this post was submitted on 17 Jul 2023
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Actually 0.99... is the same as 1. They both represent the same number
https://en.m.wikipedia.org/wiki/0.999...
If .99....9=1, then 0.999...8=0.999...9, 0.99...7=0.999...8, and so forth to where 0=1?
The tricky part is that there is no 0.999...9 because there is no last digit 9. It just keeps going forever.
If you are interested in the proof of why 0.999999999... = 1:
0.9999999... / 10 = 0.09999999... You can divide the number by 10 by adding a 0 to the first decimal place.
0.9999999... - 0.09999999... = 0.9 because the digit 9 in the second, third, fourth, ... decimal places cancel each other out.
Let's pretend there is a finite way to write 0.9999999..., but we do not know what it is yet. Let's call it x. According to the above calculations x - x/10 = 0.9 must be true. That means 0.9x = 0.9. dividing both sides by 0.9, the answer is x = 1.
The reason you can't abuse this to prove 0=1 as you suggested, is because this proof relies on an infinite number of 9 digits cancelling each other out. The number you mentioned is 0.9999...8. That could be a number with lots of lots of decimal places, but there has to be a last digit 8 eventually, so by definition it is not an infinite amount of 9 digits before. A number with infinite digits and then another digit in the end can not exist, because infinity does not end.
Maybe a stupid question, but can you even divide a number with infinite decimals?
I know you can find ratios of other infinitely repeating numbers by dividing them by 9,99,999, etc., divide those, and then write it as a decimal.
For example 0.17171717.../3
(17/99)/3 = 17/(99*3) = 17/297
but with 9 that would just be... one? 9/9=1
That in itself sounds like a basis for a proof but idk
Yes that's essentially the proof I learned in high school. 9/9=1. I believe there's multiple ways to go about it.