326 replies to this topic

[qwerty841]

i am taking precalculus right now, and thought it would be fun to see if anyone else did

[Bob Vila]

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

[PJay]

NO !

lim (x,x,1,-1)

OR

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

Edited by P.J, 25 August 2004 - 06:52 AM.

[Maexxx]

x=0.9999999999...
10x=9.999999999999...
10x-x=9
9x=9
x=1

--> 0.9999999999...=1

[PJay]

x=0.9999999999...
10x=9.999999999999...
10x-x=9
9x=9
x=1

--> 0.9999999999...=1

<{POST_SNAPBACK}>

For example :

x=0.999 Correct
10x=9.99 Correct

10x-x=9 Wrong

10x-x=8.991 NOW Correct

=> 0.9999999... ≠ 1 Correct

[Maexxx]

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)

[PJay]

I mean:

10x=9.999999
10x-x=9.99999-x    (and 9.99999-x=9)

<{POST_SNAPBACK}>

Your x = 0.9999999
then 10x = 9.999999
But 10x-x = 9.999999 - 0.9999999 = 8.9999991 WRONG


Please STOP it

[Maexxx]

yeah but 0.9999999.. is never ending so 9.9999-0.999999~9
It's correct if you look at infinite and now don't kill my brilliant formula

[Bija]

Maexx is right !!!

another example :

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

!!

[PJay]

Maexx is right !!!

another example :

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

!!

<{POST_SNAPBACK}>

you said : 0.33333333...*3=1

Please update yourself !

Edited by P.J, 25 August 2004 - 01:28 PM.

[Daruosh]

Maexx is right.
I asked some year ago from my mathematics professor.
Some mathematics belive that real 1 doesn't exist. 1 is 0.9999999999..

[PJay]

Form your mathematics professor ?

[huhn_m]

All bullshit ...

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.

[PJay]

Is "0.999... = 1" a JOKE ?

[huhn_m]

it must be since it is simply WRONG.

[Bija]

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

[Overlord]

1 is ALWYS unequal 0.999999 : agree
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

[Orwell]

Sorry PJ and huhn_m, but they are right, it is TRUE that 0.999...=1

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:

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.

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

[Daruosh]

yeah. 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???

[Bija]

yes i don't see what you find wrong in this demonstration :

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 !!

[betoe]

Lol it seems overlord, huhn_m and me voted for no. I agree with them since the first time i saw this topic. I had voted yesterday, the poll was 1 yes and 1 no (me ).

[Overlord]

i voted yes so I don't agree with you

[CrimsonCasio]

you know... i may suck at math but it seems common sense to me that 0.999 repeating is not 1, it mearly approches 1 infinately and over time becomes almost undistinuisable from 1 because they are so close:

.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?

[PJay]

yeah. 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???

<{POST_SNAPBACK}>

Hey
You said : If 0<x<1 and 0<y≤1, then x=y
Please get it to CP300

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

[Daruosh]

This absolutely a theoretical problem.
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.

[PJay]

1 - 0.99999... = 0.0000...1
What about "0.0000...1" ?

[Daruosh]

0.00000....1 is equal to 0

[PJay]

0.00000....1 is equal to 0

<{POST_SNAPBACK}>

In "0.00000....1" What about 1 ?

[Daruosh]

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.

[PJay]

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.

<{POST_SNAPBACK}>

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

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

[Daruosh]

@Peyman: na aziize baradar. man 3 ja kar mikonam (be onvane programmer va Web Designer), ama chon to mihit haye kar computeram hamishe online hast, web gadri ham mikonam.
mitoonam beporsam shoma vase chi busy hasti???

[qwerty841]

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

<{POST_SNAPBACK}>

yup

.222222222222222222... = 2/9
.999999999999999999... = 9/9 = 1

[Daruosh]

Ok. We all agree that 0.99999......... is equal to 1.
The thread is closed

[CrimsonCasio]

i dont agree... you'll have to explain it in logical rather than mathmatical terms if you want me to understand...

[PJay]

i dont agree... you'll have to explain it in logical rather than mathmatical terms if you want me to understand...

<{POST_SNAPBACK}>

I don't agree too
@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 !)

[Orwell]

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?

[Maexxx]

as I said

[huhn_m]

it might be clear in theory but it is NEVER logic.

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.

[Bija]

I don't understand that peolple still stay that 0.99999... is different that 1 after all the mathematics demonstrations given

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

[Orwell]

Interesting but...

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'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.

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.

How do you find the "last chiffre"? there is no one, and that's why we can say (and proove) that 0.9999... =1