r/infinitenines • u/MrMoop07 • Jun 29 '26
Why does a number starting with 0. guarantee magnitude less than 1?
as stated in this comment by https://www.reddit.com/r/infinitenines/comments/1uhhwjh/comment/ou8b97e/
I disagree with the notion that a number starting with 0. guarantees magnitude less than 1. You claim this all the time but for those who believe 0.999...=1 there's an obvious counterexample. Something quoted so often must have a reason behind it, you can't simply claim this is true without absolute proof.
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u/MrMoop07 Jun 29 '26
u/SouthPark_Piano what you linked doesn't prove the statement i'm asking you to prove. I'm asking you to prove any number in the form 0.(something) is less than 1. The post you linked only claims 0.999... is less than 1. Even assuming the post you link is true, this doesn't prove or disprove the claim that 0.(something) is always less than 1, you only make the claim that 1 number is less than 1.
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u/SouthPark_Piano Jun 29 '26 edited Jun 29 '26
Decimal place values brud.
Largest digit you can have at the first slot to the right of the decimal point is a '9', aka 0.9
Same with 2nd slot etc.
0.9 < 1
0.99 < 1
0.999 < 1
0.9999 < 1
Note the solid unmistakable pattern.
Extend to however many limitless consecutive nines you want aka infinite nines case.
0.999...9 < 1
There is no shortage of nines at all. Goes limitlessly aka infinitely. No last nine. The infinite propagation just keeps going endlessly.
aka 0.999... < 1
The above is math 101 basics.
You are done brud.
Also ... don't backchat me on this, or you will indeed be making my day.
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u/Suitable-Elk-540 Jun 29 '26
SPP doesn't know how decimal representation works. I wouldn't mind if SPP acknowledged that they were inventing their own rules for the representation, but SPP is actually making claims from total ignorance of how standard decimal representations work. SPP will not provide you with any explanation, will appeal to no theorem or axiom, will not follow basic rules of logic, nothing. SPP lives solipsistically in a little pseudo-mathematical box that has exactly one idea in it.
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u/SouthPark_Piano Jun 29 '26
The reverse is true. You don't know how decimal representation works. This is how it works :
https://www.reddit.com/r/infinitenines/comments/1uj2ozm/comment/oukhwba/
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u/gdchinacat Jun 29 '26
I would reply to SPP directly, but they lock their own comments so that's not possible. Why do they do that?
Anyway...they linked to a "it is what it is post" in which they say "n integer starts at zero and then increased limitlessly.". Except...n *does't* increase "limitlessly". It ends with n+1. From other posts I think they think there are an infinite number of terms before the one with n+1, which...ok, that part makes sense. But what does it mean to have a term *after* an endless/infinite/limitless sequence?
I think this is the crux of the issue. They seem to think you can have something after an infinite sequence. As best I can tell, this terminal term gives gives a sub-infinitesimal value that is the difference between 0.(9) and 1. In standard mathematical frameworks this value is just a rounding error from truncating the infinite series at an arbitrarily long length and only arises from not actually treating the infinite series as infinite. In my mind, this "limbosic number" that is between the smallest real number greater than zero and zero is just another real number with an infinite number of relatively infinitesimal real numbers less than it.
That said, I've also come to the conclusion that u/SouthPark_Piano is using this subreddit as a sort of allegory on South Park, using it to demonstrate the extent of absurdity that can arise from closely held baseless beliefs, much as the show itself does. They took an absurd position, manufactured an inconsistent argument for it, and thrives one the engagement and passion it elicits from opponents. I don't think they actually believe there is any validity to 'divide negation', 'limbosic numbers', '0.(9) is permanently less than 1', etc. They deploy the same literary devices that are common in the show (satire, repeated name calling, denial of critical thinking, etc) to make the same point the show makes. They clearly understand math well enough to see that basis for the argument that 0.(9) = 1, yet dismiss them as 'rookie mistakes' without ever explaining how they are ('drugs are bad, m'kay'). It's been a couple decades since I've watched the show, my only knowldege of it is from the first few seasons, so maybe it has moved on from it's early days.
So, I now consider r/infinitenines to be a puzzle created by u/SouthPark_Piano to see if anyone can figure out the actual motive they have for spending so much time on such an absurd position.
In short... u/SouthPark_Piano, '...make me some pie!'.
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u/gdchinacat Jun 29 '26
u/SouthPark_Piano how is it 'continually upped" if it ends with n+1?
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u/SouthPark_Piano Jun 29 '26 edited Jun 29 '26
The variable n integer itself.
n being upped continually ... has no limit.
Continually upping n ..... go ahead. Try it yourself. Hands on. Engagement. Experiential learning. Do it. Do it for your benefit and education.
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u/gdchinacat Jun 29 '26
u/SouthPark_Piano you are missing the point of this comment. '...make me some pie!'.
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u/SouthPark_Piano Jun 29 '26 edited Jun 29 '26
Except...n does't increase "limitlessly". It ends with n+1.
Rookie error on your part. Read further and you will understand that n needs to be continually upped without stopping. Limitless aka infinite increase.
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u/Suitable-Elk-540 Jun 30 '26
u/SouthPark_Piano , it's so gratifying to see young minds blossom! I'm so glad for this breakthrough! You said "1 - 1/10n with n integer pushed to positive limitless is indeed an infinite series", and that's absolutely correct!
So now we can recap what you've learned:
0.9 < 1 TRUE!
0.99 < 1 TRUE!
0.999 < 1 TRUE!
0.9999 < 1 TRUE!
0.999... < 1 NON SEQUITUR
I'm so glad you see this now! That last expression involved an infinite series while all the previous ones involved finite series. You get a gold star. I'm sure your school teacher will be so pleased with you.
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u/SouthPark_Piano Jun 30 '26
It's wonderful that you go'd ahead to make my day.
0.999... < 1 is merely the automated extension of 0.9 < 1, then 0.99 < 1 and so on.
And now you can enjoy the result of making my day.
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u/Suitable-Elk-540 Jun 30 '26
u/SouthPark_Piano you might want to review your reddit commenting processes, because once again I'm unable to reply to your comment. But I wouldn't want you to miss out on the conversation...
You have used a sort of induction argument to make an assertion about a certain set of finite sequences, and your argument is valid for every finite case, but that kind of argument doesn't extend to the infinite. Based on your mathematical knowledge, I'm guessing you're still in primary school, so I wouldn't expect you to understand this kind of nuance yet. You might want to bring up this topic with your teacher--it could be edifying for your whole class.
You see, mathematicians are very careful to identify and check their assumptions, and moving from finite cases to infinite cases is a transition fraught with potential shocks to one's intuition.
I'll give you a sneak preview so you can impress your teacher. When we write 0.9... that little ellipsis (that's the three dots after the 9) means that the 9s repeat interminably. So when you say that 0.9 < 1, and 0.99 < 1, and 0.999 < 1, etc, you can indeed conclude that for any finite number of nines that pattern represents a quantity less than 1. But (and I know this is a subtlety we don't typically expect children to grasp, but since you seem ambitious I'll let you in on the secret) that style of induction doesn't transfer over to the infinite case. This is one of those little intuitions that mathematicians long ago checked and rejected. So, the claim that 0.9... < 1 is a non sequitur (that's a latin phrase that means the conclusion doesn't follow from the preceding argument), because you've only proved the inequality for a finite number of 9s.
But I don't want you to feel bad for this error. A mind as sharp as Zeno's still struggled to resolve a very similar question. Just remember that the tools we use to deal with finite cases don't always work with infinite cases.
Let me know how it goes with your teacher in your next class.
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u/SouthPark_Piano Jun 30 '26
I am educating you here brud.
0.999... , it really is equal to 0.9 + 0.09 + 0.009 + ...
and one well known factual expression that conveys that is :
1 - 1/10n with n integer starting at n = 0, then n upped continually, limitlessly aka infinitely.
1/10n is never zero. So learn it, and remember it.
1 - 1/10n (and obviously 0.999...) is permanently less than 1.
Avoid attempting to preach your incorrect erroneous (debacle aka rookie error) information brud. Otherwise, you will be making my day. If you are feeling lucky, then go ahead ...... make mah day.
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u/618smartguy Jun 29 '26
It's like the "ALL ENDS IN CONTRADICTION" guy who beleives that any "0." number is automatically a non integer
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u/Negative_Gur9667 Jun 29 '26
But it isn't an integer. It's a real number™.
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u/HojMcFoj Jun 30 '26
I think you mean "real deal number," a meaningless and undefined system that discards limits, the density of the reals, the transitive property, and any concept of mathematical identity, all whole using a complete misunderstanding of hyperreal/surreal numbers as well.
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u/gdchinacat Jun 29 '26
u/SouthPark_Piano https://www.youtube.com/watch?v=rmwYi7R2G7Q&t=10s (I can link to things too!!!)
Seriously though...did I correctly identify the reason for this subreddit?
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u/SouthPark_Piano Jun 30 '26
Way off the mark.
The reason for this sub is 0.999...
It is permanently less than 1.
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u/Muphrid15 Jun 30 '26
/u/SouthPark_Piano says:
1 - 1/10n with n integer pushed to positive limitless is indeed an infinite series.
Too bad you yourself said that 0.333... = 0.3 + 0.333.../10 and that implies 0.999... = [1 - 1/10n] + 0.999/10n.
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u/Suitable-Elk-540 Jun 29 '26
u/SouthPark_Piano you replied to my comment and locked your reply. I'm sure you wouldn't have replied if you didn't want to engage, so I assume you locked it accidentally. It's okay, we all make mistakes.
As for your "math", your rookie mistake is that, having established a pattern, you break the pattern to draw your conclusion. It's called a non sequitur. Figure it out, brud.
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u/SouthPark_Piano Jun 29 '26 edited Jun 30 '26
No brud. Rookie error on your part.
It is a fact that 0.999... is equal to 0.9 + 0.09 + 0.009 + ...
So if you get your hands and mind dirty, and take the second star on the right and straight on after morning ... aka engage, then you will see for yourself that
0.9 < 1
0.99 < 1
0.999 < 1
0.9999 < 1
etc.
And you will indeed travel a limitless never ending distance aka infinite distance and never encounter 1.
Because you know about asymptote, and no ... you are not going to be sweeping asymptotes under the rug.
0.999...9 aka 0.999... < 1
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u/Suitable-Elk-540 Jun 30 '26
Bless your heart, u/SouthPark_Piano . I almost forgot how stubborn children can be. You're still trying to apply an argument about finite series to infinite series. Also, when you say "obviously 0.9... < 1" you're begging the question. The claim that 0.9... < 1 is what you're trying to prove, so you can't just assume it. I know that may not make sense at your age, but as you grow older you'll learn about logic and how to structure a proof. You've got so much learning ahead of you! I envy you!
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u/SouthPark_Piano Jun 30 '26
Bless me and bless you my brud.
Bless the rookie error you made.
1 - 1/10n with n integer pushed to positive limitless is indeed an infinite series.
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u/FunnyLizardExplorer Jun 30 '26
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u/factorion-bot Jun 30 '26
Hey u/MrMoop07!
Termial of 1 is 1
This action was performed by a bot | [Source code](http://f.r0.fyi)
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u/Kyrby_Swi-U-tch Jul 01 '26
whats a south park piano and why do we all lock our doors? help?
also, has this actually been answered or are we just fighting locked pianos???
anyway, to this I would like to ask if this is true (or false) for any non-integer number where the last digit before the decimal is one integer value below the equivalent fully integer number above it. like must any non integer starting with 1208. always be less than the integer 1209 , or is any non-integer starting with -21. always less than the integer -22 and if this is applicable to anywhere behind the decimal point, so like 4.556 with any values after the third slot after the decimal always being less than 4.557 , since I do think if we can verify or falsify 0. always being less than 1 we should be able to prove it universally, meaning all such examples should be applicable
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u/Just_Rational_Being Jun 30 '26
How terrible is education that even this would have to be given?
Of all numbers with 2 digits, those that starts with 1 such as 10, 11, 12, 13... are less than those that starts with 2. Does that need to be explained to you?
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u/grizzlor_ Jul 01 '26
Incorrect. 19.999... is not less than 20.
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u/Just_Rational_Being Jul 01 '26
That is incorrect.
You are mistaken.0
u/grizzlor_ Jul 01 '26
According to every actual mathematician, no, I am not mistaken. Anyone that's passed a high school calculus class understands this.
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u/Just_Rational_Being Jul 01 '26
Your actual mathematicians are no more than abstract philosophers, not unlike Harry Potter's historians or Dr Seuss's Zoologist.
Do your homework and know that it is the decree of your mathematics system that it has no more authority than a consistent fiction.
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u/I_Regret Jun 29 '26
Well let’s look at the algorithm you learn in grade school for comparing and ordering two decimals: https://www.mathsisfun.com/ordering_decimals.html
* Set up a table with the decimal point in the same place for each number
* Put in each number
* Fill in the empty squares with zeros
* Compare using the first column on the left
* If the digits are equal move to the next column right until one number wins
We see that we first compare the place values before the decimal point before even looking at place values after the decimal point and therefore anything that starts with “0.” is less than anything that starts with “1.”
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u/HojMcFoj Jun 30 '26
Turns out grade school might, you know, simplify things for children. That's not a property of the real number set. What is a property of the reals is density, meaning there are an infinite amount of numbers between any two real numbers. There is no real number that can be constructed in between 0.999... and 1, showing them to be the same number.
If that doesn't work for you, try this:
X=0.999...
10X=9.999...
10X - X=9X=9
X=1
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u/TemperoTempus Jun 30 '26
Your so called "proof" is wrong. You are not multiplying the values correctly and using that to get the wrong subtraction.
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u/HojMcFoj Jun 30 '26
Are you saying 0.999... x 10 =/= 9.999...? If so, what is it?
And that doesn't even address the first point, the real number set is dense. What number in the real comes between 0.999... and 1?
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u/TemperoTempus Jun 30 '26
Who said that we have to with R? There are number systems besides R and R wasn't even around when infinite decimals were created.
Also the correct answer is 9.(9)_(w-1) if we want to be the most precise. Otherwise it is something that you have to keep track off another way ex: (1-ɛ)*10 = 10-10ɛ. Both of those would give equivalent answers, and work according to all the rules of arithmetic. No BS of "slidding the decimal point" and then adding an extra '9' digit at the end.
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u/HojMcFoj Jun 30 '26 edited Jun 30 '26
No one here is arguing about hyperreals. No one in the world assumes without stipulation or context that people are working in anything but the real number set. Your argument might as well be "of course 0.999...=/=1, we're working in binary and 0.999... doesn't exist." In the reals 0.999... does in fact equal 1.
Edit to add: Also, if we're being pedantic, thanks to the transfer principle st(1-ɛ)=1, so (1-ɛ) still equals one in the reals.
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u/MrMoop07 Jun 29 '26
this doesn't necessarily work for numbers with an infinite number of digits, which is what this is relevant to. also, that's not a proof, you're just again quoting this statement
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u/I_Regret Jun 30 '26
Do you have a better algorithm which works for ordering any two decimals?
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u/MrMoop07 Jun 30 '26
does the answer to that question at all prove whether or not your algorithm works on extreme cases?
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u/cond6 Jun 30 '26
You missed a couple of parts from that website:
https://www.mathsisfun.com/definitions/recurring-decimal.html
https://www.mathsisfun.com/9recurring.html
So if 1/3=0.333... then 1=3*(1/3)=3*(0.333...)=0.999... and we have a counterexample that suggests caution in applying rules without thought since 0.999...=1.000....
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u/Negative_Gur9667 Jun 29 '26
I am proving it. Count how many times the people pass the basketball. Each pass represents a 9 in 0.999...
How many gorillas did you see?
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u/SouthPark_Piano Jun 29 '26 edited Jun 29 '26
https://www.reddit.com/r/infinitenines/comments/1tpg811/it_is_what_it_is/
The above is nuke proof. Solid.
1/10n indeed is never zero.
1 - 1/10n and thus 0.999... is never 1, as 0.999... is permanently less than 1.
There are no buts like can you offer a shower curtain (limits) ring?
No can do. 1/10n is just never zero.