607
Mind=Blown
(lemmy.fmhy.ml)
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
this post was submitted on 17 Jul 2023
607 points (92.4% liked)
Memes
45535 readers
204 users here now
Rules:
- Be civil and nice.
- Try not to excessively repost, as a rule of thumb, wait at least 2 months to do it if you have to.
founded 5 years ago
MODERATORS
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.
Wonderful explanation. It got the point across.
That is the best way to describe this problem I've ever heard, this is beautiful
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.
This is the kind of stuff I love to read about. Very cool
0.999...8 does not equal 0.999...9 so no
If the "…" means 'repeats without end' here, then saying "there's an 8 after" or "the final 9" is a contradiction as there is no such end to get to.
There are cases where "…" is a finite sequence, such as "1, 2, … 99, 100". But this is not one of them.
I'm aware, I was trying to use the same notation that he was so it might be easier for him to understand
Your way of thinking makes sense but you're interpreting it wrong.
If you can round up and say "0,9_ = 1" , then why can't you round down and repeat until "0 = 1"? The thing is, there's no rounding up, the 0,0...1 that you're adding is infinitely small (inexistent).
It looks a lot less unintuitive if you use fractions:
1/3 = 0.3_
0.3_ * 3 = 0.9_
0.9_ = 3/3 = 1
No, because that would imply that infinity has an end. 0.999… = 1 because there are an infinite number of 9s. There isn’t a last 9, and therefore the decimal is equal to 1. Because there are an infinite number of 9s, you can’t put an 8 or 7 at the end, because there is literally no end. The principle of 0.999… = 1 cannot extend to the point point where 0 = 1 because that’s not infinity works.
Huh... Where did you get "0.999... = 0.999...8" from? There's a huge difference here.
Read carefully. I wrote a finite number(0.999...9)
Think carefully
What does 0.99…8 represent to you exactly
If it’s an infinite amount of 9s then it can’t end in an 8 because there’s an infinite amount of 9s by definition so it’s not a real number
There is no .99...8.
The ... implies continuing to infinity, but even if it didn't, the "8" would be the end, so not an infinitely repeating decimal.
If you really wants to understand the concept , you need to learn about limits