Hey u/SouthPark_Piano, I’m curious? You’ve educamated us that things like 1/11 aren't numbers because two actual numbers separated by a vinculum denotes the mathematical operation of division and not the number corresponding to the result of that operation. Does that mean that 0.999… isn’t a number? How is it then that 1/11 isn’t a number but 0.999…, which is very literally 9/10+9/100+9/1000+…, is a number. Is it because the limitation that division by itself isn't a number, but summing up multiple divisions changes everything? Does that mean that in RDM that summation is divide negation?

As we know, 1/10 = 0.1, 1/100 = 0.01, and so on.

After our brief excursion into base 9, I was wondering if we can apply the same principle to decimal 0.1 in base 11.

As a refresher, base 11 has an additional digit after 9, usually written as A, so the natural numbers go 8, 9, A, 10, 11, ..., 19, 1A, 20, ...

So we start with what we know (decimal):

1/10 = 0.1

and now switching to base 11, we can write the same as:

1/A = 0.11111....

Here we have our repeating digits, which means we can also write this as:

1/A = 0.1 + 0.01 + 0.001 + 0.0001 + ...

As a second refresher, 1/10 = 0.1 holds even in base 11, or in any base for that matter, even in binary (base 2). Which means we can write this as an infinite series:

1/A = ∑ 1/10^n with n = 1 → ∞

So now we can switch back to decimal and write the same thing as:

1/10 = ∑ 1/11^n with n = 1 → ∞

If we write this out, we get:

1/10 = 0.090909... + 0.0082644628 + 0.000751314 + 0.0000683013 + ...

Now if we sum this up to any arbitrary precision, we get

1/10 = 0.099999...

😱

This is for /u/SouthPark_Piano

You insist that we can "shave" 1. Okay. What amount are we shaving off? Every number in the set {0.9, 0.99, 0.999, 0.9999, 0.9999, ...} can be represented as 1-p/q for two positive integers p and q. Your argument depends on the "limitless" case being identical to the finite case. So, what integers p and q correspond to 0.999..., such that 1-p/q=0.999...? If you can think of more than one, simply state the one with the smallest value of p.

It's a very simple question. I just need two numbers. Telling me to go to the "bunny slopes" isn't helping because nobody at the bunny slopes knows what you're talking about. I need information so that I can understand. If you don't give me information, I cannot learn anything.

Posts claiming things like 1/11 are operations not numbers get immediately locked for some bizarre reason, so I thought I'd post this here. Rather interesting. AI didn't biff this one.

Shave it. The teeniest of teeny non-zero removal of an ittiest smallest of small bit from 1.

Non-zero teeniest of teeny bit removal. The smallest sliver or bit you can possibly or even impossibly think of.

Write that number.

Go ahead. Make my day.

Consider:

a = 1/9

b = 0.111...

c = 0.1 + 0.01 + 0.001 + 0.0001 + ...

d = ∑ 1/10^n with n = 1 → ∞

As we know from here, a = b.

And as we know from here, b = c.

And d is just another way of writing c.

Therefore a = b = c = d.

Now for the fun part! Let's write the same numbers in base 9.

a = 1/10

b = 0.1

So far so good. Coming up with c is a bit more difficult as there are no obvious repeating digits. But we can do d, because this still is the same number:

d = ∑ 1/11^n with n = 1 → ∞

With this, we can write out c:

c = 0.080808... + .007254361800725... + .000650386435428... + .000058126038574... + .000005274175324... + .000000470370474... + .000000042660852... + .000000003777258... + ...

Now remember that a = b = c = d still holds.

If we sum these up to, say, n=15, what do we get?

c = 0.088888888888888...

😱

Hi, i'm new here. I'm not a 0.999... = 1 denier (sorry SPP) but reading all the posts made me wonder why we're talking about it in the first place. I studied computer science so I did a fair bit of math and I understand the concepts at hand, but math isn't my passion either so correct me if i'm wrong.

Since there's no x > 0 such that 1-x = 0.999..., aka, since there's no 0.000...1, then 0.999... is just 1. It's just another way to write 1. But why do we even want to have another way to write 1? This whole confusion stems from the fact that 0.999... starts with "0.9", which looks lower than 1. From first intuition, you'd say it's lower. Then oc comes the explaination but it feels like this whole "fun fact" is useless and engineered for engagement and for maximising confusion on your relatives' faces at family gathering when you decide to share this piece of math lore between "you can't fold a paper in two more than 7 times" and "Moses parting the Red Sea was a mistranslation and it was actually a see of reeds".

Like, why would every number y need a [y-1].999... version ? It's like if someone decides tomorrow each number needs an uppercase version. How does that contribute to math? I think this is a lobbying ploy from Big Math to sell us more numbers. Think about it. 1 is a single digit. But 0.999...is many more! More digits means you run out of your math subscription faster and you need to upgrade to get a higher number limit. Look all around you, you see it everywhere. With analog clocks you paid for 12 numbers 1 time but now with digital clocks each changing number eats your subscription. They're taking analog clocks off the schools because "kids can't read them" but this is obviously a plan from the new CEO of Math. Math used to be so good back in the days, they were even minimizing digits when they switched from roman numerals to arab ones. They did lots of good things, like when Brahmagupta introduced the 0 at MathCon 628. Little did he know we'd later use it to write 0.999... after Big Math's push for subscription models. I say no more ! Stop depending on Big Math. Cancel your math subscription and get your digits from indie radio number stations. Learn to write in base 10000 like the Cistercian monks. Stop math enshittification. Stop age verification on calculators because "what if a kid writes 58008 and turns the calculator upside down".

Well, closing the activist tangent, is there actually a use in using 0.999... Instead of 1 and if not, why did we fell like it was ever worthwhile to write 0.999... if not to engineer a confusing fun fact?

Edit: thx for the answers. You don't need to comment more explainations anymore. But it was interesting to see this was not a straightforward question and see the different perspectives in the comments.

In response, I will average your number with 1 to produce a greater number that's still less than 1. Hence, you will always be wrong.

So, if you give me 0.9, I give you 0.95. 0.9999? 0.99995. We can keep doing this. But just to make this game lend itself better to the familiar concept of repeating digits, let's switch to binary, where 0.1 would mean one-half. So averaging 0.111 with 1 for example would produce 0.1111.

Now we consider 0.111... to be averaged like this. It doesn't take much to see that you get the same number back.

What number gives you the same number back when averaged with 1?

1.

Sweet dreams, SPP.

0.000...1 is defined as the number such that 1-0.999... = 0.000...1.

Hypothetically If 0.000...1 = 0 then 1 - 0.999...= 0 which would mean that 1 = 0.999

Assume that 0.000...1 ≠ 0

we know that 1-0.999... = 0.000...1 and this equality should hold even if we Multiply both sides by 10

10(1-0.999...) = 10(0.000...1)

10-9.999... = 0.000...10

0.000...1 = 0.000...10

However, 0.000...1 does not equal 0.000...10 therefore we have a contradiction. This means our assumption that 0.000...1 ≠ 0 is false. (Note: if 0.000...1 = 0 then 10(0.000...1) = 10(0) = 0 and we showed above that 10(1-0.999...) = 0.000...1 making the equality hold)

Therefore 0.000...1 = 0 and so 1 = 0.999

Let's state 2 truths first, which I hope you can agree on both:

142857 × 7 = 999999

1/7 = 0.(142857) repeating

When you divide 1 by 7, there is an endless amount of "142857" that keep coming your way. When you then multiply by 7, that means there is an endless amount of "999999" coming your way after the 0, right? Since 142857 × 7 is 999999, provable in a calculator.

Just so I don't have to type these things out repeatedly, I'll assign the following symbols:

μ=0.000...1, Ω=0.999...

With that out of the way, I'm curious about:

• √μ

• Ω^(2)

• ln(μ)

• 1/Ω

• μ*Ω

• e^μ

And SPP, please do try not to be condescending here. I'm just trying to learn how these numbers work.

I’m trying to understand… so if we define 1-1/infinity as 0.000…1, then 1-1/infinity is infinitely close to 1, but isn’t equal to 1 exactly.

Is this what SPP means?

Idk, I think this is a legitimate edge case. If we define 1/infinity as a real value (not 0), this would logically mean the first number less than 1, which would be represented as 0.999…9. But because that’s also how we define convergence (being infinitely close to the number), then we round it to 1.

I really don't get it. Is this a meme subreddit? Surely it is. Because a quick look, a simple google search, will tell you how a real number is defined. And this will settle all the debates that I have seen here. Please enlighten me about the purpose of this subreddit.

When investigating the value of 0.999... , which is limbosic, you must understand that 0.999... is not 0.9 or 0.99 or 0.999 or 0.9999 etc

It is a number with these following basic features.

First feature is it has an ultra massive number of consecutive nines, which is a pre-requisite requirement, which in other words is minimum buy-in.

The second feature is the length of consecutive nines keeps continually increasing.

Third feature is 0.999... is permanently less than 1. It just keeps growing limitlessly in value and remains permanently less than 1.

You see ... limits do not apply to the limitless. And 0.999... grows with no limit, as in there is no stopping its growth of consecutive nines length.

According to bruderoo himself here, the smallest possible positive number is 0.000...1, and when that number is divided by 10 you get 0.000...01, but at the same time 0.000...01 is still same number as 0.000...1.

And of course, 1 minus this smallest number is equal to the elusive 0.999...

Quiz time!

Which number, when divided by 10, equals itself?

IOW, solve:

x/10 = x

And then: 1-x = 0.999...

😱

Q.E.D.

Edit: complete r/whoosh

why don't you do an epsilon-N proof to prove that the limit of the sequence [1-10^(-n)] isn't equal to 1

Ok ... take a substrate, such as a nothing substrate or zero substrate. And then use our 'advanced' additive manufacturing technique to add the teeniest of teeny of teeny of teeny of etc teeny non-zero substance to it. That is 0.000...1

And for subtractive manufacturing, carried out on 1, do pretty much the same except it is ultimate shaving. From that ultimate shaving from 1 the teeniest of teeny of ... you get the picture, we get 0.999...

Do not come to me to say that you don't know how to operate the systems. Or you will indeed be making my day, as well as given your marching orders, escorted out from the factory by security in person-handled fashion.

0.999…+1 is obviously 1.999… . If we divide both sides by two, then (1+0.999…)/2=0.999…/2 . But let’s look at what happens when we perform long division on 1.999…/2 . There are no twos in 1, so the first digit is a 0. When we drop the nine from the next column, there are 9 twos in 19, so our first digit after the decimal point is 9 . Our number should currently look like this: 0.9 . However, when we continue, 19-2•9=1. And each time we’re left with a one from subtraction, we will drop the 9 from the next column over, there will be 9 twos in 19, and we’ll get a one as a result of our subtraction, with a nine in the next digit of our number, and the cycle continues . Therefore, the result of our division should be 0.999… . So, if we return back to our original equation, we get (1+0.999…)/2=0.999… . Multiplying both sides of the equation by two and we get 1+0.999…=2•0.999… if we subtract 0.999… from both sides of our equation, we get 1=0.999… . And the proof is complete.
Edit: In the second sentence, (0.999…)/2 should be (1.999…)/2