Mothematician post

if 0.333... = 1/3 then 0.999... = 3/3 = 1

lots of good proofs in this thread but this one gets points for being by FAR the shortest while still being completely based on intuition and makes complete sense with an average high school mathematics background

The thing that really got 0.999...=1 to click on my mind is the fact that you can't find a number between 0.999... and 1. You might think "Just put something at the end of 0.999...," but there is no end to 0.999....

Yeah that's something that people have to get used to in maths, if the limit of a sequence exists we can just pretend to have "reached infinity" and work with like any number.

0.999... isn't a number approaching 1, it is 1.
- Yeah that's something that people have to get used to in maths, if the limit of a sequence exists we can just pretend to have "reached infinity" and work with like any number
  
  I'm not sure what you are trying to say here, and I have a background in math. I think this is just going to confuse lay people.
Yay, Dedekind cuts are finally useful!

and wants to know why 0.999... = 1

\begin{align} 0.999.... &= 9\cdot(0.1+0.01+0.001+... ) \ &= 9\cdot( 0.1^1 + 0.1^2 + 0.1^3 + ... ) \ &= 9\cdot(\sum\limits_{k=1}^\infty ( \frac{1}{10^k} ) ) \ &= 9\cdot(\sum\limits_{k=0}^\infty ( \frac{1}{10^{(k+1)}}))\ &= 9\cdot(\sum\limits_{k=0}^\infty \frac{1}{10}*(\frac{1}{10^k})) \ &= \frac{9}{10}\cdot (\sum\limits_{k=0}^\infty (\frac{1}{10^k})) \ &= \frac{9}{10}\cdot \frac{1}{(1-(\frac{1}{10}))}\ &= \frac{9}{10}\cdot \frac{10}{9} = 1 \end{align}

The crux rests on a handy result on from calculus: the sum of an infinite geometric series looks likes s = 1/(1-r), when s = \sum\limits_k=0^inf r^k, and |r| < 1.

Sorry for the latex. When will hexbear render latex? This is a bit more readable:

(aesthetic edit for our big beautiful complex analysts)

Never thought of employing the geometric series for this, that’s clever.
- ^_^ thank you!
Using i as an index 🤮
- ? Thats pretty standard though right?
- Your feedback is valid and I apologize for rendering such an ugly proof
Permanently Deleted
- A real number can't be approaching anything. It is not a function or any other sort of object that can be said to be approaching anything.
- No. 0.99 is 0.9+0.09. The proof I gave shows that 0.99999999999999999999999999999999999(...) is equal to 1.

0.999..=1 because it's funny that people get so mad about it

I agree, but the little guy does need rigour, or he will starve under crapitalism.
- I didn't have any bread to bring him but I had plenty of roses

so, start with:

x = 0.999...

now multiply each side by 10

10x = 9.9999....

now we subtract x from the left side, and 0.999... from the right, which is fine because they are equal:

9x = 9

and from there it should be fairly obvious that x is also equal to 1, which means 0.999... is also equal to 1

He's hungry
cri
Feed him

I will be posting another another proof, one using the assumption that 0.999... is the sum of the series 9/10+9/100+9/1000+... (and that 1 is the sum of the series 1+0+0+0+...), unless somebody else comes here and feeds it to the little guy first.

I suppose I will post one myself, as I do not expect anybody else to have that one in mind.

The decimals '0.999...' and '1' refer to the real numbers that are equivalence classes of Cauchy sequences of rational numbers (0.9, 0.99, 0.999,...) and (1, 1, 1,...) with respect to the relation R: (aRb) <=> (lim(a_n-b_n) as n->inf, where a_n and b_n are the nth elements of sequences a and b, respectively).

For a = (1, 1, 1,...) and b = (0.9, 0.99, 0.999,...) we have lim(a_n-b_n) as n->inf = lim(1-sum(9/10^k) for k from 1 to n) as n->inf = lim(1/10^n) as n->inf = 0. That means that (1, 1, 1,...)R(0.9, 0.99, 0.999,...), i.e. that these sequences belong to the same equivalence class of Cauchy sequences of rational numbers with respect to R. In other words, the decimals '0.999...' and '1' refer to the same real number. QED.

I like how compact this one is ;)
Permanently Deleted
- Not quite. The wording "equivalence classes of ... with respect to the relation R: aRb <==> lim( a_n - b_n) as ->inf" is key.
  
  https://en.wikipedia.org/wiki/Equivalence_class
  
  loosely, an equivalence relation is a relation between things in a set that behaves the way we want an equal sign to
  
  For an element in a set, a, the equivalence class of a is the set of all things in the larger set that are equivalent to a.
- So, under the relevant construction of the space of real numbers, every real number is an equivalence class of Cauchy sequences of rational numbers with respect to the relation R outlined in my comment. In other words, under this definition, a real number is an equivalence class that includes all such sequences that for every pair of them the relation R holds (and R is, indeed, an equivalence relation - it is reflexive, symmetric, and transitive, - that is not hard to prove).
  
  We prove that, for the sequences (1, 1, 1,...) and (0.9, 0.99, 0.999,...), the relation R holds, which means that they are both in the same equivalence class of those sequences.
  
  The decimals '1' and '0.999...', under the relevant definition, refer to numbers that are equivalence classes that include the aforementioned sequences as their elements. However, as we have proven, the sequences both belong to the same equivalence class, meaning that the decimals '1' and '0.999...' refer to the same equivalence class of Cauchy sequences of rational numbers with respect to R, i.e. they refer to the same real number, i.e. 0.999... = 1.
Had a little trouble following this as plain text, so I wrote it up in LaTeX (it'll be a bit small if you try to read it inline--you'll probably want to tap to enlarge on mobile or open the image in a new tab/click this direct link to the image):

I tried to hew as closely to your notation as I could, but let me know if you spot any errors!
- Thank you.
  
  And no, I do not notice any errors.
  
  I should re-learn Lateχ.

(I haven't done real math since forever)

Part 1: Proving If x,y are distinct positive real numbers and x < y, there is a z in R such that x < z < y.

Let y = x + e, where e is in R. e is also positive because:

x < y
x < x + e
0 < e

x < z < y becomes x < z < x + e. We can then choose z = x + 0.5e so that

x < z < y
x < x + 0.5e < x + e
0 < 0.5e < e
0.5e - 0.5e < 0.5e + 0 < 0.5e + 0.5e
-0.5e < 0 < 0.5e

Since we've shown earlier that e > 0, -0.5e < 0 < 0.5e is true no matter what e.

Part 2: Decimal Notation

Decimal notation is written as the sequence a_ma_m-1...a₁a₀.b₁b₂...b_n, where a,b are in the set {0,1,2,3,4,5,6,7,8,9}. The sequence a_ma_m-1...a₁a₀.b₁b₂...b_n itself is of the number a_m10^m + a_m-110^m-1 + ... + a₀10⁰ + b₁10^-1 + b₂10^-2 + ... + b_n10^-n.

Part 3: The actual proof through proof by contradiction

Let x = 0.999... and y = 1. This means there exists a z that is between 0.999... and 1. So let's construct z = a₀.b₁b₂...b_n. a₀ has to be either a 0 or 1 and since there is no number smaller than 1 with 1 as its first digit, a₀=0. z = 0.b₁b₂...b_n

0.999... = 9*10^-1 + 9*10^-2 + 9*10^-3 + ... while z = b₁10^-1 + b₂10^-2 + ... + b_n10^-n. If 0.999... < z, then

0.999... - z < 0
9*10^-1 + 9*10^-2 + 9*10^-3 + ... - (b₁10^-1 + b₂10^-2 + ... + b_n10^-n) < 0
(9*10^-1 - b₁10^-1) + (9*10^-2 - b₂10^-2) + (9*10^-3 - b₃10^-3) + ... < 0
(9 - b₁)*10^-1 + (9 - b₂)*10^-2 + (9 - b₃)*10^-3 + ... < 0

But as we've established in Part 2, b₁, b₂, b₃, etc have to be from the set {0,1,2,3,4,5,6,7,8,9}, meaning (9 - b_n) > 0 for any n. Therefore, (9 - b₁)*10^-1 + (9 - b₂)*10^-2 + (9 - b₃)*10^-3 + ... cannot be less than 0.

The theorem requires for x and y to be distinct positive real numbers and x < y, and since 0.999... and 1 can be trivially shown to be positive real numbers, this means that 0.999... and 1 are not distinct. In other words, 0.999... = 1.