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General Forums => Generic Discussion => Topic started by: Donaldson Tan on October 27, 2006, 09:42:45 PM

Title: Is .99999... = 1?
Post by: Donaldson Tan on October 27, 2006, 09:42:45 PM
let c = 0.99999...

10c = 9.9999...

10c - c = 9.9999... - 0.99999...

9c = 9

c = 1

0.99999... = 1
Title: Re: Is .99999... = 1?
Post by: Mitch on October 27, 2006, 11:25:12 PM
You did the math wrong, this line: 9c = 9

should of been 9c = 8.9999....
Title: Re: Is .99999... = 1?
Post by: Yggdrasil on October 27, 2006, 11:50:30 PM
0.999... = 0.9 + 0.09 + 0.009 + ...
               = 9(0.1 + 0.01 + 0.001 + ...)
               = 9(-1 + 1 + 0.1 + 0.01 + 0.001 + ...)
               = 9(-1 + ??n=0(1/10)n)
Since ??n=0rn = 1 / (1 - r) for |r| < 1,
0.999... = 9(-1 + 1/(1 - 1/10))
               = 9(-1 + 10/9)
               = 9(1/9)
               = 1
qed
Title: Re: Is .99999... = 1?
Post by: Borek on October 28, 2006, 03:43:47 AM
http://en.wikipedia.org/wiki/Infinite_geometric_series

sum = a/(1-r)

0.9 + 0.09 + 0.009 + ...

a = 0.9, r = 0.1

0.1/(1-0.1) = 1

Mitch: I recall David checked what to do to install LaTeX... It'll be handy on occasions ;)
Title: Re: Is .99999... = 1?
Post by: pantone159 on October 28, 2006, 02:35:24 PM
0.99999... means 'the limit approached as you add more and more nines to the end' which is exactly 1.0.
Title: Re: Is .99999... = 1?
Post by: Donaldson Tan on October 28, 2006, 03:55:56 PM
Mark, No.

"0.99999 = 1" is not an approximation.

That's the whole point of bringing this up.
Title: Re: Is .99999... = 1?
Post by: Donaldson Tan on October 29, 2006, 04:34:19 PM
You did the math wrong, this line: 9c = 9

should of been 9c = 8.9999....

The workings first assume 0.9999... != 1

However, the subsequent steps reveal that 0.999... = 1
Title: Re: Is .99999... = 1?
Post by: FeLiXe on October 29, 2006, 06:18:02 PM
it's a correct proof. My math professor said that a couple of times that .999.... is exactly the same as 1
Title: Re: Is .99999... = 1?
Post by: Borek on October 29, 2006, 06:20:47 PM
1/9 = 0.1111111...

9*1/9 = 0.9999999...

9*1/9 = 1
Title: Re: Is .99999... = 1?
Post by: Donaldson Tan on October 29, 2006, 06:32:23 PM
This is one of the odd but interesting results in Math..
Title: Re: Is .99999... = 1?
Post by: constant thinker on October 29, 2006, 08:25:04 PM
The concept of .9999... being equal to 1 is difficult to grasp. Seeing the proof though make me believe.

I fully agree with geodome in how it's an odd but interesting result.

When I first saw that someone prooved it equaled one, I was like what the...
Title: Re: Is .99999... = 1?
Post by: lemonoman on October 30, 2006, 10:02:08 PM
I don't understand what all the fuss is about...

And what's wrong with what Mark said??
Title: Re: Is .99999... = 1?
Post by: Borek on October 31, 2006, 02:36:56 AM
And what's wrong with what Mark said??

Nothing :)
Title: Re: Is .99999... = 1?
Post by: xiankai on December 03, 2006, 07:03:36 AM
"0.99999 = 1" is not an approximation.

if it is not an approximation,

then if c = 0.999...

10 c = 9.99...0

because the last number cannot be left out.

therefore 10 c - c = 8.999...1

9 c = 8.999...1

there are similar problems involving infinity, like how

1 + 2 + 3 ... = ?

2 + 4 + 6 ... = ?

even though both have the same number of terms, the second series is obviously twice that of the first series. however, two times infinity still equals infinity.

This is one of the odd but interesting results in Math..

so it looks like that's the way we structure the number 'infinity'.
Title: Re: Is .99999... = 1?
Post by: FeLiXe on December 03, 2006, 01:35:47 PM

therefore 10 c - c = 8.999...1

9 c = 8.999...1

That's the wrong concept of infinity. All digits are 9 and there is no last digit.

there are similar problems involving infinity, like how

1 + 2 + 3 ... = ?

2 + 4 + 6 ... = ?

even though both have the same number of terms, the second series is obviously twice that of the first series. however, two times infinity still equals infinity.

so it looks like that's the way we structure the number 'infinity'.

actually there are different kinds of infinity. for example there are more real numbers then there are whole numbers.


in fact you can easily prove that there are infinitely many types infinity.

but on the other hand from a mathematically useful point of view the amount of rational numbers is the same as whole numbers. the rational numbers are a "countable set", you can put them in order: (1, -1/2, 1/2, -1/3, 1/3, -2/3, 2/3, -1/4, 1/4, -3/4, 3/4, ...). you can define a distinct order where every rational number has its index. this would not work with the real numbers (can be proven easily too)

another question is if there is a cardinality step in between the whole and real numbers. no one has found one. but no one has proven that it does not exist either. the continuum hypothesis states there isn't one.
Title: Re: Is .99999... = 1?
Post by: xiankai 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.
Title: Re: Is .99999... = 1?
Post by: lemonoman 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)?.

Title: Re: Is .99999... = 1?
Post by: Yggdrasil 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.
Title: Re: Is .99999... = 1?
Post by: xiankai 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.
Title: Re: Is .99999... = 1?
Post by: Yggdrasil 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 ?)
Title: Re: Is .99999... = 1?
Post by: FeLiXe 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.
Title: Re: Is .99999... = 1?
Post by: xiankai 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?
Title: Re: Is .99999... = 1?
Post by: Borek 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

Please elaborate.
Title: Re: Is .99999... = 1?
Post by: FeLiXe 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.
Title: Re: Is .99999... = 1?
Post by: Yggdrasil 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.
Title: Re: Is .99999... = 1?
Post by: xiankai 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.

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

Please elaborate.

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?
Title: Re: Is .99999... = 1?
Post by: Yggdrasil 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.

For more information you can see the Wikipedia article: http://en.wikipedia.org/wiki/Number
Title: Re: Is .99999... = 1?
Post by: constant thinker on December 05, 2006, 08:25:14 PM
Jeese. I never thought this would get so much attention.
Title: Re: Is .99999... = 1?
Post by: xiankai 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?
Title: Re: Is .99999... = 1?
Post by: Borek 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.

Anybody? Mark K?
Title: Re: Is .99999... = 1?
Post by: FeLiXe on December 06, 2006, 01:35:57 PM
you can check out Wikipedia for limits http://en.wikipedia.org/wiki/Limit_%28mathematics%29
Title: Re: Is .99999... = 1?
Post by: pantone159 on December 06, 2006, 08:52:26 PM
The limit is the number that you get closer and closer to, as you advance in your sequence/sum/whatever.
Formally, I think it is defined something like this:  Pick any small number (often named epsilon) that you like.  If an infinite sum approaches a limit L, that at some point the finite sum gets within epsilon of L (and doesn't leave).  This works no matter how small epsilon is.

Although a finite sum might never reach exactly the limit L, it will get arbitrary close.  Typically, the limit is never reached exactly (without infinitely many steps).  It isn't against the rules for something to get exactly to the limit (e.g. Borek's example), it just usually doesn't happen that way.

BTW, one the most important uses of limits is calculus.  Without the concept of limit, calculus makes no sense.  For example, the derivative of a function is defined as a limit.  Take the ratio ((f(x+dx) - f(x)) / (dx) and find the limit as dx->0.  That is the derivative.  dx never actually reaches 0, the ratio doesn't work then (0/0), it only gets very close.  Limits have exact values just like derivatives do.

Title: Re: Is .99999... = 1?
Post by: Borek on December 07, 2006, 04:58:20 AM
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.

On the second thought - limit is reached for every continuous function, so it happens quite often.
Title: Re: Is .99999... = 1?
Post by: Donaldson Tan on May 10, 2007, 03:23:50 PM
Jeese. I never thought this would get so much attention.

Haha..

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.

Since "<" and ">" are not applicable, what about "=" ?
Title: Re: Is .99999... = 1?
Post by: xiankai on May 15, 2007, 07:25:07 AM
Since "<" and ">" are not applicable, what about "=" ?

the inequality signs were there to posit the existence of a number between 0.999... and 1, which does not exist as has been shown. replacing them with '=' does not really prove anything since it could be replaced by '≠' too.

mathematically 0.999... = 1, but physically i refuse to accept it.  ;)
Title: Re: Is .99999... = 1?
Post by: Inestyne on October 10, 2007, 01:20:55 PM
0.999... = 0.9 + 0.09 + 0.009 + ...
               = 9(0.1 + 0.01 + 0.001 + ...)
               = 9(-1 + 1 + 0.1 + 0.01 + 0.001 + ...)
               = 9(-1 + ??n=0(1/10)n)
Since ??n=0rn = 1 / (1 - r) for |r| < 1,
0.999... = 9(-1 + 1/(1 - 1/10))
               = 9(-1 + 10/9)
               = 9(1/9)
               = 1
qed

What? elaborate
Title: Re: Is .99999... = 1?
Post by: Yggdrasil on October 10, 2007, 08:42:10 PM
0.999... is basically the sum of a geometric series (http://en.wikipedia.org/wiki/Geometric_series) and can be written in the form:

9 (-1 + Σn=0 (1/10)n)

Since the sum of a geometric series converges to a finite number, we can use this fact to prove that 0.999... = 1.
Title: Re: Is .99999... = 1?
Post by: Inestyne on October 14, 2007, 10:01:46 PM
0.999... is basically the sum of a geometric series (http://en.wikipedia.org/wiki/Geometric_series) and can be written in the form:

9 (-1 + Σn=0 (1/10)n)

Since the sum of a geometric series converges to a finite number, we can use this fact to prove that 0.999... = 1.

bingo
Title: Re: Is .99999... = 1?
Post by: enahs on October 14, 2007, 11:03:33 PM
While mathematicians were busy arguing over if 0.99999999.... =1  is true or false, chemist and physicists invented all kinds of cool and useful stuff.

Chemist and Physicists >  Mathematicians

What do you think about that relationship?

Title: Re: Is .99999... = 1?
Post by: Yggdrasil on October 15, 2007, 12:36:53 AM
And then biologists stole the tools from chemists and physicists to do more cool stuff, get NIH funding, and win chemistry nobel prizes

biologists > chemists and physicists

:P  (j/k)
Title: Re: Is .99999... = 1?
Post by: Borek on October 15, 2007, 02:40:33 AM
When mathematicians proved that 0.99999... = 1 chemists and physicians were unscrupulously using this results in their calculations; they never admitted where did the knowledge came from.

Biologists did not understand the proof nor its significance, so they happily ignored it :P

mathematicians > chemists + physicists + biologists

 :D
Title: Re: Is .99999... = 1?
Post by: pantone159 on October 15, 2007, 11:18:25 AM
It is less true for chemists and biologists, but physicists would have accomplished hardly anything without well-developed mathematics.
 
Title: Re: Is .99999... = 1?
Post by: Yggdrasil on October 15, 2007, 08:50:36 PM
And chemists would have accomplished hardly anything without well developed physics and biologists would have accomplished hardly anything without well developed chemistry.  Although the latter is probably more true than the former.
Title: Re: Is .99999... = 1?
Post by: Maz on October 16, 2007, 10:23:54 AM
lol, well I guess I'll have to chime in here.

Try to remember, chemistry is simply a fringe branch of physics (q.m. mixed with stat/thermal physics).  Biology is a fringe branch of chemistry.  Mathematics is worthless without relating to the physical world.  Enter Galileo, Kepler, Newton....Most early "physicists" were mathematicians by training who decided to stop futzing about with the useless abstract and started working in the useful abstract. 

So the point is mathematicians ~ physicists > chemists > biologists > social "scientists" (bah)

 ;D
Title: Re: Is .99999... = 1?
Post by: Mitch on October 17, 2007, 01:37:28 AM
Mitch > Your Mamma
Title: Re: Is .99999... = 1?
Post by: Dersan00 on December 08, 2009, 09:53:43 PM
The proof I've always seen associated with this involves fractions.

(1/3) = .3333...
3 x (1/3) = (3/3) = 1
3 x .3333... = .9999...

(3/3) = .9999....

Who the hell invented math anyways? I mean subtraction is the addition of negative numbers, division is multiplication of fractions, and all math can be done with just 1's and 0's.

CRAZY.
Title: Re: Is .99999... = 1?
Post by: juanrga on November 26, 2011, 03:35:28 PM
let c = 0.99999...

10c = 9.9999...

10c - c = 9.9999... - 0.99999...

9c = 9

c = 1

0.99999... = 1

Fastest proof

A=1; B=0.99999...

A - B = 0.00000...
Title: Re: Is .99999... = 1?
Post by: Benzene Martini on October 16, 2015, 10:54:20 PM
In theory, no.

In reality, it depends.

It is all relative.
Title: Re: Is .99999... = 1?
Post by: SirReal on October 16, 2015, 10:56:12 PM
You can't make that argument since the series of 9's after the decimal point is infinite.
Title: Re: Is .99999... = 1?
Post by: Benzene Martini on October 16, 2015, 11:07:58 PM
You can't make that argument since the series of 9's after the decimal point is infinite.

Yes you can. You just have to know how to present the data.
Title: Re: Is .99999... = 1?
Post by: SirReal on October 18, 2015, 03:00:05 PM
But 1/3 is = .33333, it is the infinite sum of:

Σ(n=1 :rarrow:∞) (3/10^n)  right?

So doesn't that mean that the limit of this series will indeed approach 1/3, and then the math is valid?  I think the problem is that the series is being treated as a number.  However, I am not a mathematician.  I will have to consult my calculus 2 professor from several terms ago, she is quite literally a genius who does research on algebraic topology.  Perhaps she will be able to shed light on this dilemma