2010年7月21日 · The usual demonstration consists of getting someone to agree that $\frac13=0.33333 . . . $ and then multiply it by $3$ to get $0.99999 . . . $. At this point they might be convinced, but might equally feel puzzled or duped.

well 0.3333 is 1/3, so 1/3 + 1/3 +1/3 = 1 not 0.99999 because if you do by fractions you have (1+1+1)/3 = 3/3 = 1 exactly not rounded. the problem is that computers with floating point do this and then people that do not know math get confused.

2016年1月26日 · I accept that two numbers can have the same supremum depending on how you generate a decimal representation. So $2.4999\\ldots = 2.5$ etc. Can anyone point me to resources that would explain what the

The first proof assumes that 1/3=0.33333... which would need to be proven, especially if one doesn't understand why 0.99999...=1. The first and the third proof also assume computation works with numbers with infinitely many digits, which would only be the case if these numbers indeed converge to a rational one.

2020年12月27日 · I would say it suffices. It’s a bit clunky since adding an infinite amount of things can lead to paradoxes and fallacies if you aren’t careful, but 0.99999... does really equal 1 in the real numbers (I specify the reals because there is a class of numbers called the surreal numbers, where 0.99... is not 1, if you are interested in googling it)

2017年8月11日 · But it always ends up telling me that $0.9999999... = 1$. Is there any mistake in my logic? And I also realized this applied to other recurring decimals ending in $9$. E.g: $0.5999999...=5.4/9 = 0.6$ . So is there a way to write $0.999999...$ as a fraction so you can differentiate it from $1$?

2023年8月1日 · The issue seems to be accepting 0.99999=1. This can be viewed as a limit but in fact can be shown without needing limits at all. Here is one such approach: Let x = 0.99999…. Then 10x=9.99999… Then 10x - x = 9.99999 - 0.99999 Therefore 9x = 9 And x = 1. Note this is a strict equality and not the end point of some form of limit.

The problem most people have with 0.999... is that they don't actually understand what the “...” means. The number 0.999... is defined to be the limit of the sequence 0.9, 0.99, 0.999, 0.9999, and the limit of that sequence is 1. To see why, consider that you tell me that the difference 1 - 0.999... is some non-zero number x.

2023年9月18日 · 1 - .99999 = .00001 As a matter of precision, however far you take this pattern, the difference between 1 and a bunch of 9s will be a bunch of 0s ending with a 1. As we do this thousands and billions of times, and infinitely, the difference keeps getting smaller but never 0, right? You can always sample with greater precision and find a difference?

The number $0.9999\cdots$ is in fact equal to $1$, which is why you get $\frac{9}{9}$. See this previous question.