A (correct proof) of 0.99=1 | Sequences and Limits | Math Basics | Math2Go

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In this video we will look into the well known fact that 0.99 period ("repeating") equals 1! As alot of proofs out there are mathematically wrong, we will take a different approach and prove the statement using sequences and limits.

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Noo! In the sequence x_n we are not taking the limit of x facing infinity, but the limit of *n* going towards infinity!

pseudoexpertise
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I see further problems with the definition of limits and how math makes infinity finite.

I am using middleschool basics to show that ".99..." does not exactly equal 1.

Real numbers are unique and precise.

Decimal place notation has been used to represent numbers for over 20 centuries.

In decimal place notation, integers do not have nonzero digits to the right of the decimal point. ".99..." should not represent a integer.

Real numbers are a continuum that consists of zero, natural, integers, rational and Irrational numbers.

A rational number is the ratio of 2 integers. There are no two integers that result in ".99...".

The limit, while by definition, has to be a real number, the limit does not have to exactly equal the entity, function or sequence it refers to. ".99..." can't be a limit. (Nobody has said it was.)

The main problem is how math treats infinity. Infinity by simple definition:
Incomplete, (never ending)
Inconsistent, (different "sizes")
and imprecise (has no boundries)

It is not possible to represent every real number in decimal (or any number base) placed number notation. What is the number adjacent to 1. The nature of the real numbers being unique and precise and they are a continuum, there is no way to conceive of a number next to 1 in base placed number notation.

If you raise ".99..." to increasing n powers you get
.99...980...01" not simply ".99..." or 1

johnlabonte-chul