# .999999999999... = 1 ?

### #1

Posted 24 August 2004 - 11:15 PM

### #2

Posted 24 August 2004 - 11:19 PM

### #3

Posted 25 August 2004 - 07:50 AM

**NO !**

**lim (x,x,1,-1)**

OR

**9999999999.../10000000000... = 0.9999999999...**

**Edited by P.J, 25 August 2004 - 07:52 AM.**

### #4

Posted 25 August 2004 - 08:00 AM

10x=9.999999999999...

10x-x=9

9x=9

x=1

--> 0.9999999999...=1

### #5

Posted 25 August 2004 - 01:46 PM

For example :x=0.9999999999...

10x=9.999999999999...

10x-x=9

9x=9

x=1

--> 0.9999999999...=1

x=0.999

**Correct**

10x=9.99

**Correct**

10x-x=9

**Wrong**

10x-x=8.991

**NOW Correct**

=> 0.9999999... â‰ 1

**Correct**

### #6

Posted 25 August 2004 - 02:04 PM

x=0.999 Correct

10x=9.99 Correct

excl.gif 10x-x=9 excl.gif Wrong

10x-x=8.991 rock.gif NOW Correct

I mean:

10x=9.999999

10x-x=9.99999-x (and 9.99999-x=9)

### #7

Posted 25 August 2004 - 02:10 PM

I mean:

10x=9.999999

10x-x=9.99999-x (and 9.99999-x=9)

Your x = 0.9999999

then 10x = 9.999999

But 10x-x = 9.999999 - 0.9999999 =

**8.9999991**WRONG

**Please STOP it**

### #8

Posted 25 August 2004 - 02:19 PM

It's correct if you look at infinite and now don't kill my brilliant formula

### #9

Posted 25 August 2004 - 02:19 PM

another example :

0.99999999...=0.33333333...*3=1/3*3=1

!!

### #10

Posted 25 August 2004 - 02:24 PM

Maexx is right !!!

another example :

0.99999999...=0.33333333...*3=1/3*3=1

!!

you said :

**0.33333333...*3=1**

Please update yourself !

**Edited by P.J, 25 August 2004 - 02:28 PM.**

### #11

Posted 25 August 2004 - 02:31 PM

I asked some year ago from my mathematics professor.

Some mathematics belive that real 1 doesn't exist. 1 is 0.9999999999..

### #12

Posted 25 August 2004 - 02:35 PM

### #13

Posted 25 August 2004 - 03:03 PM

1 is ALWYS unequal 0.999999. If you do not agree let's do the folowing deal.

You give me 1 Euro and I give you 0.99 Euro (limited by the smallest

piece of money. Else we could also do it with 0.99999999) in return.

No lets do this 10^100 times. When you are right then this mustn't matter

to you. When I am you'll be a poor looser

x=0.9999999999...

10x=9.999999999999...

10x-x=9

9x=9

x=1

--> 0.9999999999...=1

WRONG!

Multiplication is, in this case with indefinite numbers NO

equivalent rewriting of a formula. You can't do this since you

loose the exactness.

Your result is based on NO exactness and if you are not exact you can

also write that you simply round the last number and then get 1.

### #14

Posted 25 August 2004 - 03:23 PM

### #15

Posted 25 August 2004 - 03:28 PM

### #16

Posted 25 August 2004 - 03:35 PM

you said : 0.33333333...*3=1

Please update yourself !

PJ i don't understand

and i've seen last year maexx's demonstration in mathematics so 0.999999...=1

### #17

Posted 25 August 2004 - 05:41 PM

1 is unequal 0.99999999999999999999999 : agree

1 is unequal 0.999... : disagree

0,99999.....99999 != 1 : agree

0,99999.....99999 = 1 : stupid

0,99999.....99999.... = 1 : agree

0,99999.....99999.... != 1 : disagree

0,99999... = 1

0,99999... = lim 1 - 10^-n | n->+inf

= lim 1 - lim 10^-n

= 1 - 0

= 1

or

0,99999.... = 0,9 + 0,09 + 0,009 + 0,0009 + ...

= sum(9*10^-n | for n=1 to +inf)

= 9*sum(10^-n | for n=1 to +inf)

= 9* lim ( (1/10)^(n+1) - (1/10)) / (1/10 - 1) | n->+inf

= 9* (lim (1/10)^(n+1) - lim 1/10) / (lim (1/10 - 1))

= 9* (0 - 1/10) / (-9/10)

= 1

### #18

Posted 25 August 2004 - 05:51 PM

I have the same demonstration as the others

10* 0.99999... (= 9.9999....)

- 1* 0.99999... (= 0.9999...)

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

9*0.9999... = 9

Thus 0.9999... = 1.

EDIT:

there is no problem multiplying some number with infinite decimals: it not, you couldn't write something like 10*sqrt(2) for example...WRONG!

Multiplication is, in this case with indefinite numbers NO

equivalent rewriting of a formula. You can't do this since you

loose the exactness.

Your result is based on NO exactness and if you are not exact you can

also write that you simply round the last number and then get 1.

### #19

Posted 25 August 2004 - 05:52 PM

0.999......... is equal to 1

refer to your school books.

@PJ: az daste to peyman, ebbin che joori mardoom sare kar gozashti???

### #20

Posted 25 August 2004 - 06:05 PM

x=0.9999999...(infinite)

10x=9.999999...(infinite)

10x=9+0.99999...(infinite)

10x=9+x

9x=9

x=1

so 0.9999999... (infinite)=1 !!

### #21

Posted 25 August 2004 - 09:16 PM

### #22

Posted 25 August 2004 - 09:21 PM

### #23

Posted 25 August 2004 - 09:25 PM

.9 = close to 1

.99 = closer to 1

.999 = even closer to 1

.9999...(insert lots of 9's here)...999 = really close to 1

it would go on like that closing the distance exponentially, yet i think by the very definition of exponential growth/decay it can never reach 1 because the distance it covers lessens each time the number gets closer to 1.

am i making sense to anyone but myself?

### #24

Posted 25 August 2004 - 09:48 PM

Heyyeah. It's very simple problem.

0.999......... is equal to 1

refer to your school books.

@PJ: az daste to peyman, ebbin che joori mardoom sare kar gozashti???

You said : If 0<x<1 and 0<yâ‰¤1, then x=y

Please get it to CP300

@ Daruosh: Lotfan saket ! Mibini ke felan "qwerty841" sare karemun gozashteh va to !

### #25

Posted 25 August 2004 - 09:57 PM

Everyone with school education (and higher) knows that 0.9999... in equal to one. There is a algorithm that to convert fraction numbers into floationg point and revers. I don't want to explain and proove the algorithm here. But its obvious (very very) that 0.9999... is equal to 1.

If you don't want to agree, no one can make you (or force you) to agree.

### #26

Posted 25 August 2004 - 10:03 PM

What about "0.0000...1" ?

### #27

Posted 25 August 2004 - 10:04 PM

### #28

Posted 25 August 2004 - 10:08 PM

### #29

Posted 25 August 2004 - 10:15 PM

Stop playing game with digits and words If you like to discuss about philosophy of limits and numbers, start a new thread. I can overcome you if you are interested.

### #30

Posted 25 August 2004 - 10:20 PM

Hey Daruosh, I'm very busy. I don't have any free time !Hey man

Stop playing game with digits and words If you like to discuss about philosophy of limits and numbers, start a new thread. I can overcome you if you are interested.

Please STOP it

@Daruosh: To ke hamishe tu inja pelasi va karo zendegi nadari, vali man na !

### #31

Posted 26 August 2004 - 12:42 AM

mitoonam beporsam shoma vase chi busy hasti???

### #32

Posted 26 August 2004 - 12:51 AM

yupi took it (do you have mrs. baker?)

.222222222222222222... = 2/9

.999999999999999999... = 9/9 = 1

### #33

Posted 26 August 2004 - 01:03 AM

The thread is closed

### #34

Posted 26 August 2004 - 02:19 AM

### #35

Posted 26 August 2004 - 08:24 AM

I don't agree tooi dont agree... you'll have to explain it in logical rather than mathmatical terms if you want me to understand...

@Daruosh: agar click koni, mifahmi ! (dar zemn, kare web ke dige arzeshi nadare va har ki balade va hoghughesh ham ziad nish. mage Noghtechin ro nadidi ? man ye pam to un vare abe va ye pam ham inja !)

### #36

Posted 26 August 2004 - 09:40 AM

Why is 0.9999... = 1?

In modern mathematics, the string of symbols 0.9999... is understood

to be a shorthand for ``the infinite sum 9/10 + 9/100 + 9/1000 +

...''. This in turn is shorthand for ``the limit of the sequence of

real numbers 9/10, 9/10 + 9/100, 9/10 + 9/100 + 9/1000, ...''. Using

the well-known epsilon-delta definition of the limit (you can find it

in any of the given references on analysis), one can easily show that

this limit is 1. The statement that 0.9999... = 1 is simply an

abbreviation of this fact.

0.9999... = sum_(n = 1)^(oo) (9)/(10^n) = lim_(m --> oo) sum_(n = 1)^m

(9)/(10^n)

Choose varepsilon > 0. Suppose delta = 1/- log_(10) varepsilon , thus

varepsilon = 10^(-1/delta). For every m > 1/delta we have that

sum_(n = 1)^m (9)/(10^n) - 1 = (1)/(10^m) < (1)/(10^(1/delta)) =

varepsilon

So by the varepsilon - delta definition of the limit we have

lim_(m --> oo) sum_(n = 1)^m (9)/(10^n) = 1

Not formal enough? In that case you need to go back to the

construction of the number system. After you have constructed the

reals (Cauchy sequences are well suited for this case, see

[Shapiro75]), you can indeed verify that the preceding proof correctly

shows 0.9999... = 1.

An informal argument could be given by noticing that the following

sequence of ``natural'' operations has as a consequence 0.9999... = 1.

Therefore it's ``natural'' to assume 0.9999... = 1.

x = 0.9999....

10 x = 10 * 0.9999...

10 x = 9.9999...

10 x - x = 9.99999... - 0.9999...

Thus 0.9999... = 1.

An even easier argument multiplies both sides of 0.3333... = 1/3 by 3.

The result is 0.9999... = 3/3 = 1.

Another informal argument is to notice that all periodic numbers such

as 0.46464646... are equal to the period divided over the same number

of 9s. Thus 0.46464646... = 46/99. Applying the same argument to

0.9999... = 9/9 = 1.

Although the three informal arguments might convince you that

0.9999... = 1, they are not complete proofs. Basically, you need to

prove that each step on the way is allowed and is correct. They are

also ``clumsy'' ways to prove the equality since they go around the

bush: proving 0.9999... = 1 directly is much easier.

You can even have that while you are proving it the ``clumsy'' way,

you get proof of the result in another way. For instance, in the first

argument the first step is showing that 0.9999... is real indeed. You

can do this by giving the formal proof stated in the beginning of this

FAQ question. But then you have 0.9999... = 1 as corollary. So the

rest of the argument is irrelevant: you already proved what you wanted

to prove.

References

R.V. Churchill and J.W. Brown. Complex Variables and Applications.

5^(th) ed., McGraw-Hill, 1990.

E. Hewitt and K. Stromberg. Real and Abstract Analysis.

Springer-Verlag, Berlin, 1965.

W. Rudin. Principles of Mathematical Analysis. McGraw-Hill, 1976.

L. Shapiro. Introduction to Abstract Algebra. McGraw-Hill, 1975.

I've found that here: http://ejo.univ-lyon...th-faq/0.999999

Is this clear now?

### #37

Posted 26 August 2004 - 09:49 AM

### #38

Posted 26 August 2004 - 02:01 PM

And you should considder that

lim -> inf. of 0.9999 .... = 1 (true)

is NEVER equal to

0.9999.... = 1

according to your theory

9/0 is infinite

since it would be the same like

lim x -> zero of 9/x.

But we all know that this is not defined.

You can not assume that a lim. is the same as the equal sign.

You can write:

lim -> inf. 0.9999.... =1

but NEVER

0.9999... = 1 Because this is simply WRONG.

As I said this is a matter of exactness.

And you loose exactness by multiplying an undefined number.

And it the difference of the last chiffre (in the indefinite) is only 1 then this

IS important. It is not for numbers like sqrt(2) since rounding those does not affect the whole number but only the last few numbers.

It is a fact of LOGIC thinking and not of ACADEMIC thinking that they are

unequal.

### #39

Posted 26 August 2004 - 03:27 PM

it is not a question, it is a fact, as 2+2=4, 0.99999999(...)=1

with the infinite there is no 1 at the end of 1-0.9999999... because it would say the number is finished

so 1-0.99999..=0

moreover you can convert decimal numbers whith known sequences (as 0.454545 or 0.242424..) into fractions

for example

x=0.242424..

100x=24.2424...

100x=24+x

99x=24

x=24/99

with the same demonstration

x=0.999999

10x=9.9999999

10x=9+x

9x=9

x=1

as we have already said

### #40

Posted 26 August 2004 - 03:33 PM

You're wrong. 9/0 is not defined because it IS equal to lim x->0 of 9/x, and this limit doesn't exist. This lim = inf for x->0 positive, and =-inf for x->0 negative, so it cannot exist.according to your theory

9/0 is infinite

since it would be the same like

lim x -> zero of 9/x.

But we all know that this is not defined.

You can not assume that a lim. is the same as the equal sign.

How do you find the "last chiffre"? there is no one, and that's why we can say (and proove) that 0.9999... =1And it the difference of the last chiffre (in the indefinite) is only 1 then this

IS important. It is not for numbers like sqrt(2) since rounding those does not affect the whole number but only the last few numbers.

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users