This little guy craves the light of knowledge and wants to know why 0.999… = 1. He wants rigour, but he does accept proofs starting with any sort of premise.

Enlighten him.

13 points

if 0.333… = 1/3 then 0.999… = 3/3 = 1

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10 points

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

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9 points

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

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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…

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Yay, Dedekind cuts are finally useful!

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Deleted by creator
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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.

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3 points
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Im trying to make the distinction between a function that approaches a value as it’s input grows, for instance a sequence seen as a function on the domain of the natural numbers, and the value itself.

I have seen a lot of people view 0.999… as a number that “approaches one”, so formally speaking as the sequence (0.9, 0.99, 0.999, …) and not the number itself which that sequence approaches which they would agree is 1.

The “its the largest number which is less than 1” type of thinking.

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10 points
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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)

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Never thought of employing the geometric series for this, that’s clever.

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6 points
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^_^ thank you!

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3 points

Using i as an index 🤮

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? Thats pretty standard though right?

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6 points
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in computer programs, yes

not so much in analysis

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The symbol ‘i’ is usually reserved for the imaginary unit.

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2 points

Your feedback is valid and I apologize for rendering such an ugly proof

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2 points

And doesn’t this mean 0.99 approaches 1

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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.

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5 points

No. 0.99 is 0.9+0.09. The proof I gave shows that 0.99999999999999999999999999999999999(…) is equal to 1.

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3 points

K

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0.999…=1 because it’s funny that people get so mad about it

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I agree, but the little guy does need rigour, or he will starve under crapitalism.

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6 points
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I didn’t have any bread to bring him but I had plenty of roses

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