1

# 0.99999.....=1

Jpsy

1
Fri 6 Dec, 2013 01:36 am
@Kolyo,
In fact, now that I think about, using common sense, .9 repeated does not = 1. With out going into 5 pages of obscure mathematics, just clear your head, and think about what you are saying. 1 = 1. .999999999999999999999999999999999999999999999999999999999999999 even if you wrote an infinity of 9s they would still all be 9s. They would not = 1. It's difficult and fun to try to rap your head around infinity, but if you just step back and think about it, I think you will see .99 repeated does not = 1. It only equals .99 repeated. Only 1 = 1.
It sticks out clearly in my mind that the limit of a function as x approaches infinity may be a whole number like 1, but I believe the definition of a limit states the function comes as close as possible to approaching that number without ever actually reaching that number. I think the answer to this question will be answered just by looking up the definition of a limit.
0 Replies

Jpsy

1
Fri 6 Dec, 2013 03:44 am
@Kolyo,
First Proof:
Quote:
Let x = .999...
Then 10x = 9.999...
9x = 10x - x = 9.999... - 0.999... = 9
So x = 1.

It's hard to argue with that proof, it was creative that you thought that up, but I can't help thinking there is a flaw in that proof. Something is wrong with the internal logic of that proof. Tomorrow I will graph that equation and manipulate it.

x= 3.33 repeated
10x = 33.33 repeated
10x-x= 33.333 - 3.333
x(10-1)= 30
9x = 30
x = 30/9
x= 3.33 repeated
well it worked for 3.33 repeated

x= .99998
10x= 9.9998
10x- x = 9.9998-.99998
9x = 8.99982
x=8.99982/9
x=.99998

Well it works, but I can't help thinking your breaking some mathematical rule in that equation, by just manipulating one number in an equation like that. Maybe I'm wrong.

x = 0
10x = 0
10x-x = 0
9x = 0
x = 0/9 = 0 huh

x= 1/3= .33 repeated
1/9 (x)= 1/27
1/9 x - x =1/27- 1/3
-8/9(x) = -8/27
x = 1/3
huh

x = .99 repeated
1/10 (x) =.099 repeated
1/10 (x) - x = -.899 repeated with a 1 at the end
-.9 x = -.89999999999999999999999999...99991

I don't believe this equation is solvable, that may provide a hint. The fact that the logic this equation is based upon doesn't work in this instance, may prove that the logic is flawed. you can't have the 9s go on for infinity and than add a 1. This equation may only be valid with numbers that are capable of being represented by a fraction.

It just seems like .9999 goes towards infinity difference between this number and 1 become vanishingly (I just made that word up, I'm like Sarah Palin) infinitesimally small but it doesn't quite close in on 1.
with .9 you are .1 away from 1.
with .99 you are .01 away from 1.
with .9999999999999999999999999999999999999999
you are .0000000000000000000000000000000001 away from 1.
There will always be that 1 in the difference no matter how far you take the .9s

Anyway I commend you Kolyo. We'd probably both be better off debating this, rather than debating social and political topics.
markr

1
Fri 6 Dec, 2013 02:36 pm
@Jpsy,
"x = .99 repeated
1/10 (x) =.099 repeated
1/10 (x) - x = -.899 repeated with a 1 at the end"

First, you can't have a 1 at the end of an infinite sequence of 9s. There is no end.
Second, x/10 - x = -0.9:

0.0 999999999999999999999
-
0.9 999999999999999999999
---------------------------------------
The infinite sequences of 9s (after the spaces) cancel each other.
0 Replies

Kolyo

1
Fri 6 Dec, 2013 04:48 pm
@Jpsy,
Jpsy wrote:

It's hard to argue with that proof, it was creative that you thought that up, but I can't help thinking there is a flaw in that proof. Something is wrong with the internal logic of that proof. Tomorrow I will graph that equation and manipulate it.

I didn't think that proof up but rather learned it in my Foundations of Math class in college. I don't trust it either. There's too much hand-waving involved. I provided it because I thought some people might like it better than the other two more rigorous proofs, both of which involve some sophistication.
0 Replies

Kolyo

1
Fri 6 Dec, 2013 04:52 pm
@Jpsy,
Jpsy wrote:

Well, now I see why you got so mad at my comment about mathematics. You are good at math.

You should also know that I'm not a woman, myself, but have loved many a mathematicienne in time.

That's why I get disconsolate when people say things that might stop women from studying the subject.
0 Replies

raprap

1
Sat 7 Dec, 2013 11:25 pm
Boy,! If ya'll are having such a hard time realizing that 0.999999999.......=1
I'd hate to see you with 7*(0.142857.........)=1
The dots (....) mean an endless repeating sequence

BTW
Let X=0.142857..........
So 1,000,000X=142857.142857.......
Then 1,000,000X-X=999,999X=142,857
And X=142,857/999,999=142,857/(7*142857)=1/7
so 7(1/7)=7/7=1

If you want to repeat this technique with any number, try 3, 7, 11, 13, 17, 19, 23 or any other prime number--it all works the same.

Rap
Brandon9000

2
Sat 7 Dec, 2013 11:41 pm
The assertion in the title is a well known fact accepted by mathematicians the world over for ages. I learned it in school nearly a half century ago.
0 Replies

Jpsy

1
Sun 8 Dec, 2013 01:50 am
@raprap,
But how can an inifinte number of 9s suddenly become 1? It defies logic (as mathematics and physics often do (but only in appearance). It will always be just shy of 1.
0 Replies

Jpsy

1
Sun 8 Dec, 2013 02:01 am
@raprap,
I even came up with an equation to make my point.
the difference between 1 and .99 repeated is the following.

lim as x goes to infinity 1/(10^x) = 0

but the definition of a limit is a number that the equation approaches but never actually reaches.

1-.9 = .1
1/(10^1)= .1

1-.99 = .01
1/(10^2) = .01

1-.999 = 1/(10^3) = .001

Brandon9000

1
Sun 8 Dec, 2013 03:10 am
@Jpsy,
Yes, you're smarter than generations of mathematicians. What you left out of your post is that this is an infinite number of 9s.
Jpsy

1
Sun 8 Dec, 2013 04:37 am
@Brandon9000,
I'm not trying to be a smartass. I'm using my own intuition and just having fun. I will go back and read your proofs on the previous 5 pages later.

Quote:
lim as x goes to infinity 1/(10^x) = 0

Tell me what's wrong with that limit. Are you following it.

that equation describes the difference between .99 repeated ( repeated means TO INFINITY) and 1, does it not. Am I wrong point it out.

the number for x represents the number of nines after the decimal place.

1/(x^10)

1-.9 = .1
1/(10^1)= .1

1-.99 = .01
1/(10^2) = .01

1-.999 = 1/(10^3) = .001

1-.99999999 = 1^(10^8)=.00000001

(there are eight 9s) (there are seven os and a 1)

So that equation works as far as I can tell.

lim as x goes to infinity for the equation 1^(10^x)= 0

A converging limit is an equation (as x goes to infinity) that approaches a number but never reaches it.

I'm not trying to be a smart ass or claiming to be a math genius. Tell me where I'm making an error. I will check out your explanations for why .99 repeated = 1 later. Remember I'm just doing this for fun, I'm not trying to claim I'm Newton. I've never heard that .99 repeated = 1 so I'll take your word for it, but for my own curiosity tell me what's wrong with my limit.

Jpsy

1
Sun 8 Dec, 2013 04:57 am
@Jpsy,
To clarify further

1- .99 repeated (repeated mean the 9s repeat for infinity)

should equal

lim as x goes to infinity for the equation 1^(10^x)= 0

Kolyo

1
Mon 9 Dec, 2013 03:31 pm
@Jpsy,
Jpsy wrote:

I even came up with an equation to make my point.
the difference between 1 and .99 repeated is the following.

lim as x goes to infinity 1/(10^x) = 0

but the definition of a limit is a number that the equation approaches but never actually reaches.

1-.9 = .1
1/(10^1)= .1

1-.99 = .01
1/(10^2) = .01

1-.999 = 1/(10^3) = .001

Right. As you keep adding 9's you get closer and closer to 1 but never reach it. Basically you're dealing with a sequence .9, .99, .999 and it gets closer and closer to 1 but never gets there. However .999... represents that very value that the sequence approaches but never reaches. Thus its value is 1.

If .999... were PART of the sequence it would be less than 1. But it's not part of the sequence. It is the number that the terms approach but never reach, the limit, 1.
Jpsy

1
Mon 9 Dec, 2013 04:41 pm
@Kolyo,
Yes, that makes sense to me. But many of the people who have responded on this thread have said that mathematicians have actually proved .99 repeated = 1. I really don't understand how, but I will go back and read your proof and their proofs. I don't know why but this stuff is kind of fun.
Kolyo

1
Mon 9 Dec, 2013 05:23 pm
@Jpsy,
But I'm one of the people saying 0.999... = 1.

The terms in the sequence 0.9, 0.99, 0.999, ... never reach 1, but "0.999..." is the limit of the sequence (not a term in it), and it equals 1.
0 Replies

Brandon9000

1
Tue 10 Dec, 2013 06:09 pm
@Jpsy,
Jpsy wrote:

..A converging limit is an equation (as x goes to infinity) that approaches a number but never reaches it...

No, that is not the definition of a limit. Have you had calculus? A limit is the number that an infinite sequence converges toward.
0 Replies

raprap

1
Tue 10 Dec, 2013 08:47 pm
@Jpsy,
Interesting you're making the argument that
0.9=1-(1/10^1)
0.99=1-(1/10^2)
0.999=1-(1/10^3)
0.9999=1-(1/10^4)
so
0.99999........=1-(1/10^n) with n=infinity

now lets do a little limiting

What is
lim n->infinity [1-(1/10^n)] = 1-lim n->infinity (1/10^n)=1-0=1
ergo
0.99999........=1

Rap
0 Replies

1
Wed 11 Dec, 2013 07:25 am
@Jpsy,
Jpsy wrote:

Yes, that makes sense to me. But many of the people who have responded on this thread have said that mathematicians have actually proved .99 repeated = 1. I really don't understand how, but I will go back and read your proof and their proofs. I don't know why but this stuff is kind of fun.

There are a number of ways to prove that .999... equals 1. There's even a Wikipedia page dedicated to it.

My favorite is this:

1/3 = .333...

3(1/3) = 3 (.333...)
3/3 = .999...
1 = .999...
Jpsy

1
Thu 12 Dec, 2013 03:10 am
1/3 = .333...

3(1/3) = 3 (.333...)
3/3 = .999...
1 = .999...

Such a simple and elegant proof.

It's been 15 years since I've touched a calculus book.

I guess the only way I can understand this is by looking at the proofs you guys present, because your proofs are pretty straight-forward and correct, but I'll never understand it verbally or visually. It kind of reminded me of the famous Greek philosopher's race problem. If each step you take is 1/2 between where you are and the finish line will you ever finish the race?
I thought the answer was no because you can continue to divide by 2 forever and therefore never reach the finish, but I guess from what you are showing me the answer could be yes.

Does 1/2 +1/4+1/8+1/16... = 1

infinity is a strange concept to deal with.

Can anyone explain verbally why .99 repeated = 1.
I still can't wrap my head around it.

Like I explained 1- .99999 with any amount of 9s, the difference is always going to have a 1 in the answer, except when the 999s go on for infinitely (which you've proven). Can anyone explain in words how and why that one disappears to leave you with 1-.99 repeated = 0. What is it about infinity that causes that .99 repeated to converge to 1?

Jpsy

1
Thu 12 Dec, 2013 03:36 am
@Jpsy,

Actually, a woman answers my question here
0 Replies

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