1/3 +1/3 +1/3 = 1 but .333 +.333+ .333 = .9999999999?

Question

whats wrong with the decimal system

Thomas C- nop 10x- x wold equal 8.99999999999999991 not 9

Answer 1

There's nothing "wrong" with the decimal system. It's just that it offers two different ways of writing the same value. 1 = 0.9999... (with the 9s repeating forever). This shows the limitations in working with a base-10 system.

This subject has been "debated" here ad nauseum, but the bottom line is that 1 = 0.9999...(again, repeating). NOT "approximately". They're EQUAL.

You should be more clear in what you wrote though. 1/3 is not equal to "0.333", "0.3333" or even "0.3333333". It's equal to "0.333....", where the elipsis at the end here implies "the threes go on forever". Likewise, a decimal point followed by an INFINITE amount of 9s is precisely equal to 1. There is no number small enough that falls in between the two.

>>Thomas C- nop 10x- x wold equal
>>8.99999999999999991 not 9

No, he's correct. Again, you're falling for the mistake of treating these as finite strings of numbers. When you multiply 0.999... by 10 you get 9.999.... There is no "0" at end or a less number of 9s to the right of the decimal because you're dealing with infinity, which has no end.

Answer 2

Actually, .333 < 1/3, and .333 + .333 + .333 = .999, not .9999999999 . Now, it is true that 1/3 = 0.3333..., and that 0.3333... + 0.3333... + 0.3333... = 0.9999... = 1, but note the ellipses -- they mean that the decimal repeats infinitely many times, without terminating. This is very important, since one statement is true, whereas the other is bunkum.

So why is this you ask? Of course, anyone can write a proof that 0.9999... = 1, but somehow I doubt you'll be convinced, mainly because you've also seen "proofs" that 1=0, and those invariably contained some subtle flaw, so as long as your intuition protests the identity 0.9999... = 1, you will think that there is some subtle flaw with the proofs that 0.9999... = 1 and that you are simply unable to find it, the assurances of mathematicians that 0.9999... IS equal to 1 notwithstanding. So, let's work on the intuition, shall we?

Question: What does 0.9999... actually mean?

Answer: 0.9999... means 9/10 + 9/100 + 9/1000 + 9/10000... . This is just the definition of our decimal system, and should be fairly uncontroversial.

Question: What is the sum of infinitely many numbers?

Answer: Ah, this one's trickier. Presumably, you know what the sum of two numbers is, and if you wish to find the sum of three number, say a + b + c, it can be found by simply taking the sum of the first two, and adding that to the third (i.e. as (a+b) + c). Of course, you could also interpret a+b+c as a+(b+c) -- the associative property guarantees that it doesn't really matter. And of course, by adding them up one at a time, you can find the sum of any four numbers, any five numbers, any six numbers, and in general any finite number of numbers. But what of infinitely many numbers? If we try to add them up one at a time, we will never finish (that what infinitely many means). So some other method of determining the sum is required.

Let us suppose that, as in this case, the numbers being summed are decreasing in magnitude, as they are here. Then if we take the sum of the first n numbers (where n is finite, so this sum is well-defined), we should be closer to the true sum than we were at the start. And if we take the sum of the first m numbers, where m is greater than n (but still finite), then we should be closer still. And let us suppose, as is the case here, that as n increases, the sum of the first n numbers becomes and stays arbitrarily close to some number L. Then it is reasonable to suppose that the sum of all those numbers IS L -- certainly this is more intuitive than the alternative, namely that the sum of the first n elements should get closer and closer and closer to L, but then when you take the sum of all of them, the sum suddenly jumps to some unrelated number. Of course, the latter is not logically inconsistent -- precisely because you cannot sum infinitely many numbers by adding them one at a time, the sum of ALL the numbers need not necessarily have anything whatsoever to do with the sum of any finite subset. Nonetheless, since we find it intuitively plausible that it should, mathematicians have adopted the following definition for the sum of an infinite series:

[k=1, ∞]∑x_k ≝ [n→∞]lim [k=1, n]∑x_k

In other words, we define the sum of an infinite series to be the limit, as n approaches infinity, of the sum of the first n numbers in that series. Like all definitions in mathematics, this is a stipulative definition, not a lexical one. In other words, this definition _assigns_ a meaning to the sum of an infinite series, it doesn't try to report on one that already exists (indeed, because you can't add the numbers one at a time, there IS no prior meaning for the sum of an infinite series).

So, what does this have to do with .9999...? Well, consider the sum of the first n numbers in the series 9/10 + 9/100 + 9/1000 + 9/10000...:

.9 (n=1)
.99 (n=2)
.999 (n=3)
.9999 (n=4)
.99999 (n=5)

As you can plainly see, the sequence of partial sums becomes arbitrarily close to 1. And indeed, if you give me a positive real number ε, no matter how small it is, I can find a natural number N such that whenever n>N, the sum [k=1, n]∑9/10^k is withing a distance of ε from 1. Which means, according to the formal definition of limits, that [n→∞]lim [k=1, n]∑9/10^k = 1. And per the definition of the sum of an infinite series, we have [k=1, ∞]∑9/10^k = [n→∞]lim [k=1, n]∑9/10^k, therefore it follows that 0.9999... = [k=1, ∞]∑9/10^k = 1.

Now, I can already hear the "but wouldn't it always be just a little less than 1?" objection forming on your lips, so save it. You're thinking that every stage of the summation must be accomplished by adding some number of the form 9/10^k, and that since 9/10^k is always less than the difference between the current partial sum and 1, that the sum will never actually reach 1. For finite n, indeed [k=1, n]∑9/10^k < 1, because all of these sums really are obtained by adding 9/10^n to [k=1, n-1]∑9/10^k. But as I have tried to explain to you during the preceding discussion, the sum [k=1, ∞]∑9/10^k is NOT obtained by adding a number to [k=1, ∞-1]∑9/10^k -- indeed, the latter expression is nonsensical. No, [k=1, ∞]∑9/10^k is obtained by taking the _limit_ of [k=1, n]∑9/10^k as n→∞, and because of that does not have to be less than 1, even if all of the sums [k=1, n]∑9/10^k are.

The minute you realize that the sum of infinitely many numbers is obtained by taking limits, rather than by repeated addition, is the minute this stops being counterintuitive.

Of course, your next objection will be something about "how can two different decimals refer to the same number?" And the answer is "the same way 1/2 and 2/4 can refer to the same number." Decimals are a representation of real numbers, not themselves real numbers. Admittedly, many people don't know what the real numbers actually are, since they may have never been given anything more specific than "numbers which can be given by a decimal representation." Presumably, teachers do this because they think the real answer "R is the unique (up to isomorphism) dedekind-complete ordered field" are too abstract for their students. Of course, after seeing so many people get hung up over the idea that one number can have multiple decimal representations, I would actually prefer they give students an incomprehensible correct definition than a comprehensible definition that's wrong.

If you're wondering how many numbers have multiple representations, the answer is all the decimal fractions -- i.e. all numbers of the form p/(2^j 5^k), where p is an integer and j and k are nonnegative integers. These are precisely the numbers which can be represented using a finite decimal expansion. And with the exception of 0, each of them has an alternate expansion ending in an infinite sequence of nines (i.e. 2.75 = 2.7499999..., 12 = 11.9999999..., -16 = -15.999999..., and so forth). 0 is a little unusual, in that its alternate representation is -0, rather than something point 999999.... It is a straightforward (albeit tedious) exercise to prove that these are the only exceptions, and that formal operations on decimal sequences will always yield A correct decimal representation of the correct answer (although which representation is given may depend on the order in which the operations are performed).

In sum, there is absolutely nothing wrong with the decimal system, and once you understand what the sum of an infinite series actually refers to, it is obvious that 0.9999... = 1, so the fact that formal operations on decimal sequences yield that result is not even surprising. If you still want more information, check out the relevant wikipedia articles.

Answer 3

Proof 1 - Sum of fractions
1/3 = 0.3333~
1/3 = 0.3333~
1/3 = 0.3333~
------------------
3/3 = 1 = 0.9999~



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Proof 2 - Conversion to fraction
n = 0.9999~
10•n = 9.9999~

10•n - n = 9•n
10•n - n = 9.9999~ - 0.9999~

9•n = 9
n = 1

∴ n = 1 = 0.9999~



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Proof 3 - Infinite Geometric Series
0.9999~ = 0.9 + 0.09 + 0.009 ... =
9•10^-1 + 9•10^-2 + 9•10^-3 + 9•10^-4 ... =

∞
∑ 9 • (1/10)^n
n=1

Theorem - Infinite geometric series can be evaluated:
∞
∑ c • (r)^n = c•r / (1 - r)
n=1


Therefore:
9•(1/10) / (1 - 1/10) = (9/10) / (9/10) = 1

The infinite series that represents 0.9999~ can be simplified to 1



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Proof 4 - Argument from Philosophy - The definition of the real numbers as a continuum

"If two numbers, x and z, are not equal such that x < z, there exists a third number, y, such that x < y < z"

This means that if two numbers are different, there is a number in between them.

If 0.9999~ does not equal 1, what number could possibly exist between them?

No number exists that is greater than 0.9999~ but less than 1.

This is another way of saying that no number exists that can be added to 0.9999~ such that the sum will not be greater than 1.

∴ the two are equal.




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Proof 5 - From Averages

The average of two numbers, m and n, is found by adding them and dividing by 2.
[m + n] / 2

The average, A, is greater than the lesser number and smaller than the greater number
m < A < n

Unless the average is made between a number and itself:
A = a = [ a + a ] / 2


Assume that 0.9999~ < 1

Find the average of 1 and 0.9999~
[ 1 + 0.9999~ ] / 2 =
[ 1.9999~ ] / 2 =
0.9999~

The average is equal to the smaller of the two numbers, A = m. Therefore A = n, also, and thusly m = n.

This proof has close ties to proof 4


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Proof 6 - By Arithmetic
Any number, n, minus itself, n - n, will equal zero.

Take 1 and subtract 0.9999~

1 - 0.9999~ = 0.0000~ = 0

Therefore, 1 = 0.9999~



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Proof 7 - As a Limit.

Infinite sums cannot be evaluated as easily as the limit of a finite sum approaching infinity.


0.9999~ = 1 - 0.0000~...1


lim [ 1 - 0.0000000000...1 ] =
n→∞ |__n digits__|


lim [ 1 - 10^-n ] =
n→∞

lim [ 1 ] - lim [ 10^-n ] =
n→∞ n→∞

1 - 0 =

1


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Proof 8 - Comparing Limits
The limit of two functions are equal only if they evaluate to an equal value at the limit

if the limits are equal:

lim [ f(n) ] = lim [ g(n) ]
n→c n→c

Then the values are equal:

f(c) = g(c)


The limit of 1 is equal to the limit of the decimal expantion of 0.9999~. As more 9's are affixed, the limit approaches 1

Becaues the limits approach the same value, they have the same value:
1 = 0.9999~


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Proof 9 - A Fallacious Proof - Lack of contrary evidence

The presumption that 0.9999~ is less than 1 is taken intuitively and as a result of our recognition of decimal place values.

Consider that no proof exists, aside from intuition, that concludes 0.9999~ is less than 1.






The magic of math, huh?

Answer 4

.333+.333+.333 = .999

1/3 is a rational number, it goes on forever.

So, there is no physical way to represent this fraction with the decimals.

This is why we round our numbers and why if you add enough digits to your calculator, it will get the picture and round for you.

There is a mathematical proof showing how .9999... is equal to 1. Ask a professor in the department of Mathematics.

Answer 5

.333 isn't equal to 1/3, as 1/3 equals .3333333333333333333333333333333333333333333333333333333333333333333 forever, and even that isn't exactly equal. There is no decimal equal to 1/3, .333 is just as close as you can get, and .99999999999999 is as close as you'll get to 1.

Answer 6

There's no way to get closer to 1/3. The repeating decimal rounds itself up eventually.

Answer 7

that is because .333 isn't exactly a 1/3

Answer 8

.99999999 etc == 1. Here's the proof:

Let X = .99999...
Then 10X - X = 9.9999999.... - 0.999999999.... = 9
So 10X - X = 9
and 9X = 9.

Dividing each side by 9 gives
X = 1

Answer 9

Check out wikipedia why 0.99999... is exactly = to 1.