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0.999~=1

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For those who haven't seen this debate before, it was a fad during Summer 2003 driven by debate over whether 0.999999...(repeating) is equal to 1, containing multiple 500 topics with neither side backing down on board 8 alone (to say nothing of LUE).

One might ask how a simple Mathematical question could provide so much debate. The answer is that the debate, from a mathematical perspective was something akin to "Oranges have skins you can peel, but the inside is solid" "what do you mean? Coconuts are nothing like that!!" yet everyone assumed they were talking about the same thing--numbers are just numbers, right?

Before delving into Psychology, the fundamental axioms of the Real Numbers, and the failings of the western education system, if you just want an answer, go with 0.99~=1, and walk away.

For the curious/masochistic...read on:

Helpful IntuitionEdit

While these aren't necessarily proofs, they do give a good idea of why 0.99~=1 is a logical thing to believe.

  • Long Division
  • Now, think about this: Say you're about to say, "Hi." But say that every time you're about to say it, you get punched and can't. So you try again, and get punched again and can't say it. And this continues forever. Theoretically, there should be a "Hi" at the end of all the punches. However, as the punches continue for eternity, you will never say "Hi."
  • because Blizzard said so

Common "Proofs"Edit

1/3 = 0.33~
3*1/3 = 3*0.33~
1 = 3/3 = 0.99~
10*0.99~ = 9.99~
10*0.99~ - 0.99~ = 9.99~ - 0.99~
9(0.99~) = 9
0.99~ = 1
[1]
Short version:
The sum of an infinite geometric is given by S = a/(1-r).
0.99~ = sum (0.9*(0.1)^n)
Hence a=0.9, r=0.1, and the Sum S = 0.9/(1-0.1) = 1
For any real numbers A < B, there exists a C such that A < C < B.
However, there are no real numbers between 0.99~ and 1, because the 9s 
keep on going forever.
Therefore, 0.99~ = 1.
Given the sequence n1 = 0.9, n2 = 0.99, n3 = 0.999, it is clear that the 
limit of ni as i goes to infinity is equal to 0.99~.
However, the limit is also equal to 1.
Therefore 0.99~ = 1.
Definition: An infinite string of decimal digits is defined to be an 
   infinite sum
Definition: An infinite sum is defined to equal the limit of the sequence
   of finite partial sums
Definition: If a sequence s_n has the property that for any epsilon > 0, 
   there is an N such that for all n > N, | s_n - L | < epsilon, then
   the limit of the sequence s_n is defined to be L

0.999~ = sum of 9*10^(-n) from n = 1 to infinity
0.999~ = limit of (sum of 9*10^(-n) from n = 1 to m) as m goes to infinity
0.999~ = limit of (9/10*(1 - (1/10)^m)/(1 - 1/10)) as m goes to infinity
0.999~ = limit of (1 - (1/10)^m) as m goes to infinity

For epsilon > 0, let m = ceiling(log(1/epsilon)), so

| (1 - (1/10)^m) - 1 | = (1/10)^m = 1/10^m < 1/10^log(1/epsilon) = epsilon

Therefore, by definition of the limit,

0.999~ = 1

DissensionsEdit

There are people who, given the above proofs, continue to argue. There are a number of fairly big deviations they must make from the norm, however. In particular:

1/3 =/= 0.33~Edit

(1/3 = 0.33~ can be proven with limits).

10*0.99~ =/= 9.99~Edit

In particular, that there is no repeating decimal representation of 10*0.99~. (10*0.99~ = 9.99~ can also be proven with limits).

The sum of an infinite geometric series =/= a/(1-r)Edit

(Much like the special case 1/3 vs 0.33~, S=a/(1-r) can be proven with limits).

There is a number in between 0.99~ and 1Edit

Otherwise we're left with two cases:

  • CASE 1: (0.99~ + 1)/2 >= 1
0.99~ + 1 >= 2
0.99~ >=1
0.99~ = 1
  • CASE 2: (0.99~ + 1)/2 <= 0.99~
0.99~ +1 <= 2*0.99~
1 <= 0.99~
1 = 0.99~

Notice that (0.99~ + 1)/2 is another number that we can't represent through normal repeating decimal representation. In fact, by induction there must be an infinite such non-decimal-representable numbers in between 0.99~ and 1 for them to be different.

Unfortunately none of this helps us answer the question; The way you prove there are no numbers in between 0.99~ and 1 is by using their equality. The way you find numbers between 0.99~ and 1 is by using their inequality.


Similarly, epsilon-delta proofs don't really help here. If (1-0.99~)/2 > 0, then the sequence 0.9, 0.99, 0.999, ... does not converge to 1, since we let epsilon be (1-0.99~)/2. If the difference is 0, on the other hand, then of course the sequence converges to 1. One could always require the epsilons to be from the set of rational numbers, mind you, in which case the sequence would be convergent.

convergent sequences do not have a limitEdit

It has often been noted that peole who have taken calculus almost never argue the claim--they've used limits so often that they just don't question them.

Nonetheless, whether you define the real numbers as sequences of rational numbers with equality defined to be mutual convergence, as dedekind cuts of the rational numbers, or as the least upper bounds of bounded rational subsets, you can prove that convergent subsequences have a limit.

I don't believe this Least Upper Bound Dedekind Cut Rational Number Sequences stuffEdit

Okay.

What do you mean "Okay"?Edit

We've gotten to the core of the definition of Real Numbers. Real Numbers as a whole require an additional assumption which fractions, square roots, pi, etc, do not. If you want to define a different set of Real Numbers, you can. Mathematicians are known to do things like this regularly with Noneuclidean geometry (it's not abnormal to find mathematicians working in geometric spaces where the angles of a triangle don't add up to 180 degrees/pi radians). HOWEVER, your definition will not be the Real Numbers which are used by mathematicians, scientists, and the rest of the world. Also, arithmetic becomes much, much harder.

A world where 0.99~ = 1Edit

In short, this forms an ordered field (in particular, see the bullet point on the formal Laurent series).

For the rest of this section, I will use the following notation.

x = 1-0.99~

First notice that as a part of being a field, 1/x is a number. This is not Infinity (or not a traditional definition of infinity since 2/x =/= 1/x. It is, however, larger than every rational number).

Second, notice that calculations in general become much harder. For instance, which is larger?

1/0.99~

or

2-0.99~

The answer should become a bit more obvious to those with Taylor series experience once we translate it to our new notation:

1/(1-x)

or

1+x

In general, when comparing polynomial fractions and trying to figure out which is larger, you can use the following....

A(x)/B(x) vs G(x)/H(x)

translates to...

A(x)*H(x) vs G(x)*B(x)

Why should I use the Real Numbers instead of this system?Edit

For a few reasons

  • This system makes for much harder calculations.
  • Everyone else uses the Real Numbers. Compare this to, say, Windows--a lot of people would strongly disagree about Windows being the best operating system, but it runs all the PC games, and you can link your computer to your friend's computer and they speak the same language (this sentence written on a Linux box, for the record; thank God the real numbers can't be infected by spyware). Compare with C++. Lots of programmers don't think it's the ideal language, yet it's been around for so long that it has the most efficient compilers, and nobody wants to throw out their old code.
  • Nothing in the real world has been measured to more than...twelve decimal places I believe it is, let alone infinite decimal places. The suggested number system has no real practical use.

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