0.999~ = 1

[Cos Dashit]

Is it possible to accept the algebraic proof, yet simultaneously reject the fraction proof? A sort of doublethink?

The reasoning being that 1/3 does equal .333~, it merely represents the numbers; something akin to 22/7 and pi (although obviously less extreme)?

[Kuroneko]

Is it possible to accept the algebraic proof, yet simultaneously reject the fraction proof? A sort of doublethink?

Yes. This is even quite natural if one perceives one proof as having a correct conclusion but incorrect reasoning. More loosely, it's likely that one can doublethink just about anything.

The reasoning being that 1/3 does equal .333~, it merely represents the numbers; something akin to 22/7 and pi (although obviously less extreme)?

Well, it's possible to think that, but it wouldn't be correct. Both 1/3 and .333~ are exact representations of the same number, not at all like the relationship between 22/7 and π.

[Cos Dashit]

The reasoning being that 1/3 does equal .333~, it merely represents the numbers; something akin to 22/7 and pi (although obviously less extreme)?

Well, it's possible to think that, but it wouldn't be correct. Both 1/3 and .333~ are exact representations of the same number, not at all like the relationship between 22/7 and π.

But .333~ * 3 = .999~, yet 1/3 * 3 = 1? Is there a proof that shows that the two numbers are equivalent?

Wait, is this circular logic?

[Kuroneko]

But .333~ * 3 = .999~, yet 1/3 * 3 = 1?

Correct.

Is there a proof that shows that the two numbers are equivalent?

Didn't you just give one above? Equality is transitive. Unless you question that 1/3 is .333~, which can be proven by the method given below (replace 9 by 3).

Wait, is this circular logic?

Well, it takes a few mathematical details for granted, but it isn't strictly speaking incorrect. A more rigorous approach may be as follows: .999~ = Sum_k[ 9/10^k ] is by definition the limit of the sequence of partial sums {9/10, 9/10+9/100, 9/10+9/100+9/100, ..., S_n, ... }, where S_n = 9 Sum_{0<k<n}[ 1/10^k ]. Letting x = 1/10, we have S_n = 9[x-x^n]/[1-x]. The limit is clearly 9x/[1-x] = 1, as x^n→0 as n→∞. Therefore, .999~ = 1.

[dworkin]

Of course outside of pure math there is the point where you have to go,

"Shit, it's basically 1" and accect that it's good enough for government work.

[CmdrWilkens]

The other proof that .999~=1 is more of a logic proof and that is any two distinct numbers (i.e. any case where A /= B) there exists a number C such that A <C<B or A>C>B.

[Shadow WarChief]

The reasoning being that 1/3 does equal .333~, it merely represents the numbers; something akin to 22/7 and pi (although obviously less extreme)?

Well, it's possible to think that, but it wouldn't be correct. Both 1/3 and .333~ are exact representations of the same number, not at all like the relationship between 22/7 and π.

But .333~ * 3 = .999~, yet 1/3 * 3 = 1? Is there a proof that shows that the two numbers are equivalent?

10X-X=9X

9X/9=X


So, with X=.999~

.999~ * 10 = 9.999~

9.999~-.999~=9.000

9/9=1=X

X= .999~ =1

[Durandal]

Let go of the assumption that every number in existence must have a unique decimal representation, and the equality of 0.999... and 1 will make sense.

Oh, and the other proof is to show that there exists no number which is between 0.999... and 1, that is, there is no number which is both greater than 0.999... and less than 1.

[Admiral Valdemar]

Oh, and the other proof is to show that there exists no number which is between 0.999... and 1, that is, there is no number which is both greater than 0.999... and less than 1.

Sure there is: 0.9999999999999999999999999.

in real-world terms, it won't always equate to exactly 1 such as travelling at 0.99999 of c. You'll never hit lightspeed, but you'll get damn close and keep getting closer with diminishing returns, like halving a number indefinitely.

[Winston Blake]

Oh, and the other proof is to show that there exists no number which is between 0.999... and 1, that is, there is no number which is both greater than 0.999... and less than 1.

Sure there is: 0.9999999999999999999999999.

in real-world terms, it won't always equate to exactly 1 such as travelling at 0.99999 of c. You'll never hit lightspeed, but you'll get damn close and keep getting closer with diminishing returns, like halving a number indefinitely.

Huh? 0.9999999999999999999999999 < 0.999~. You can't reach (0.999~)c because that's identical to reaching c. Why shouldn't 0.999~ = 1 in this case?

[Admiral Valdemar]

Huh? 0.9999999999999999999999999 < 0.999~. You can't reach (0.999~)c because that's identical to reaching c. Why shouldn't 0.999~ = 1 in this case?

I was being facetious there, hence the smiley.

The lightspeed example was just to show how real-world incidents can differ, since no one would say you were doing c even if up to 0.9999999...999 and so on.

[Gil Hamilton]

And then the chemistry major goes "Well, 0.999... is close enough to 1 that I can't measure it with this here graduated cylinder, so I'm just going to write 1 here to make my math easier."

[Admiral Valdemar]

True. Again, in real-life we can have accuracy to only so many decimal places. Anything at 0.999 litres is essentially 1 litre.

[Rye]

This is something that I remember bugged me in primary school. I think I resolved it by merely conclusing fractions were superior.

Another thing that bugged me about the 0.999... thing was that if you halved it, surely you'd get 0.4999...5 and that "actually" being 0.5 confused me to all hell. Where did those 5 extra numbers come from?

[GrandMasterTerwynn]

For today's almost-freakish coincidence, 0.999... is today's featured Wikipedia article.

[Admiral Valdemar]

I'm pretty sure this thread was influenced by that.

[Kuroneko]

Oh, and the other proof is to show that there exists no number which is between 0.999... and 1, that is, there is no number which is both greater than 0.999... and less than 1.

Of course, that approach is based on the assumption that every real number has some (not necessarily unique) decimal representation, but that isn't hard to prove.

Another thing that bugged me about the 0.999... thing was that if you halved it, surely you'd get 0.4999...5 and that "actually" being 0.5 confused me to all hell. Where did those 5 extra numbers come from?

Hmm? If ones halves 0.9999~, one should get 0.4999~, not 0.4999~5. The latter isn't even a real number. And yes, 0.4999~ = 0.5.

[Surlethe]

Oh, and the other proof is to show that there exists no number which is between 0.999... and 1, that is, there is no number which is both greater than 0.999... and less than 1.

Suppose, for contradiction, that there does exist some real number k such that 0.999... < k < 1. We have 0.999... as the limit of the partial sums of 9(1 + 0.1 + 0.01 + 0.001 + ... + (1/10)^n ) as n tends to infinity. That means, given an ε>0, there is some N such that, for n>N, we have 0.999 ... - 9(\sum_{j=1}^{n}(1/10)^n)<ε (taking the modulus is unnecessary since each term of the sum is greater than zero). Since k > 0.999 ... , there must exist a fixed positive R such that k - 0.999 ... = R. So, we have 1 - 0.999 ... > k - 0.999 ... = R > 0. We now have, considering the nth partial sum, 1 - 9(\sum_{j=1}^{n}(1/10)^n) = 1 - 9[(1/10)-(1/10)^n]/[1-(1/10)] = 1 - (9/10)/(9/10) + 9((1/10)^n)/(9/10) = (1/10)^(n-1). Clearly, given an ε>0, there is some N such that 0 < (1/10)^(n-1) < ε when n-1>N. Therefore, 1 - 0.999 ... < ε for an arbitrarily small ε; but this gives ε>R>0, for a fixed R: a contradiction.

This proof smells circular, but isn't; it's just unnecessary to take the extra step, since in the course of it you show, essentially, that 1 = 0.999 ... ; however, it suffices if one still needs extra convincing that the ε proof might not provide.

[Surlethe]

And then the chemistry major goes "Well, 0.999... is close enough to 1 that I can't measure it with this here graduated cylinder, so I'm just going to write 1 here to make my math easier."

Interestingly enough, this is closer to the mathematical approach than one might think. Once you've been exposed to (and understand) the technical definition of convergence (i.e., a sequence a_n is convergent to some A iff given an ε>0 there is an N such that n>N implies |a_n - A|<ε), it's actually quite similar to the idea of ease that you outlined. While the mathematical idea is a little more streamlined, since you can control the error term to an arbitrary degree, it is, in essence, the same thing: since as you go down the sequence 0.999 ... , the difference between it and 1 becomes as small as you want, we can justifiably set 0.999 ... = 1, even though they're not precisely the same thing, since at no finite n does the partial sum 0.9999 ... 9 = 9(\sum_{j=1}^{n}(1/10)^n) actually equal 1.

[Rye]

Hmm? If ones halves 0.9999~, one should get 0.4999~, not 0.4999~5. The latter isn't even a real number. And yes, 0.4999~ = 0.5.

I know that's meant to be the answer, and there are probably ways beyond me that show it, and I can accept that ultimately, yes, 0.999... is equal to 1, though I don't intuitively grasp it. The issue I'm having, though, is if you divide 0.9 by 2, you get 0.45, right? So any number in the 0.999... sequence will divide into a variant, 0.9999 / 2 = 0.49995 for instance, and no matter how I try, I can't intuitively see why that won't always be the case, even if we can never get there, the last number should be a 5, due to the way a 9 divides.

[Braedley]

Ah, but to the dismay of many computer programmers, such as myself, 0.99999... as seen by a computer does not always equal 1. Case in point (not fabricated, I swear):

Code: Select all

; IDL Version 6.2, Microsoft Windows (Win32 x86 m32)
; Journal File for Michael Braedley@MIKE
; Working directory: C:\Documents and Settings\Michael Braedley
; Date: Wed Oct 25 20:50:34 2006
 
test=59.999999999D/40000.
print, test*40000
;       60.000000
print, test*40000 eq 60
;   0

A similar situation happened to me at work today. A simple round solved that problem though (damn off by one errors due to floating point rounding errors). The error was also much more subtle.

[Cos Dashit]

For today's almost-freakish coincidence, 0.999... is today's featured Wikipedia article.

Yes, my friend and I had an argument on .999 a year ago, and this reminded him of it. He informed me, and the discussion continues.

Well, it's possible to think that, but it wouldn't be correct. Both 1/3 and .333~ are exact representations of the same number, not at all like the relationship between 22/7 and π.

But .333~ * 3 = .999~, yet 1/3 * 3 = 1? Is there a proof that shows that the two numbers are equivalent?

10X-X=9X

9X/9=X


So, with X=.999~

.999~ * 10 = 9.999~

9.999~-.999~=9.000

9/9=1=X

X= .999~ =1

I've read the algebraic proof; I also accept this proof, until something convinces me otherwise. I was wondering about the fraction proof, and by what means we know that 1/3 = .333~.

[Cos Dashit]

Ah, but to the dismay of many computer programmers, such as myself, 0.99999... as seen by a computer does not always equal 1. Case in point (not fabricated, I swear):

Code: Select all

; IDL Version 6.2, Microsoft Windows (Win32 x86 m32)
; Journal File for Michael Braedley@MIKE
; Working directory: C:\Documents and Settings\Michael Braedley
; Date: Wed Oct 25 20:50:34 2006
 
test=59.999999999D/40000.
print, test*40000
;       60.000000
print, test*40000 eq 60
;   0

A similar situation happened to me at work today. A simple round solved that problem though (damn off by one errors due to floating point rounding errors). The error was also much more subtle.

Good point, but computers have to round off at some decimal place. The 59.999999999 in the computer wasn't 59.999~.

[Surlethe]

I know that's meant to be the answer, and there are probably ways beyond me that show it, and I can accept that ultimately, yes, 0.999... is equal to 1, though I don't intuitively grasp it. The issue I'm having, though, is if you divide 0.9 by 2, you get 0.45, right? So any number in the 0.999... sequence will divide into a variant, 0.9999 / 2 = 0.49995 for instance, and no matter how I try, I can't intuitively see why that won't always be the case, even if we can never get there, the last number should be a 5, due to the way a 9 divides.

The cheap way to resolve this is to concede 0.9999... = 1, so (0.9999 ...)/2 = 1/2 = 0.5 = 0.49999 ... .

The other way of looking at it is this: we can write the 0.499999 ... as a sequence: {0.4, 0.49, 0.499, 0.4999, 0.4999, 0.49999, ...}. Clearly, this goes to 0.5. Now, write a new sequence {0.4, 0.45, 0.495, 0.4995, 0.49995, ...}. Let us compare the the nth terms in each sequence (0.4999 ... and this new one): 0.49999...(n times)...9 and 0.499...(n-1 times)...95. The difference between the two is 0.00...(n+1 times)...4. This means that the first term minus the second one is less than 4(1/10)^(n+2). Because as n increases without bound, this becomes vanishingly small, we can safely say that 0.999 ... /2 = 0.4999 ... = 0.5.

The simplistic reason for this equality is that the terminal 5 is not enough to change where the sequence seems to be headed: as 9s crowd up behind it, they overwhelm the five, until it is so insignificant it doesn't matter. That's what it means for the difference to go to zero as n becomes big (i.e., as you move out further and futher in your sequence): because the 5 is getting pushed out one decimal place each time, the difference between the two sequences decreases by an order of magnitude.

[Kuroneko]

The issue I'm having, though, is if you divide 0.9 by 2, you get 0.45, right? So any number in the 0.999... sequence will divide into a variant, 0.9999 / 2 = 0.49995 for instance, and no matter how I try, I can't intuitively see why that won't always be the case, even if we can never get there, the last number should be a 5, due to the way a 9 divides.

Ask yourself: in the case of 0.999~/2, which digit would be the 5? You'll find that this hypothetical digit would not correspond to any finite place (ordinal number of the digit). By definition, decimal expansions are sums of powers of 10, with the exponent corresponding to the place of the digit--0.123 = 1(1/10) + 2(1/10)² + 3(1/10)³, etc., with infinite decimal expansions resolved by a limit of partial sums. In that case, in what sense does an expression luike "0.4999~5", with infinitely many 9's before the 5, represent a real number? In other words, what would be the exponent of 1/10 corresponding to that digit?

Addendum: I'm not sure in what sense this makes fractional representations are superior as compared to the decimal 0.999~ = 1. Sure, they allow exact representation of all rational numbers, but almost all numbers are not rational, and the fractional representations is no sense unique either, e.g., 1/2 = 2/4. The fact that decimal representations are sometimes non-unique seems to be a poor reason, although of course one can have other reasons depending on the task at hand.