0.9
0.99
0.999
0.9999
etc
extend to limitless
0.999...9 aka 0.999...
An infinite quantity of finite numbers from this family.
The extreme member 0.999...9 aka 0.999... is indeed also less than 1 in both value and magnitude.
It's a done deal.
1 is approximately 0.999...
From a recent post:
The number of nines in 0.999... is limitless, aka infinite aka inexhaustible.
It does not matter how many nines there are, 'infinite' quantity or not.
The "0." prefix guarantees magnitude less than 1.
The only way you can get a 1 from a (0.999... + x) operation is to add a /10^n scaled down version of '1' (ie. the 'x') to a limbo nine in 0.999...9 aka 0.999...
You will never get a '1' from 0.999... itself, because afterall, 0.999... is equal to 0.9 + 0.09 + 0.009 + ...
which is 1 - 1/10^n for the case integer n pushed to limitless aka infinite. And 1/10^n is never zero is an unbreakable fact.
So 0.999... is never 1 because 1/10^n is never zero.
From a recent post:
0.999...
It never has been 1 and it never will be 1.
Decimal place value governs that.
It does not matter how many consecutive nines there are for 0.999... , infinite length of consecutive nines or not.
0.999... is indeed expressable as the infinite summation 0.9 + 0.09 + 0.009 + ... aka 1 - 1/10^n with n integer pushed to limitless aka infinite.
And 1/10^n is never zero.
It means with zero uncertainty that 1 - 1/10^n for the case n integer pushed to limitless , is never 1.
It means 0.999... is never 1.
From a recent post:
0.999... = 0.999...9 = 0.9 + 0.09 + 0.009 + ...
= 1 - 1 /10^n with n n integer starting at n = 1, then n increased continually, limitlessly aka infinitely where the pre-requisite number of nines that begins to qualify 0.999...9 is the largest number of consecutive nines to the right of "0." that you can or cannot generate with your brain, and from there --- n continues to increase limitlessly aka infinitely.
1/10^n is never zero.
1 - 1/10^n is never 1.
0.999...9 aka 0.999... is never 1.
From a recent post:
0.999... and 1.
They don't represent the same value fortunately.
In the same way that it is necessary to add a 0.1 to 0.9 to get 1,
and need to add 0.01 to 0.99 to get 1,
it is necessary to add a limbo 1 to 0.999... to get 1.
Otherwise, if you can't add a limbo 1 to any limbo nine in 0.999... , then it's a case of tough luck, because 0.999... isn't 1 in the first place. And to get a 1, you need to add a limbo 1. Which then gets us back to the beginning, ie. 0.999... isn't 1 to begin with anyway.
From a recent post:
Start writing the digits of 0.999...
Begin with 0.9999999
and keep tacking on those nines, as it is well known that 0.999... is equal to 0.9 + 0.09 + 0.009 + ...
Keep going until you prove you can get a result of 1 without adding any version of 1/10^n to your progressively increasing value.
From a recent post:
You better understand that if you reckon mistakenly that 1 is 0.999... , aka 1 being written in two eays, then you better the hell show the 'other way' of writing the square root of 1.
From a recent post.
f(x) = 1 + x + x^2 + x^3 + ... + x^n-1 + x^n
= sum of terms x^n with n integer starting at n = 0 then increased continually limitlessly aka infinitely.
For f(x) = 1 + [ x + x^2 + x^3 + ... + x^n-1 + x^n ] , n keeps increasing.
x.f(x) = [ x + x^2 + x^3 + x^4 + ... + x^n-1 + x^n ] + x^n+1
f(x) - x.f(x) = 1 - x^n+1
(1 - x)f(x) = 1 - x^n+1
f(x) = ( 1 - x^n+1 )/(1 - x)
For case x being fractional and having magnitude greater than zero AND less than 1, f(x) is never equal to 1/(1-x) because for fractional x with magnitude greater than zero and less than 1, x^n+1 is never zero.
f(x) being approximately 1/(1-x) , that we can accept.
This person attempted to 'defend' very unsuccessfully their rookie error by proving they made a rookie error.
https://lcamtuf.substack.com/p/09999-1
They wrote :
1 - 1/10^n , and wrote that you can't plug n equal 'infinity' into it.
Their rookie error is in their misunderstanding of infinite n.
Infinite n means continually upping n without limit.
So when n integer starts at n = 1, making n limitless means pushing n value higher and higher, at ultra extreme infinite rate if youS like. Continual increase. Limitless increase.
There is no shortage of integer n, and you can and will keep increasing n until the cows never come home.
1/10^n is never zero for anyone or anything.
1 - 1/10^n (hence 0.999...9 aka 0.999...) is permanently less than 1 because 1/10^n is permanently greater than zero.
While it is just a figure of speech and I don't support materialistic wannabes etc,
from a recent post:
https://www.reddit.com/r/infinitenines/comments/1t6fwmn/comment/okhqnfc/
The "0." prefix guarantees magnitude less than 1.
0.9 is less than 1, gap 0.1
0.99 is less than 1, gap 0.01
0.999 is less than 1, gap 0.001
0.999...9 aka 0.999... is less than 1, gap 0.000...1
0.999... is equal to 0.9 + 0.09 + 0.009 + ...
conveyed accurately as
1 - 1/10^n with n integer beginning at n = 1, then n increased continually, limitlessly aka infinitely.
1/10^n is permanently greater than zero. 1 - 1/10^n is permanently less than 1.
0.999... is permanently less than 1 because 1 - 1/10^n is permanently less than 1, because 1/10^n is never zero.
1 = [ 1 - 1/10^n ] + 1/10^n for the case n integer starting at n = 1, then n increased continually, limitlessly aka infinitely, gives you the golden factual equation:
1 = 0.999...9 + 0.000...1
aka
1 = 0.999... + 0.000...1
aka 1 - 0.999... = 0.000...1
The gap 0.000...1 is never zero.
1 and youS cannot get away with your blunder aka debacle aka rookie error.
It does not matter if there are no more nines (or not) to 'fit' to the right of the "0." prefix.
0.9 must have something added to it to get a 1 RESULT. You need to add 0.1 to get a result of 1.
0.99 must have 0.01 added to it to get a result of 1.
And so on.
0.999... must have something added to it to get a 1 result. Having only nines to the right of the decimal point does absolutely not mean than 1 is equal to 0.999...
No addition of a limbo kicker to 0.999... , no 1 result. That is, no-can-do.
Let that be an important lesson for youS, those that made the rookie errors in blindly and incorrectly and shamefully and embarrassingly shot yourselves in the foot with the 1 equalling 0.999... basic of basics mistake.
This is actually known. But the rookie error makers keep trying unsuccessfully to do a cover-up.
0.999... is 0.9 + 0.09 + 0.009 + ...
The limitless aka infinite sum described by one well-known expression:
1 - 1/10^n with n integer starting at n = 1, and n increased continually, limitlessly aka infinitely.
As the numbers of n is limitless, and correspondingly the numbers of 1/10^n is limitless, then the limitless has no limit at all.
Also, obviously, 1/10^n is a scaling down operation that has a non-zero result for scaling down of numbers aside from zero.
So 1 - 1/10^n is never 1 because 1/10^n is never zero.
So 0.999... is never 1.
0.999... is indeed not 1.
And this is what Terence needs to understand. And it is what most of youS rookie error makers need to remember permanently.
1 is NOT 0.999...
1 is approximately 0.999...
The "0." prefix in 0.999... guarantees magnitude less than 1.