/wiki/Equality_(mathematics) I still havent gotten through all the article, but to properly proove that 0.9999 = 1 you need to know the definition of equality
So it depends on what the definition of is is?
6÷2(1+2)=1