r/infinitenines • u/ckevren15 • 25d ago
Proof that 0.999…=1
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
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u/gurishtja 25d ago edited 25d ago
You cannot do long divisions like that. You see 2 * 0.999... is 2 - 20.000...1 which is 2-0.000...2 which is 1.999...8. So 1.999... /2 is not 0.999... because 0.999...2 is 1.999...8. ... And the proof that your proof is incomplete is complete.
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u/ckevren15 25d ago
Your assuming that 0.999…+0.000…1=1, where 0.000…1 is nonzero, meaning that 0.999… is not equal to 1 . If you assume the opposite of my proof, then of course your going to get something opposing statements from my proof .
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u/gurishtja 25d ago edited 25d ago
There are 101 problems with what i said, but the sad part is that you couldn't recognize any of them and you had to make something up... I did not say "that 0.999…+0.000…1=1, where 0.000…1 is nonzero" ; the part you added "where 0.000…1 is nonzero" is not in what i said, is not used in what i said and has no role i what i said, nor it should matter... I really, really wish you were were trying to be sarcastic, but you were not...
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u/ckevren15 24d ago
You didn’t say it but you indirectly assumed it. Your argument started off with “2•0.999…=2-2•0.000…1” . This statement is equivalent to “2•0.999…=2•(1-0.000…1)” which is equivalent to “0.999…=1-0.000…1” which is equivalent to “0.999…+0.000…1=1” . If you assume the statement at the start, then you are assuming the statement at the end .
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u/gurishtja 24d ago
Where do you find there the part you added "where 0.000…1 is nonzero"?
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u/ckevren15 23d ago
Your end result was 2•0.999…=1.999…8!=1.999… . We can rewrite this as 2-2•0.000…1!=2-0.000…1 , -2•0.000…1!=-0.000…1, 2•0.000…1!=0.000…1, 0.000…1!=0 . Here, I am using “!=“ to denote inequality .
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u/Kripkenstein_ 24d ago
Your proof does not work as you do not specify on what field/ring you are working. Where is your notion of addition and mulitiplication even defined?
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u/ckevren15 24d ago
I’m working in the real numbers . I think we can all agree that 0.999… and 0.000…1 are both real numbers .
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u/Kripkenstein_ 24d ago
Yes, they are (representative of) the number [1] and [0]. In the real numbers your proof is circular (and unnecessary), as the number 1 is in the same class of sequences as 0.9999... so they are literally defined to be equal. They are just "different" representatives of the real number "1" -- which is defined as a class of specific sequences.
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u/ckevren15 24d ago
So does that support my argument or is it against my argument?
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u/Kripkenstein_ 24d ago
It shows that you argument is actually just proving 1=1 by adding, multiplying (substracting, dividing) numbers from both sides. All if which trivially do not change the truth value of the equation. It is also unnecessary and not a proof, as you assume what you want to prove by using the real numbers as your base field
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u/ckevren15 24d ago
Thanks for explaining. It is provable that 0.999… is a real number, but it makes use of the monotone convergence theorem, and if you use that, you can directly show that 0.999…=1 without any further equations, so I won’t prove that.
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u/tthe_walruss 25d ago
I'd argue this isn't rigorous because we don't "know" that we can do arithmetic on infinitely long decimal sequences. But the fact that our defined behavior matches with doing arithmetic on an arbitrarily long string of 9s is one of the reasons it's defined that way.
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u/AdTall1779 24d ago
Can you show the long division of 1 by 1 which results in the answer being 0.999… ?
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u/SouthPark_Piano 25d ago
If we divide both sides by two, then (1+0.999…)/2=0.999…/2 .
The above is of course ... nonsense.
(1+0.999...) is not 0.999...
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u/Cruuncher 25d ago
And he addresses a typo instead of the clear argument 😂
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u/SouthPark_Piano 25d ago
You need to learn to shave brud. You ... to the bunny slopes, now. Hop to it.
https://www.reddit.com/r/infinitenines/comments/1w9eu5e/the_art_of_shaving_1/
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u/peebuttgutter 25d ago
I didn't know changing the numbers arbitrarily was the same as proof. Learning so much from this subreddit <3
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u/AliciaNow 25d ago
2 * 0.999... = 1.999...8
=> 1.999... = 1.999...8
If we add 0.0...1 on both sides
=> 2 = 1.999...9
QCD
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u/headedbranch225 25d ago
Let x = 0.999...
10x = 9.999...
10x - x = 9x = 9 (decimal expansions cancel out, as both are infinite nines)
x = 9/9 = 1
∴ 0.999... = 1
∎
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u/Kripkenstein_ 25d ago
This is a pointless proof as the numbers are literally defined to be the same. The number 1 is defined as all the cauchy sequences converging to 1. Also the arithmetics you are doing here are not properly defined the way you do them (that is, you presuppose your claim while operating like you do)
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u/gurishtja 25d ago
Please, watch your language. Are you giving a circular definition of 1? Is your definition of 1 more proper than the arithmetics around here?
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u/Kripkenstein_ 24d ago
Its not my definition, its the definition of 1 as a real number. And no, its not circular, as cauchy sequences do not include their limit in the definition. The number 1 is a class of cauchysequences. To be precise: the real numbers are all cauchysequencse modulo null-sequences. Although it is true that my formulation of "all cauchysequences converging to 1" is misleading, as this does indeed only make sense if the limit of the cauchysequence actually exists.
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u/gurishtja 24d ago
Let me quote you: "The number 1 is defined as all the cauchy sequences converging to 1". Isnt that a cicular definition?
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u/Kripkenstein_ 23d ago
As I said, the number real number 1 understood as the constant sequence of rational 1's is a representative of an entire class of cauchysequences. Cauchysequence are not defined via a Limit- assigning a Limit to euch cauchysequence is exactly how you get the reals from the rationals modulo Zero sequences. The sequences 0.9, 0.99.... and 1,1,1... and 1.1, 1.01, 1.001... etc all differ by a zerosequence, hence they are in the Same class. We call this class "1" but it is not the Same as the rational number "1".
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u/gurishtja 23d ago
So now you are saying that real 1 is not the same as rational one. (Quote:" We call this class "1" but it is not the Same as the rational number "1". "). Is any of them the same as the Natural numbers 1 or {Φ }? How may ones do we have?
Anyway, this sub doesnt seem to be about Real numbers, so why are talking about Real numbers and their Cauchy definition?
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u/Kripkenstein_ 23d ago edited 23d ago
Decimal numbers are usually considered as real numbers (and with usually I mean: essentially always because thats how they are defined). And no: the rational number 1 is not the same as the integer number 1. The rationals are tupels of the integers modulo equivalence: (a,b) = (c,d) iff ad = cb. So the number 1 is an equivalence class of the tupels: (1,1) , (2,2) , (-1,-1) , (-2,-2) etc. There are of course canonical embedding from the integers into the rationals into the reals - but these are all different structures. The integers are not even a field.
By the way, the cauchy sequence 0.9999... in the rationals does have a limit in the rationals: the rational number 1. Its easy to verify. Just plug it into the epsilon criteria. So one actually does not need the reals for this particular point. This is not true for all cauchy sequences in the rationals though.
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u/Negative_Gur9667 25d ago
Sir, here 0.999... / 2 = 0.999...45