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Mind=Blown (lemmy.fmhy.ml)
submitted 3 years ago by to c/memes@lemmy.ml
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[–] [S] 0 points 3 years ago (7 children)

If .99....9=1, then 0.999...8=0.999...9, 0.99...7=0.999...8, and so forth to where 0=1?

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  • [–] 51 points 3 years ago* (last edited 3 years ago) (4 children)

    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.

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  • [–] 2 points 3 years ago (1 child)

    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

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  • [–] 9 points 3 years ago (1 child)

    0.999...8 does not equal 0.999...9 so no

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  • [–] 2 points 3 years ago (1 child)

    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.

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  • [–] 6 points 3 years ago

    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

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  • [–] 4 points 3 years ago (1 child)

    Huh... Where did you get "0.999... = 0.999...8" from? There's a huge difference here.

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  • [–] [S] -4 points 3 years ago* (last edited 3 years ago) (1 child)

    Read carefully. I wrote a finite number(0.999...9)

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  • [–] 5 points 3 years ago

    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

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  • [–] 4 points 3 years ago

    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.

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