*12.8:*

**co.quora/math/****here's why I have a problem with (0.999...) = 1:**

. if the number (0.999...) = (1 - 1/infinity) = 1;

then the set [0,1) = {0, ... 1- 1/infinity} = {0, ... 1} = [0..1];

but then we have [0,1) = [0,1] so did we want to mean that?

12.17: wiki:

The equality of 0.999... and 1 is closely related to

the absence of nonzero infinitesimals in the real number system,

the most commonly used system in mathematical analysis.

Some alternative number systems, such as the hyperreals,

do contain nonzero infinitesimals;

and then the symbol "0.999..." admits the interpretation

of falling infinitesimally short of 1.

The equality 0.999... = 1 has long been accepted by mathematicians

because they are concerned with real numbers not hyperreals.

*12.27: me:*

0.999... is not a real number; it's a hyperreal;

because it is equal to 1 -1/infinity (the infinitesimal);

making it infinitely close to 1 but not real;

that's why 0.999... can't be equal to 1;

1 is a real; 0.999... is a hyperreal.

12.12:

**do you have a proof for (0.999...) = 1;**

or do you disagree with it?

the "..." indicates the 9's extend forever.

I'm saying that 0.99999999999999999999999...

is the largest number before 1.0;

but everybody on quora disagrees with the obvious!

12.15:

[a friend] Mon, Dec 14, 2015 at 10:11 PM,

I think that it equals 1. The proof that I believe is that

.33333333... = 1/3

And

3 x .3333333.... = .9999999......

So

3 x 1/3 =1

**me:**

I suspect that what you have proven is that

0.33333333... is not an exact representation of 1/3;

but it boggles the mind to say something like

the infinitely last digit should be 4 not 3.

there is no infinitely last digit!

therefore I suspect there is no way to say

1/3 exactly in base 10.

. let me try on you the same problem from a different angle:

[0,1) is set notation indicating

all the real numbers between 0 and 1 excluding 1.0;

what is the largest number in that set if not 0.999... ?

12.17:

**1/3 is not quite the same as 0.333...:**

1 = 3* 0.333... plus an infinitesimal:

0.333.. + 0.333... + 0.333... + x = 1;

0.999... + x = 1;

x = 1 - 0.999...

= 1 - (1- 1/infinity)

= 1-1 + 1/infinity

= 1/infinity

12.18: Professor Fred Richman:

. The cancellable elements are precisely the terminating decimals

because 0.999... + x = 1 + x for all nonterminating x.

Is 1/3 = 0.333...? Clearly if a sum is cancellable,

then each addend is cancellable,

so there is no decimal number x such that x + x + x = 1.

That is, 1/3 is not a decimal number.

More generally, no nonterminating decimal number x

can satisfy an equation of the form mx = n

with m and n positive integers.

16,17:

**special properties of infinitesimals:**

http://www.purplemath.com/modules/howcan1.htm

-- this page has many proofs which all use accepted math;

so you can learn a lot about the caveats of accepted math.

. the key to working with 0.999... is expressing it algebraically:

x = 0.999... = 1 - 1/infinity;

where 1/infinity represents the infinitesimal

an indefinitely small quantity.

. for instance

for any 2 reals there is another real between them;

but that applies only to numbers that aren't the sum of an infinitesimal;

because 1 - 1/infinity is right before 1.0.

. also with infinitesimally summed numbers

addition and multiplication doesn't work as expected:

**. the accepted algebraic proof is this:**

x = 0.999...

10x = 9.999...

subtract those equations to get:

9x = 9.000...;

x = 9/9 = 1;

**but the correct algebra is this:**

x = 0.999... = 1 - 1/infinity;

10x = 10 - 10/infinity;

subtracting them gets:

9x =

10 - 10/infinity -(1 - 1/infinity);

= 9 - 10/infinity + 1/infinity;

= 9 - 9/infinity;

= 9 - 1/infinity

= 8.999...;

x = 8.999.../9 = 0.999....

or x= (9 - 1/infinity)/9

= 1 - 1/infinity*9

= 1 - 1/infinity

= 0.999....

**here's a problem with addition of infinitesimals:**

from the previous example we have:

9x = 8.999...;

but if I say 8.999... = 8 + 0.999...;

then I get x = 1 again:

9x = 8 + 0.999...;

9x = 8 + x;

(9-1)x = 8;

x = 1;

it turns out that for x to remain 1 - 1/infinity;

8.999... = 7.999... + 0.999...;

then 10 x - x = 7.999... + x;

10x - x -x = 7.999...;

x = (7.999...)/8

= (8 - 1/infinity) /8

= 1 - 1/infinity*8

= 1 - 1/infinity = 0.999...

so why did 9 -1/infinity = 7.999... + 0.999...

= (8 - 1/infinity) + (1 - 1/infinity) ?

instead of 8 + 0.999...

= 8 + (1 - 1/infinity)?

it's as if the infinitesimal subtraction is applying to the entire number;

so if you make a number the sum of its integer and fraction parts

the infinitesimal subtraction needs to happen to both parts.

9 -1/infinity = (8+1) - 1/infinity

= (8- 1/infinity) +(1 - 1/infinity);

Kalid Azad`Do Infinitesimals Exist?:

Our current number system assumes the long-standing

Archimedean property:

if a number is smaller than every other number, it must be zero.

If we assume infinitely small numbers don’t exist,

we can show 0.999… = 1.

Karin Usadi Katz, Mikhail G. Katz 2008:

*A strict non-standard inequality .999... ≺ 1*

So long as the number system has not been specified,

the students' hunch that .999... can fall infinitesimally short of 1,

can be justified in a mathematically rigorous fashion.

12.18:

Professor Fred Richman:

. we see that in the traditional definition of the real numbers,

the equation 0.999... = 1

is built in at the beginning.

That is why anyone who challenges that equation

is, in fact, challenging the traditional

formal view of the real numbers.

. Clearly 0.999... = 1 + 0¯,

so 0¯ is a sort of negative infinitesimal.

On the other hand,

you can't solve the equation 0.999... + X = 1

because, in [the model of real numbers],

the sum of a traditional real with any real

is a traditional real.

H. Jerome Keisler`Elementary Calculus: An

**Infinitesimal Approach**

/Hyperreal Numbers

. in the real number system, only zero is an infinitesimal;

hyperreal numbers {R*} contain infinitesimals that are non-zero.

if a-b is infinitesimal, we say they are infinitely close to one another.

1/infinitesimal = infinite.

**the extension principle:**

reals are a subset of hyperreals;

the order relation of reals is a subset of that of hyperreals.

. 1/infinity is greater than zero

yet less than every real number;

for every real function, f,

there is the natural extension of f, f*,

that includes hyperreals.

[. there is some confusion about the term infinitesimal;

because it should always mean non-zero but infinitely small;

but in the real number system,

zero is considered infinitesimal.]

if you mean non-zero but infinitely small;

then you say {positive, negative} infinitesimal.

**transfer principle:**

. every statement about the reals

is also true for the hyperreals.

[ I can think of a counter example to this:

2-1/infinity = 1.999.... = 1 + 0.999...;

but that result leads to a proof that 0.999... = 1;

so for hyperreals the negative infinitesimal

needs to apply to both parts:

1.999.... = 0.999... + 0.999... .]

. for x in [0,1), x^2 is less than x;

so (0.999...)^2 = (1- 1/infinity)^2 is less than (1- 1/infinity);

[ and that makes sense intuitively,

but what number is it now? 0.999....8 ?

the 9's were going out forever, so where do you truncate to place the 8?

and if you try to distribute it:

(1- 1/infinity)^2 = 1^2 - (1/infinity)^2

and infinity * infinity is still an infinity.

so infinitesimals are not transferring to real functions so well.

according to the natural number exponent functions

0.999... is the same as 1.

]

. the following operations on infinitesimals

are themselves infinitesimals, given in increasing order

(there is some infinitesimal difference between them):

(1/infinity)^3, (1/infinity)^2,

1/infinity*100, 1/infinity, 100/infinity,

(1/infinity)^1/2, 1/infinity + (1/infinity)^1/2

[ no proof is given as how various infinities can vary in size;

but perhaps there are algebra tricks where it matters,

when you can get 2 infinities to cancel each other...]

. infinitesimal /infinity is infinitesimal;

[ (1/infinity) /infinity

= (1/infinity)* (1/infinity)

= (1/infinity)^2 ]

infinitesimal/infinitesimal^2 = infinity;

( 1/infinity ) / ( (1/infinity)(1/infinity) )

= 1 / 1/infinity

**standard part principle:**

. every finite hyperreal number

(eg, 0.999... = 1- 1/infinity);

is infinitely close to exactly one real number

(eg 1- 1/infinity is infinitely close to 1).

12.18:

**what is the Archimedean Property?**

stackexchange`Archimedean Property:

. for all e in reals > 0, there is some n in integers,

such that 1/n is less than e.

This essentially means that there are no

infinitesimally small elements in the real line,

no matter how small e gets

we will always be able to find an even smaller

positive real number of the form 1/n.

You may want to note that the Archimedean Property of R

is one of the most important consequences of

its completeness (Least Upper Bound Property).

In particular, it is essential in proving that

1/n converges to 0, an elementary but fundamental fact.

[ but in the hyperreal system,

it converges to the infinitesimal, not zero.]

planetmath`Archimedean property:

Let x be any real number.

Then there exists a natural number n such that n>x.

[ there is no number greater than infinity;

but infinity is not a real number.]

**Corollary:**

If w is a real number greater than 0,

there exists a natural n such that 0≺1/n≺w.

[ there is no number between zero and 1/infinity

but 1/infinity is hyperreal not real.]

**Corollary:**

If x and y are real numbers with x≺y,

there exists a rational number r such that x≺r≺y.

[ 0.999... is hyperreal not real;

so there is no rational between 0.999... and 1 .]

12.27:

**(0.999...) is a hyperreal:**

. the key to refuting 0.999... = 1

is explaining that 0.999... is not a real number;

the numbers that are a sum of an infinitesimal

are allowed only in the hyperreal number system

not the real number system.

**co.quora/math/problem with (0.999...) = 1:**

to Uri Granta, Anders Kaseorg, Sridhar-Ramesh, Mark-Eichenlaub

0.999... is not a real number; it's a hyperreal;

because it is equal to 1 -1/infinity (the infinitesimal);

making it infinitely close to 1 but not real;

that's why 0.999... can't be equal to 1;

1 is a real; 0.999... is a hyperreal.

Professor Fred Richman of Florida Atlantic University

(http://math.fau.edu/richman/)

wrote a paper on the following topic which can be read at his site:

http://math.fau.edu/richman/HTML/999.htm

In short, they are different ways of saying the same thing.

me:

the paper actually says this(0.9* = 0.999....):

"Clearly 0.9* = 1 + 0¯,

so 0¯ is a sort of negative infinitesimal.

On the other hand,

you can't solve the equation 0.9* + *X* = 1

because, in cut *D*,

the sum of a traditional real with any real

is a traditional real."

so, 0.999... is not a real number;

it's a hyperreal; because

it is equal to 1 -1/infinity (the infinitesimal);

making it infinitely close to 1 but not real;

that's why 0.999... can't be equal to 1;

1 is a real; 0.999... is a hyperreal.