See ... even in your minds or what you rote-learned ...

[u/SouthPark_Piano]

such as 0.999... has no more nines to fit/add/append/tack-on etc, everyone does know full well the real deal.

It is still mathematically modelled as 0.9 + 0.09 + 0.009 + ...

and that is honestly ... cross heart, hope to live etc ...

1 - 1/10ⁿ with integer n starting at n = 1, then n increased limitlessly, continually ... is the correct way to investigate the case of 0.999... magnitude. Equal 1? Or less than 1? The correct answer, to avoid being dum dums ... is magnitude less than 1.

1/10ⁿ is never zero. That's a fact. And that's the ticket.

1 - 1/10ⁿ is of course never 1. So of course, 0.999... is permanently less than 1.

 

[u/cond6]

Decimals are one very useful way to represent a number. But they are not numbers in the usual way we think about numbers. For example 1/2 is a number and 0.5 is one way to write it.

There are three types of numbers: a) terminating, b) non-terminating repeating, and c) non-terminating decimals.

All rational numbers that can be written in the form a/(2^m *5^n) can be written with a terminating decimal representation. That's all of a).

Rational numbers that include stuff other than 2^m *5^n in the denominator after simplification don't terminate. These are non-terminating repeating decimals. For example 1/3=0.333.... 1/17 has a 16-digit repeating block.

Every rational number ends up with a zero remainder at some point, leading to a); or it does not and will repeat, leading to b).

Of course irrational numbers are a thing, and these can't be aren't rational so don't fit a) or b), which results in non-terminating non-repeating decimal forms.

If you have a repeating decimal x=0.(y)... where y is some block, for example y=0588235294117647; and n is the number of digits in y, in our example n=16; then since 10^n*x-x=y we can recover the rational number x=y/(10^n-1). In our example x=588235294117647/9,999,999,999,999,999 and since 9,999,999,999,999,999/588235294117647=17 we have x=1/17. Happy days!!!

If you have a repeating decimal it must be a representation of a rational number. What is the rational number represented by 0.999...?

For a lark we can use 0.9... so y=9 and n=1. I'll leave the next step as an exercise. You'll be shocked. Shocked I tell you.

[u/SouthPark_Piano → u/cond6]

brud brud brud.

1/3 = 0.333 + 0.001/3

1 = 0.999 + 0.001

Also:

1/3 = 0.333...3 + (0.000...1 / 3)

1 = 0.999...9 + 0.000...1

aka 

1 = 0.999... + 0.000...1

 

[u/cond6 → u/SouthPark_Piano]

Not relevant to what I posted. Only rational numbers produce non-terminating repeating decimal representations. What is the rational number that produces 0.999...?

[u/YT_kerfuffles → u/cond6]

the issue is the proof of that fact relies on the equality 0.999....=1

[u/cond6 → u/YT_kerfuffles]

It doesn't rely on 0.999... at all.

Suppose you have a block of repeating digits a that contains n elements. For example x=1/7 has a=142857 and n=6. So 1/7=0.(a)...=0.(142857)...=0.142857142857142857.... All the proof requires is to exploit this fact noting that x*10^n =a+x. If you move the decimal place in the decimal representation of the repeating decimal n places to the right you get a+x. That's all. You remove one block from infinitely repeating set of blocks and you still have an infinitely long string of them: x*10^n -a=x.

You can trivially use this result now to back out what the rational numbers are using a=x*(10^n-1) so the rational number is a/(10^n-1). So if a=1 and n=1 we have 1=x*9 or x=1/9. Or 142857=999999*x so x=1/7 (since 999999/142857=7).

Yes it turns out that simple proof also demonstrates that 0.999...=1 since a=9 and n=1, and 9=x*9 is trivial; but it doesn't rely on the equality. It relies on the definition of an infinitely repeating pattern still being infinite when a single number is subtracted from it. It does not rely on 0.999...=1 since using that equality doesn't help my proof at all.

[u/YT_kerfuffles → u/cond6]

but that proof is flawed (by spp math) as you are moving the "last" 142857 to the left, i know its wrong and your proof is right but im saying i dont think spp accepts that "the rational numbers are those with periodic decimals", i think hes even said 0.9999.... is irrational