.999 repeating equals 1

Discuss, and show your work.

I find it unlikely in the extreme that everyone here thinks they are equal. But I can buy that nobody is willing to try and argue they aren't equal...

1-(10^-infinity) is just 1-(1/infinity). It is un interchangeable.

1/9 = 0.111 repeating
Multiply both sides by 9
9/9 = 0.999 repeating
This is just an example of this that I found interesting

0.999 to infinity will still not terminate to 1.

9.999 to infinity simple won't terminate to the round number 10. Only 9's

We have a taker after all. Here's the simplest proof...and though some mathematicians don't like this proof, they will still tell you that the two values are equal.

1/3 = .333 repeating

2/3 = .666 repeating

3/3 = .999 repeating = 1

https://en.wikipedia.org/wiki/0.999.

Makes sense to me. If 3.3333... and 6.6666... equal 1/3 and 1/6, 0.9999... definitely could round to 1.

Oops I meant 0.3333... and 0.6666...

10-(1/infinity) is not 0.999 ad infinitum. It is just 10-(1/infinity) UNCHANGEABLE.

They are like, in different uninterchangeable Number Demensions. Same as 10-i (where i is the square root of -1) it can not be broken down any further.

.999 repeating and 1 are equal by nomenclature. If you actually tried to run out .999 forever to reach 1 by hand, you would be right...

By convention, writing out .999 repeating (ala with "..." or with the line over it) represents the infinite series of 9s, already "completed", so to speak...which equals 1.

To address Zachy42, there is no rounding involved, it's already 1. Though there would be if you wrote a finite number of nines without using the ellipses or a line over the 9s.

An infinity number of 9's written without a decimal point, say n 9's will never be 10^n, always just (10^n)-1

Case closed!

You are mistaken. You cannot write an infinite number of nines to begin with, but that aside, writing this:

.999...

does not actually represent an ongoing process of writing an infinite number of nines.

The case *is* closed. .999 repeating equals 1. The *number* "0.999..." does not represent a process. It represents a number, and that number is 1. You can write "1" as 1, or you can write "1" as ".999...". Two different representations of the same number.

Saying that ".999..." doesn't equal 1 is like saying that 3/2 doesn't equal 1.5 because the 3 hasn't been divided into 2 yet ...

Look at it this way:

1 - (any nonexistent number) = 1

There is no real number between .999 repeating and 1.

(make sure you watch the entire second video if you are going to use it in an argument)

https://en.wikipedia.org/wiki/0.999.

https://files.eric.ed.gov/fulltext/EJ961516.pdf

9×∑k=1∞10−k:=9limn→∞×∑k=1n10−k=9×(110+1100+11000+…)

https://www.themathdoctors.org/frequently-questioned-answers-0-999-1/

http://backreaction.blogspot.com/2009/05/is-1-0999999.html

https://polymathematics.typepad.com/polymath/2006/06/no_im_sorry_it_.html

Absolutely correct. To add on to what you’ve said, 3/3 is equivalent to one therefore .999 repeating is equivalent to one. 

I actually asked a friend of mi e this question this week to see what he‘d say. Pretty good at math. Said no, blah blah. Proved it to him and he was amazed. Would you say in practice it is not equal but in theory it is?

Fremble's algebraic argument was the one I learned in the sixth grade.