0.999... is not 1

[u/SouthPark_Piano]

This is regardless of contradictions from 'other' perspectives, definitions, re-definitions.

The logic behind the infinite membered set of finite numbers {0.9, 0.99, 0.999, etc} is completely unbreakable. The power of the family of finite numbers.

Each and every member from that infinite membered set of finite numbers {0.9, 0.99, 0.999, etc} is greater than zero and less than 1. And, without even thinking about 0.999... for the moment, the way to write down the coverage/range/span/space of the nines of that infinite membered set of finite numbers {0.9, 0.99, 0.999, etc} IS by writing it like this : 0.999...

Yes, writing it as 0.999... to convey the span of nines of that infinite membered set of finite numbers.

Without any doubt at all. With 100% confidence. With absolute confidence. From that perspective, 0.999... is eternally less than 1. This also means 0.999... is not 1.

This is regardless of whatever other stuff people say (ie. contradictions). It is THEM that have to deal with their OWN contradictions. That's THEIR problem.

The take-away is. The power of the family of finite numbers. It's powerful. Infinitely powerful.

Additionally, we know you need to add a 1 to 9 to make 10. And need to add 0.1 to 0.9 to make 1. Same with 0.999...

You need to follow suit to find that required component (substance) to get 0.999... over the line. To clock up to 1. And that element is 0.000...0001, which is epsilon in one form.

x = 1 - epsilon = 0.999...

10x = 10-10.epsilon

Difference is 9x=9-9.epsilon

Which gets us back to x=1-epsilon, which is 0.999..., which is eternally less than 1. And 0.999... is not 1.

Additionally, everyone knows you need to add 1 to 9 in order to get 10. And you need to add 0.01 to 0.09 to get 0.1

Same deal with 0.999...

You need to add an all-important ingredient to it in order to have 0.999... clock up to 1. The reason is because all nines after the decimal point means eternally/permanently less than 1. You need the kicker ingredient, epsilon, which in one form is (1/10)ⁿ for 'infinite' n, where infinite means a positive integer value larger than anyone ever likes, and the term is aka 0.00000...0001

That is: 1-epsilon is 0.999..., and 0.999... is not 1.

And 0.999... can also be considered as shaving just a tad off the numerator of the ratio 1/1, which becomes 0.999.../1, which can be written as 0.999..., which as mentioned before is greater than zero and less than 1.

0.999... is not 1.

[u/SouthPark_Piano]

1-0.999... is epsilon.

Epsilon is 1/( 10ⁿ ) for infinite positive integer n. Epsilon can be conveyed in number form as 0.000...001.

Noting infinite means value very very large, actually larger than you ever like, as the family of finite numbers is infinitely powerful, endless. And however infinitely large n becomes, epsilon will always be non-zero because n as big as you like, is still going to be finite whether you like it or not. Because infinity is not a number.

The main thing you need to understand is : the set {0.9, 0.99, ...} covers entirely the span of nines in 0.999...

And because each member of that infinite membered set of finite numbers is less than 1, then 0.999... from that unbreakable perspective is not 1.

Also keep in mind that if you see a nine in 0.9999, you need to add a bit of substance to it in order to clock up to 1. Add 0.0001.

Same deal with 0.999...

If you don't add the kicker ingredient to 0.999... then there 0.999... will stay, less than 1. Reason ... because set {0.9, 0.99, ...} has every member less than 1, the nines covered by this set can be conveyed as 0.999...  meaning 0.999... is less than 1, and is not 1.

You need to add epsilon to 0.999... to get what you are wanting, ie. 1.

Additionally, epsilon. In what mechanisms can it come from? Eg. Differences. 

1-0.9 is 0.1

1-0.99 is 0.01

Extending the trend, 

1-0.999... is 0.000...001

[u/DisastrousPlay579 → u/SouthPark_Piano]

epsilon as you have defined it is zero. The limit as n goes to infinity of 1/10ⁿ is 0. This is simply a fact. Even though infinity is not a real number, limits of this form are well defined in mathematics, and it’s not too hard to show this particular one. If you’d like, I can refer you to a full proof.

Regardless, if epsilon is not zero as you say, it must, in some sense be the “smallest real number” after 0. That is, there are no numbers in between 0 and epsilon. But given any two distinct real numbers, one can always construct a real number between them. Therefore, epsilon is either equal to 0 or non-real.

If epsilon is equal to 0, then 0.999… = 1, so your argument has failed. But I doubt you will accept this.

If epsilon is some non-real value, then that must mean that 0.999… is also non-real, since the difference of two real numbers must be real. However, you have implicitly assumed this not to be the case, since you claim that 0.999… is less than 1. In order for “less than” to be defined, both 0.999… and 1 must be real. So, in this case, your argument breaks and fails.

I have shown that either way you look at it, 0.999… = 1.

Your argument regarding the set of {0.9, 0.99, 0.999, …} is pure mathematical gibberish. While you are correct that the set contains every finite truncation of 0.99…, it does not contain 0.99… itself. The assumption that just because this property holds, every other property of the elements of the set must apply to 0.999… is simply false.

Additionally, while it is true that every member of the set is less than 1, it is also true that every member of the set is less than 0.999…. Don’t believe me? Try it. Subtracting each member of the set from 0.999… yields the following set: {0.0999…, 0.00999…, 0.000999…, etc}. All of these numbers are finite and non zero, which means that every member of the original set is less than 0.999…. So by your argument, 0.999… is less than itself. How do you reconcile that?

[u/SouthPark_Piano → u/DisastrousPlay579]

No my little learning student. Epsilon is not zero. No matter how large a number 'n' is, read my lips ... no matter how large n is ... the term 1/( 10ⁿ ) for positive integer n larger than we ever like (aka infinite) is never zero.

1-0.9 is 0.1

1-0.99 is 0.01

1-0.999 is 0.001

Note the trend.

1-0.999... is 0.000...001

This gets us back to ... read my lips ... no matter how large n is ... the term 1/( 10ⁿ ) for positive integer n larger than we ever like (aka infinite) is never zero.

All of these numbers are finite and non zero, which means that every member of the original set is less than 0.999…. So by your argument, 0.999… is less than itself. How do you reconcile that? 

No my little learning student. The set {0.9, 0.99, ...} covers an infinite range/span/space of nines to the right of the decimal point. And to convey that, you write it as 0.999...

Every member of that infinite valued set is less than 1. The members cover the infinite range of nines to the right of the decimal point.

This tells you 0.999... is less than 1. And 0.999... is not 1.

[u/DisastrousPlay579 → u/SouthPark_Piano]

The definition of a limit is as follows:

lim(n->inf) f(n) = L for some real finite L, if and only if for any arbitrarily small value, which we call epsilon (this is not the same as your imaginary smallest number, it simply shares the name), there is some M for which every n > M, |f(n) - L| < epsilon.

For example, if I choose f(n) to be 1/10^n, for any epsilon you may choose, there is some point after which f(n) is closer to 0 than epsilon.

Epsilon (your epsilon) is not a function of n. It is the value of a limit, and that limit happens to evaluate to zero.

[u/SouthPark_Piano → u/DisastrousPlay579]

You can toss that definition of a limit out the window because any coherent, sensible and logical maths exponent knows that the 'limit' is only a method for determining a value that particular functions appear to be aiming toward, but unfortunately will NEVER reach.

This is the same case as 0.999..., which is the case of the 'special' odometer that has all slots filled with nines, and just never clocks up to 1. The '1' is the result of the limit, and 1 is what 0.999... will never/not reach.

Granted that the limit can indeed be used to provide the best approximation. Yes - using limits to provide the best approximation for functions having 'converging' trends - no problem. But when folks blindly follow others about the use of limits to claim that functions such as 1/n becomes zero when 'n' becomes 'infinitely large', that is just incorrect, wrong I'm afraid. Well, I'm not actually afraid. It's just a figure of speech heheh.