r/infinitenines

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such as 0.999... has no more nines to fit/add/append/tack-on etc, everyone does know full well the real deal.

It is still mathematically modelled as 0.9 + 0.09 + 0.009 + ...

and that is honestly ... cross heart, hope to live etc ...

1 - 1/10^n with integer n starting at n = 1, then n increased limitlessly, continually ... is the correct way to investigate the case of 0.999... magnitude. Equal 1? Or less than 1? The correct answer, to avoid being dum dums ... is magnitude less than 1.

1/10^n is never zero. That's a fact. And that's the ticket.

1 - 1/10^n is of course never 1. So of course, 0.999... is permanently less than 1.

 

If we agree on 0.33333... = 1/3, then we can prove it [that .9 repeating is equal to 1] another way too:

0.33333... = 1/3 | •3

<=> 0.99999... = 3/3

<=> 0.99999... = 1

This video is worth a watch

Any ideas?

100/10 + 1000/100 + 10000/1000 + etc = A

91/10 + 991/100 + 9991/1000 + etc = B

1. What is A?

2. What is B?

3. What is A - B?

There is someone on this thread that is trying to peddle the nonsense that 0.999... is defined as the largest number less than 1.

Can we define a number by some arbitrary property? Yes, absolutely.

For example we define the number sqrt(2) as the number x>0 such that x^2=2.
We define the golden ratio ψ=a/b for two numbers a>b>0 if a/b=(a+b)/a. This is also equivalent to the solution to the quadratic equation ψ^2 = ψ+1. 

Both of these numbers are roots of quadratic equations. Can we define numbers that aren’t roots of equations? Yes, the transcendentals like pi and e are.
As everybody knows pi is defined as the ratio circumference/diameter.
e is defined in at least two ways: the formula for the annual interest where the rate r is compounded n periods per year is (1+r/n)^n, then e^r= lim_{n→∞}(1+r/n)^n. Alternatively f(x)=e^x can be defined as the function that equals its own derivative f(x)=f’(x). In both cases e=lim_{n→∞} Σ_{k=1}^n 1/k!.
pi solves the integral pi=int_{-1,1}(1-x^2)^(-1/2)dx and can be expressed as a variety of infinite sums.
Both are transcendental numbers that take their value from the way the number is literally defined.

So can we define a number x as the largest number less than 1? 

Firstly, this is not a solution to some polynomial equation, nor is it defined in some sensible fashion like pi or e. 

But ignoring this insurmountable problem, we run into the issue that there simply is no number that can possible satisfy that definition. There is a trivial proof by contradiction that has been made by many of this sub (but some with learning difficulties refuse to accept the proof and keep peddling disingenuous nonsense). If x purports to be the largest number less than y the then x<y and for all numbers z<y we must have x≥z. Since x<y it is the case that x<(x+y)/2<y directly contradicting the claim that x is weakly greater than all numbers less than y. There is no way to even define a largest number less than another. (Archimedes drops his mic.)

Why? Because the set of numbers z<y is open (or half open depending on the LHS of the interval). There are always numbers between any z and y.

A number u is an upper bound of a set S if u≥s for all s∈S. The least upper bound is the smallest of the upper bounds (if it exists). If x was the largest number less than one it would be the least upper bound for the set of all numbers less than one. However the least upper bound of that set is one itself. One is an upper bound of all numbers less than one. Then the proof we just did for some x<1 as a candidate upper bound proves that one is the least upper bound.

Of course the fact that 0.999...=1 and 1 is the least upper bound of all numbers less than one mean that 0.999 is the LUB. However even so it's not the largest number strictly less than 1 for obvious reasons.

pretty sure we can all agree "..." means "infinite" and "infinite" means "never ending". so how can there be an end after the "...". just found that funny

And in general, for any given expression in the form of (0.999...+n)/(n+1), what do you think the answer is?

I've seen SPP agree in multiple places that 1/3 = 0.333... and that 1/3 * 3 = 1 by what SPP calls "divide negation." I've also seen SPP agree that 0.333... * 3 = 0.999... but somehow still finds that 0.999... ≠ 1. SPP, I will provide what you'd consider a false proof below, and I want you to say where I made any mistakes.

1 = 1/3 * 3 (1) (This is "divide negation.")

= 0.333... * 3 (2) (You've agreed to this previously.)

= 0.999... (3) (You've also agreed to this.)

∴ 1 = 0.999...

SPP, I want you to identify the line(s) with an error. However, based on what I've seen you say in other places, I don't see where you could say that there is an error. To make what I'm asking clear, TELL ME THE LINE(S) WITH AN ERROR. I'm not looking for some explanation on why you think 0.999...≠1; I'm looking for you to tell me where I went wrong in this particular proof my telling me the line number and why what I did in that line is wrong. You should not be bringing nonsense about 0.999... or 0.333... growing endlessly or how 1/10^n is never 0.

There is a certain troll who has overtaken this forum. The prevalence of their leading posts has encroached on SPP's territory. It's caused SPP to engage less, which is not a good thing. This troll is prone to blocking after they reply, so the proper course of action is to engage with the troll until they block you to exhaust their resources. If they block everyone they might as well get out of here. Be aware that they have probably edited early comments to change the narrative, not realizing people can fill in the blanks afterwards. The best course of action with this troll is to absolutely engage with them until they give up. Them blocking you temporarily is a sign that you're doing the right thing. Please let's all engage this troll.

We need to get rid of this troll to allow SPP to post what he wants. SPP is being overshadowed.

edit: I doubt this troll will reply at all to this post. lol jk that ensured the first reply to this post :) you're too predictable.

I already explained it a few times. So far nobody could bring a good counter argument. So maby we can use this post here to discuss!

SPP, you supported the analogy that your 0.000..1 can be compared to epsilon. We don’t know much about your real deal math, but regardless 0.000…1 is not an approach for working with infinitesimals. The transfer principle implies all expressions which are valid in the real numbers are also valid in the hyperreals (which infinitesimals are part of). 0.000…1 is not well-defined in the reals nor do we have reasons to believe it is in the hyperreals. A representation of a value cannot simultaneously terminate and not terminate (e.g 0.999…9). 0.000…1 isn’t even a real value because what is its index value supposed to be? 0.000..1 remains ill-defined in the hyperreals, and is not an infinitesimal (it isn’t a positive real number because it can’t be defined as a real number and it’s also interpretable as 0 if the zeros are never meant to terminate). It’s painful to reason about what positive real values are smaller than 0.000…1, as although some infinitesimals are smaller than other infinitesimals, none of those hypothetical values are well-defined (similar to 0.000…1 itself). I am in doubt that you can settle if the last digit of √(your 0.000…1) is 1 or something else, and if you can’t do so evidentially then the square root function in your system is very narrow. The surreals and other sets/proper classes don’t affirm the well-definition of the 0.000…1 you have proposed, and your own real deal math cannot demonstrate that 0.000…1 is a formal infinitesimal.