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### Topic: Is .99999... = 1?  (Read 49878 times)

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#### xiankai

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##### Re: Is .99999... = 1?
« Reply #15 on: December 03, 2006, 09:17:00 PM »
That's the wrong concept of infinity. All digits are 9 and there is no last digit.

i was under the impression infinity only meant a number so large that there was no limit;

for example the arithmetic progression of the following series:

Sn = 1 + 3 + 5 + ... + (2n-1)

where Sn --> ? as n --> ?

here we take an approximation that n = Sn, an odd result that isnt surprising since that is the nature of ?.

and i think an infinite number does not have to have all digits as 9 (unless you are speaking in context) because...

o there are many ways to obtain infinity
o all infinities are not the same
o infinity is not one single number that is the biggest out there

but the part about having no last digit, i think you have a point there.

i was earlier illustrating how that when u perform approximation, then the '...' and '...' of c and 9c are the same in the sense they are infinite, thus arises the weird result as obtained earlier.

if approximation was not used and '...' are assumed to be fixed numbers, then there wouldn't be a problem.
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#### lemonoman

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##### Re: Is .99999... = 1?
« Reply #16 on: December 03, 2006, 10:02:07 PM »
i was under the impression infinity only meant a number so large that there was no limit;

True...but as soon as you take 1/? , it becomes a representation of 0 (or so we're debating here, I believe)

I think the original point is that the difference between 1 and 0.99999... is 'infinite'ly small...and an infinitely small difference implies a difference of 0 ... hence 1 and 0.99999... are equivalent.

Put another way, as you add extra 9s to 0.99999.... , the difference between 0.99999.... and 1 becomes 1/10 as small....so an infinite number of 9s implies a difference of (1/10)?.

#### Yggdrasil

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##### Re: Is .99999... = 1?
« Reply #17 on: December 03, 2006, 10:18:48 PM »
But 0.999... does have an infinite number of digits so it is exactly equal to 1.  0.999... with 10^100 nines will be approximately equal to 1, but 0.999... with an infinite number of nines is exactly equal to 1.

#### xiankai

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##### Re: Is .99999... = 1?
« Reply #18 on: December 04, 2006, 02:58:12 AM »
taken another, way, 0.999... is smaller than 1 by a number that is infinitely small, but that does not make it 1 because the number remains to be added.

which begs another question, is infinity a number? if so, is it an irrational number, or what? if not, is it a limit?

i think that infinity merely represents a limit, which assumes approximations.

when i say limit, i mean an upper range of numbers, but not such as infinity has a fixed range.

because i believe the crux in the paradox lies in how we define infinity, and the numerical operations that are applied to it.
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#### Yggdrasil

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##### Re: Is .99999... = 1?
« Reply #19 on: December 04, 2006, 03:49:19 AM »
If 0.999... is smaller than 1 then there should be a number (actually an infinite number of numbers) between them.  But, what number could fit between 0.999... and 1? (i.e. does there exist an x such that 0.999... < x < 1 ?)

#### FeLiXe

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##### Re: Is .99999... = 1?
« Reply #20 on: December 04, 2006, 07:08:48 AM »
infinity is not part of the real numbers.

you can define countable infinity as the amount of natural numbers there are. The sum of any countable set of numbers like 1+3+5+7+9+11+... is also countable infinty (I am pretty sure). It is important to notice that that sum is not a natural number.

uncountable infinity is the amount of real numbers (or greater cardinality).

infinity is something much more basic than a limit. So I don't think it is ever defined using a limit.
« Last Edit: December 04, 2006, 07:30:18 AM by FeLiXe »
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#### xiankai

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##### Re: Is .99999... = 1?
« Reply #21 on: December 04, 2006, 10:22:28 AM »
If 0.999... is smaller than 1 then there should be a number (actually an infinite number of numbers) between them.  But, what number could fit between 0.999... and 1? (i.e. does there exist an x such that 0.999... < x < 1 ?)

what if 0.999... was the number just before 1? even if a number doesnt fit in between the two, 0.999... can be smaller than 1.

--------------------------------------------------

for 0.999... = 1,

1 - (1/10?) = 0.999... = 1

where 1/10? is approximated (or represented like lemonoman says)  to 0,

or
?
?     9/10n = 0.999... = 1
r=1

where the infinite geometric progression converges to 1, which is an upper limit.

thus i really think it is an approximation or limit. although there are infinite ways of obtaining the number 0.999... the only way i can imagine is by arithmetic operations, all of which requires approximations or limits.

but if it is more basic then that, like the biggest number out there, well, i'll be interested in what people think of infinity apart from my view.

one last thing i have noticed; as 0.999... is not a real number, is it equal to a real number, 1?
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#### Borek

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##### Re: Is .99999... = 1?
« Reply #22 on: December 04, 2006, 11:57:29 AM »
what if 0.999... was the number just before 1? even if a number doesnt fit in between the two, 0.999... can be smaller than 1.

Let's assume x is a real number, y is a number 'just before' x, so that there is no number between. (x+y)/2 > y and (x+y)/2 < x. So (x+y)/2 fits between x and y - but we assumed there is no such number. So the assumption was faulty.

one last thing i have noticed; as 0.999... is not a real number

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#### FeLiXe

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##### Re: Is .99999... = 1?
« Reply #23 on: December 04, 2006, 01:00:14 PM »
I don't know the exact definition of real numbers but I think .999999... is defined as the sum of an infinite geometric series of rational numbers:

.9999.... = (.9 + .09 + .009 + .0009 + ...) = .9 (1 + .1 + .01 + .001 +  ...) = .9 (1 / (1-.1)) = .9 (1 / .9) = 1

then it is just a different representation of 1

It is the sum of infinitely many terms. But the series converges and therefore it has a finite real value. It is a real number.
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#### Yggdrasil

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##### Re: Is .99999... = 1?
« Reply #24 on: December 04, 2006, 05:13:08 PM »
?
?     9/10n = 0.999... = 1
r=1

where the infinite geometric progression converges to 1, which is an upper limit.

thus i really think it is an approximation or limit.

You have the wrong idea of limits and convergence of infinite serries.  The series converges; therefore, it has a well defined, exact value.  The sum on an infinite series is not an approximation, it is an exact value.

#### xiankai

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##### Re: Is .99999... = 1?
« Reply #25 on: December 04, 2006, 08:43:51 PM »
Let's assume x is a real number, y is a number 'just before' x, so that there is no number between. (x+y)/2 > y and (x+y)/2 < x. So (x+y)/2 fits between x and y - but we assumed there is no such number. So the assumption was faulty.

x = 1, y = 0.999...

(x + y)/2 > x
1.999... > 2

(x+y)/2 < x
1.999... < 2

there is really no such number.

--------------------------------------------------

the number involves infinity and,

infinity is not part of the real numbers.

--------------------------------------------------

You have the wrong idea of limits and convergence of infinite series.  The series converges; therefore, it has a well defined, exact value.  The sum on an infinite series is not an approximation, it is an exact value.

let Sn be the sum of the said geometric progression
Sn = 9/10 (1 - 1/10n) / ( 1 - 1/10)
Sn = 1 - 1/10n

the geometric progression involves the assumption that 1/10n --> 0 as n --> ?. i am thinking the arrows imply a limit, because no matter how big the denominator is, there will still be a number, but for most purposes it is almost equal to zero. thus i do not think it is an exact value.

--------------------------------------------------

It is the sum of infinitely many terms. But the series converges and therefore it has a finite real value. It is a real number.

you can define countable infinity as the amount of natural numbers there are. The sum of any countable set of numbers like 1+3+5+7+9+11+... is also countable infinty (I am pretty sure). It is important to notice that that sum is not a natural number.

what is the difference between natural numbers and real numbers? are natural numbers a subset of real numbers?
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#### Yggdrasil

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##### Re: Is .99999... = 1?
« Reply #26 on: December 04, 2006, 11:13:04 PM »
the geometric progression involves the assumption that 1/10n --> 0 as n --> ?. i am thinking the arrows imply a limit, because no matter how big the denominator is, there will still be a number, but for most purposes it is almost equal to zero. thus i do not think it is an exact value.

But, a the limit has a well-defined, exact value so the sum is an exact value.  Part of the problem with the name "limit" is that from the connotations of the word limit, some students believe that limits represent approximations.  But, limits are not approximations, they are exact values.  A whole branch of mathematics, analysis, is built on the fact that limits can have well-defined, exact values.

Quote
what is the difference between natural numbers and real numbers? are natural numbers a subset of real numbers?

Here's a brief overview of some of the common numerical structures used in mathematics:

The real numbers, denoted R, while intuitively easy to define and understand, have a very complex formal definitions in mathematics.  Therefore, I think it will suffice to say that the real numbers represent any finite (i.e. not infinite) quantity you can think of that does not involve i (sqrt of -1).  In a physical sense, the real numbers are measurements you can obtain from an instrument.  For example, lets say you are measuring the displacement of a particle.  You can obtain values which are negative or positive.  The values can be integers (e.g. 1, 3, 104), non-integers (e.g. 1.5, 4.32), and even numbers whose decimal representations do not end (e.g. pi, e, sqrt(2)).  The real numbers form a mathematical structure called a field.

The real numbers are a subset of the complex numbers, denoted C.  Complex numbers involve, so-called imaginary numbers (square roots of negative numbers).  They are of the form:

a + bi

where a and b are real numbers and i denotes sqrt(-1).  Note that for b=0, you recover the real numbers.  Like the reals, the complex numbers form a field.

There are various subsets of real numbers.  Natural numbers, denoted N, are "counting numbers."  In a physical sense, a natural number would be what one would answer to the question "how many people are in this room?" or "how many molecules occupy 1L?"  Since you cannot have half of a person or a fraction of a molecule, the answers must be natural number such as 1, 10, 504, 6.02x10^23.  Similarly, you cannot have a negative number of people or molecules, so the natural numbers are restricted to positive numbers.  Some people consider 0 a natural number, but some people do not.

The natural numbers are a subset of the integers, denoted by Z.  Integers are better known as "whole numbers" and they are formed by taking the natural numbers and their additive inverses (i.e. their negatives).  The integers form a mathematical structure known as a ring.

From the definition of integers, we can define other subsets of real numbers, such as the rational numbers, denoted Q.  Rational numbers are any number which can be represented by a ratio of two integers.  Therefore, the rational numbers include the integers (ratio of an integer with 1), and all fractions.  The rationals also form a field.  The irrational numbers (which I will denote Qc are any number which is not rational (i.e. has no representation as a ratio of two integers).  Irrational numbers are numbers with never-ending, non-repeating decimal representations, such as pi, e, and sqrt(2).  Note: not all numbers which never-ending decimals are irrational -- for example, 0.333... = 1/3 is rational despite having a never-ending decimal.  In fact, any number with a repeating, never-ending decimal is rational.

Together the rational numbers and irrational numbers partition the real numbers.  In summary:

N < Z < R = Q U QC < C

where < denotes subset.  Note that there are other, more complex structures which can be built on top of these basic numerical structures.  For example, on top of a field structure, you can build a vector space.  For example, the Cartesian coordinate plane (x,y) is the vector space R2, because every element of R2 can be described by two real numbers.

« Last Edit: December 04, 2006, 11:18:33 PM by Yggdrasil »

#### constant thinker

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##### Re: Is .99999... = 1?
« Reply #27 on: December 05, 2006, 08:25:14 PM »
Jeese. I never thought this would get so much attention.
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#### xiankai

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##### Re: Is .99999... = 1?
« Reply #28 on: December 06, 2006, 06:17:35 AM »
what is a limit, then?

is a limit meant to be reached? because somehow i was thinking they werent supposed to. just like an asymptote of a curve, the function approaches the limit arbitrarily close, but never gets to it. after all, infinity is not on the real number line.

and how is it possible for infinity to be valid in numerical operations, when it itself is not a real number?
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#### Borek

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##### Re: Is .99999... = 1?
« Reply #29 on: December 06, 2006, 07:37:47 AM »
what is a limit, then?

Number, that you can get as close to as you can imagine. Even then it can be proved that you can get closer. That's the limit definition

Quote
is a limit meant to be reached?

Good question. I can imagine sequences that reach the limit (lim x for x -> 1 is just 1 and it reaches the limit) and sequences that don't reach the limit (lim 1/x for x -> ? gets as close to 0 as possible, but never reaches the number). Limit definition doesn't state anything about the limit being reachable or not. But don't rely on me when it comes to math.

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« Last Edit: December 06, 2006, 08:55:24 AM by Borek »
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