Earlier quoted context omitted.
0.999... + 0.000...1 = 1 0.000...1 = 1/∞ 0.999... = 1 - 1/∞
What exactly is the value of the number that ends in a 1 but has an infinite number of 9s before it?
0.999...1 = 1 - 1/∞ - 8/∞ = 1 - 9/∞0.999...= 1
421–430 of 647 posts
Re: 0.999...= 1
#422Earlier quoted context omitted.
(STATEMENT OF PERSONAL IGNORANCE [SOPI]: Anyone who actually understands this stuff please correct my mistakes below. Thanks.) In the real numbers, which are not always simple or intuitive, 0.99... = 1. That's true and I seem to understand the proof. But the real numbers aren't the only system that might be sitting behind "0.99..." and "1" when I write those symbols down and talk intuitively to people in my family. T…
0.999... + 0.000...1 = 1 0.000...1 = 1/∞ 0.999... = 1 - 1/∞ 1/∞ is zero or not?
If you accept that
1/∞ = 0
Then you accept that
∞ * 0 = 1
But the definition of 0 is exactly that anything multiplied by it must be 0. So this cannot be true.
To take a more verbal route: you cannot take nothingness and repeat it. Repeating (or multiplying) nothingness (or 0) is fundamentally nonsense.
Programmer explanation: one cannot loop through `null` even once, let alone a large number.
Re: 0.999...= 1
#423Earlier quoted context omitted.
Yep. Agree 100%. It is like the blue dress. I think the problem is the repeating function. Infinite things are non-intuitive and should be presented differently. Even here on HN you still see people confused about "convergence" and "identity". 0.999... doesn't CONVERGE, it literally is 1. I suspect this persists even with students that have had second year college calculus that discusses convergent series and sums.
Fine. Define 0.999... as the limit of the series sum(n=1 ... N)(10^-n), as N-> infinity. This is standard high school calculus. "Number" and "series" and "limit" and "convergence" don't all mean the same thing. However this number is defined as the limit of a convergent series. So the question really is meaningful. (One clue that this question is meaningful is the amount of space introductory calculus textbooks use t…
Thanks to a commenter who pointed out that my sum above should be
sum(n=1 ... N)(9*10^-n).
I can't, uh ... fully endorse that comment, which is not entirely accurate and doesn't answer my question. But I sure did miss the '9'.
Re: 0.999...= 1
#424Re: 0.999...= 1
#425A formally rigorous proof of this (in Metamath) is here: http://us.metamath.org/mpeuni/0.999....html Unlike typical math proofs, which hint at the underlying steps, every step in this proof only uses precisely an axiom or previously-proven theorem, and you can click on the step to see it. The same is true for all the other theorems. In the end it only depends on predicate logic and ZFC set theory. All the proofs have…
That is cool! I've been thinking about building a site that lets you explore a big graph of math proofs for a long time, since doing my pure math degree. Glad to see someone else has done something like it.
Re: 0.999...= 1
#426A formally rigorous proof of this (in Metamath) is here: http://us.metamath.org/mpeuni/0.999....html Unlike typical math proofs, which hint at the underlying steps, every step in this proof only uses precisely an axiom or previously-proven theorem, and you can click on the step to see it. The same is true for all the other theorems. In the end it only depends on predicate logic and ZFC set theory. All the proofs have…
I wonder if this particular example does the opposite. It's too rigorous for a normal human. It seems humorously rigorous to state things like 10 is not equal to 0, and 10 is a real number and 1 is less than 10, not to mention the number of seemingly redundant repetitions showing how each of the numbers discussed is a complex number, separately an individually. Does defining 1+9=10 make the proof more believable?
Rigorous proofs like this are for mathematicians and computers, I doubt they help anyone believe who doesn't believe.
I'm not sure how best to help someone who doesn't believe, but it could take arguments with stronger intuition, or just allowing the person to demonstrate with their own proof. It probably depends on the person, and why they don't believe it.
Re: 0.999...= 1
#427Earlier quoted context omitted.
> 1 - 0.999... = 0.000... proves that 0.999... = 1. If people accept the former, and that the RHS of the former is in fact 0, they've already also accepted that 0.999…=1. I don't see what the discussion is at that point
If people don't accept the former, they can take out a pencil and paper to compute it themselves. After a few digits it will become obvious. I don't have a direct computation for making the latter obvious, just indirect ones like 1 - 0.999... and 3 x 0.333...
> If people don't accept the former, they can take out a pencil and paper to compute it themselves. After a few digits it will become obvious.
How can they compute 1-0.999… when they clearly have no idea what 0.999… is?
Re: 0.999...= 1
#428Earlier quoted context omitted.
None of these, except 0.999… and 1 are well-known standard objects in this setting. You have to define what you mean.
I defined it: 0.000...1 = 1/10^∞ = 1/∞
That's not a definition. Neither 10^∞ nor 1/∞ is defined in any standard system, so you'll have to define those too if you want to use them to define 0.000…1.
Re: 0.999...= 1
#429The proof relies on the assertion that the supremum of an increasing sequence is equal to the limit. This is mathematical dogma, and should be introduced as such. Once that is accepted, it becomes obvious. This is illustrative of what I see as a fundamental problem in mathematics education: nobody ever teaches the rules. In this case, the rules of simple arithmetic hit a dead end for mathematicians, so they invented…
What? 0.999... = 1 is not dogma. Please don't spread misinformation. And at least read the link before commenting on something.
Did you read what @jl2718 posted? Namely:
> the supremum of an increasing sequence is equal to the limit
-- this is not misinformation (and to anyone familiar with some introductory analysis, correct[1]). Of course, calling it "dogma" is a bit inflammatory, but not technically wrong. It's kind of of a made-up rule to help us work with infinities (particularly in ℝ -- but it happens all the time in set theory, as well).
But to agree with GP, touting it as "intuitive" or "mind-blowing" is indeed silly.
[1] http://www.math.toronto.edu/ilia/Teaching/MAT378.2010/Limits...
Re: 0.999...= 1
#430There is a nice characterization of decimal expansions in terms of paths on a graph:
Let C be the countable product of the set with ten elements, i.e. {0, 1, 2, ..., 9}. The space C naturally has the topology of a Cantor set (compact, totally disconnected, etc). Furthermore, for example, in this space the tuples (1, 9, 9, 9, ...) and (2, 0, 0, 0, ...) are distinct elements.
The space C can also be described in terms of a directed graph, where there is a single root with ten outward directed edges, and each child node then has ten outward directed edges, etc. C can be thought of as the space of infinite paths on this graph.
A continuous and surjective map from C to the unit interval [0, 1] can be constructed from a measure on these paths. For any suitable measure, this map is finite-to-one, meaning at most finitely many elements of C are mapped to a single element in the interval. For example there is a map which sends (1, 9, 9, ...) and (2, 0, 0,....) to the element "0.2".
The point is that all decimal expansions of elements of [0, 1] can be described like this, and we can instead think of the unit interval not as being composed of numbers _instrinsically_, but more like some kind of mathematical object that _admits_ decimal expansions. The unit interval itself can be described in other ways mathematically, and is not necessarily tied to being represented as real numbers. Hope this helps someone!