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0.999...= 1

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Re: 0.999...= 1

#641

Post by gspr »

TheRealPomax wrote: post #639
gspr wrote: post #624

Earlier quoted context omitted.

My point is that it cannot be clear what assuming (1) entails when you aren't properly defining the quantities involved. You have to answer in clear and mathematical language what the quantity in (1) is defined as. What is the definition of "0.666…7"? As it currently stands, assumption (1) is similar in nature to me saying "gnarfgnarf is an imaginary number". It's completely meaningles unless I define what I mean by…

No, it isn't. The idea that saying "gnarfgnarf is an imaginary number is completely meaningless" is the opposite of true: if you assert that gnarfgnarf is an imaginary number, that is the definition we'll be using for the remainder of whatever proof we use that in . Anywhere the proof now talks about gnarfgnarf, we're talking about something that is an imaginary number, and has to follow all the rules that imaginary…

I'm really sorry, I jumped to conclusions as to what you were saying without properly reading the comment you linked to!

Re: 0.999...= 1

#642

Post by nybble41 »

drran wrote: post #640
nybble41 wrote: post #612

Earlier quoted context omitted.

You could just as easily say that 0.000…54234 can be written as 1/∞. Surreal Numbers is a bit of a detour in this case. The premise of the idea that 1 - 0.999… could be written 0.000…1 is the mistaken concept that there is some point "after an infinite number of steps" where the expansion of 0.999… stops and you can leave the remaining 1. The expansion never stops and there is no final remainder. The result is 0.000……

x = 0.999... x = 9/10 + 9/10^2 + 9/10^3 ... 9/10^inf x = 9/10 + (9/10 + 9/10^2 + 9/10^3 ... 9/10^(inf-1))/10 x = 9/10 + (x - 9/10^inf)/10 x - 9/10 = (x - 9/10^inf)/10 10x - 9 = x - 9/10^inf 9x - 9 + 9/10^inf = 0 9x = 9 - 9/10^inf x = 1 - 1/10^inf x = 1 - 0.000...1

    x = 9/10 + 9/10^2 + 9/10^3 ... 9/10^inf
    x = 9/10 + (9/10 + 9/10^2 + 9/10^3 ... 9/10^(inf-1))/10
This is exactly the issue I was referring to. You're assuming the sequence stops "at infinity" but infinity is not a concrete number of steps, it's the absence of any end condition. Subtracting one step from "no end condition" is nonsense. The sequence (0.999… - 0.9)×10 does not end earlier than 0.999…; these are exactly the same sequence, repeating 9s without end. The difference between them is zero in every digit, with no trailing 1.

Re: 0.999...= 1

#643

Post by [deleted] »

gspr wrote: post #596
alpple wrote: post #313

Why do we accept .999... as a valid notation. Why not only allow 1 to denote this concept?

> Why do we accept .999... as a valid notation. Why not only allow 1 to denote this concept? That would be adding a special rule for purely cosmetic reasons. It's typically not done in mathematics, where concise rules are usually more cherished than special-casing things.

[deleted]

Re: 0.999...= 1

#644

Post by drran »

nybble41 wrote: post #642
drran wrote: post #640

Earlier quoted context omitted.

x = 0.999... x = 9/10 + 9/10^2 + 9/10^3 ... 9/10^inf x = 9/10 + (9/10 + 9/10^2 + 9/10^3 ... 9/10^(inf-1))/10 x = 9/10 + (x - 9/10^inf)/10 x - 9/10 = (x - 9/10^inf)/10 10x - 9 = x - 9/10^inf 9x - 9 + 9/10^inf = 0 9x = 9 - 9/10^inf x = 1 - 1/10^inf x = 1 - 0.000...1

x = 9/10 + 9/10^2 + 9/10^3 ... 9/10^inf x = 9/10 + (9/10 + 9/10^2 + 9/10^3 ... 9/10^(inf-1))/10 This is exactly the issue I was referring to. You're assuming the sequence stops "at infinity" but infinity is not a concrete number of steps, it's the absence of any end condition. Subtracting one step from "no end condition" is nonsense. The sequence (0.999… - 0.9)×10 does not end earlier than 0.999…; these are exactly t…

The sequence doesn't end earlier, but it does start later by one element, so they are not exactly same sequences. One infinite sequence has one more element than another infinite sequence, so difference is 1/inf.

Re: 0.999...= 1

#645

Post by nybble41 »

drran wrote: post #644
nybble41 wrote: post #642

Earlier quoted context omitted.

x = 9/10 + 9/10^2 + 9/10^3 ... 9/10^inf x = 9/10 + (9/10 + 9/10^2 + 9/10^3 ... 9/10^(inf-1))/10 This is exactly the issue I was referring to. You're assuming the sequence stops "at infinity" but infinity is not a concrete number of steps, it's the absence of any end condition. Subtracting one step from "no end condition" is nonsense. The sequence (0.999… - 0.9)×10 does not end earlier than 0.999…; these are exactly t…

The sequence doesn't end earlier, but it does start later by one element, so they are not exactly same sequences. One infinite sequence has one more element than another infinite sequence, so difference is 1/inf.

No, they start at the same element (they're both 0.999… and thus start with 9/10) and have the same (infinite) "number" of elements. If you lined them up digit by digit there is never a point where one digit is a 9 and the other is a 0.

Re: 0.999...= 1

#646

Post by nybble41 »

nybble41 wrote: post #645
drran wrote: post #644

Earlier quoted context omitted.

The sequence doesn't end earlier, but it does start later by one element, so they are not exactly same sequences. One infinite sequence has one more element than another infinite sequence, so difference is 1/inf.

No, they start at the same element (they're both 0.999… and thus start with 9/10) and have the same (infinite) "number" of elements. If you lined them up digit by digit there is never a point where one digit is a 9 and the other is a 0.

Incidentally, this confusion is part of the reason why actual infinite sequences are written with a trailing ellipsis (9/10 + 9/10² + 9/10³ + …) and not a final term involving infinity. There is no final term—not even 9÷10^∞. So the correct way to write the separation of the leading term in your formula is:

    x = 9÷10 + 9÷10² + 9÷10³ + …
    x = 9÷10 + (9÷10 + 9÷10² + …) ÷ 10
Note that the number of elements in the sequence is the same no matter how many leading terms you write, so long as the pattern doesn't change. The notation { 1, 3, 5, 7, 9, … } and { 1, 3, 5, … } both refer to exactly the same set; the first notation is merely a bit more verbose. Similarly, the parenthesized portion of the second formula above is exactly equal to x, despite being written with two explicit leading terms rather than three.

Re: 0.999...= 1

#647

Post by heinrichhartman »

astrobe_ wrote: post #320
heinrichhartman wrote: post #302

0.9999 = 1 is a consequence of the way we define rational and real numbers and limits. There are alternative definitions of numbers where this equality does not hold: Non Standard Analysis https://en.wikipedia.org/wiki/Nonstandard_analysis being the most famous one. But for the sake of argument, let's just define numbers as sequences of digits with a mixed in period somewhere: MyNumber := { a = (a_1, a_2, ...) -- lis…

1 = 0.9... is the consequence of purposely ambiguous and questionable notation. That's an old teachers' trick to make students talk and listen about mathematics.

This has nothing to do with notation. It's perfectly possible to define infinite sequences without using dots. In particular if they are constant. In the above case:

    a_i = 9 for i \in \IZ and i = 0
Where \IZ are the integers.
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