Is 1 = 0.9999......

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edfcmc

Senior member
May 24, 2001
531
0
71
x = 0.9999...
10x = (10)0.9999...
10x - x = (10)0.9999... - 0.9999...
9x = 10
x = 10/9

hahahahaahaa
 

Krk3561

Diamond Member
Jun 12, 2002
3,242
0
0
Originally posted by: Kyteland
We're having a debate at work. Is 1=0.99999..... repeating. I say that this holds but one of my coworkers claims that multiplication breaks down for an infinitely repeating number.

x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.

What do you think?

9 x .99999999999999999999999999999999999999999999999999999999999999999999999999 does not equal 1. Any calculator (except for a very accurate computer) you put that in will give you 9 because it cant process that many digits, so it will round up
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Originally posted by: SilentRunning
Silverpig what is the fraction which is equal to 0.999999........ and don't say 1/1 because dividing 1 by 1 equals 1 with a remainder 0.


If you can represent 0.999999..... with a fractional equivalent then I will concede that 0.99999.... is not an irrational number.

Did you not see how I showed that 1/1 = 0.999... via long division?
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Originally posted by: SilentRunning
OK, part of the proof


For -1<r<1 , the sum converges as n approaches infinity ,in which case

Sure seems to be stating an approximation to me. Converge, does not equate to equal to.

They are stating that r^(n+1) approaches zero as n approaches infinity

That is all for now, must do something constructive.

That's all fine and dandy but where in 0.999... do you see a sum? How can a number approach anything? What does a single number converge to?

All rational numbers are repeating decimals.

2/1 = 2.000000000000000000...

5/4 = 1.250000000000000000...

Do you agree?
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Originally posted by: MadRat
Hence, objective values cannot be used to define subjective ones.

What, pray tell, is an objective value, and what is a subjective value?
 

SilentRunning

Golden Member
Aug 8, 2001
1,493
0
76
Originally posted by: silverpig
Originally posted by: SilentRunning
OK, part of the proof


For -1<r<1 , the sum converges as n approaches infinity ,in which case

Sure seems to be stating an approximation to me. Converge, does not equate to equal to.

They are stating that r^(n+1) approaches zero as n approaches infinity

That is all for now, must do something constructive.

That's all fine and dandy but where in 0.999... do you see a sum? How can a number approach anything? What does a single number converge to?

All rational numbers are repeating decimals.

2/1 = 2.000000000000000000...

5/4 = 1.250000000000000000...

Do you agree?

First off the response was to the equations presented by Haircut.

If you divide 1 by 1 it equals

1.00000...
or 0.9999999... + 10^-(n) as n ---> infinity (the remainder will not just disappear)

In your above examples the remainder = 0 it does not approach zero

This really comes down to what you were taught I guess. Either you were taught:

1. that infinitely small numbers are equal to zero.
2. that an infinitely small number could be approximated by zero but was not equal to zero.

I was taught number 2 apparently you were taught number 1. Case closed
 

aux

Senior member
Mar 16, 2002
533
0
0
Originally posted by: Alphathree33

...
Again, properties of the real number system are not approximate. They are axioms. These things work by definition, not by proof or by punching a lot of numbers on your calculator and scratching your head.

Actually, we can define the real number system by Dedekind cuts (as done in Rudin's Principles of Mathematical Analysis, on a side note -- it's a shame that many people with math degrees have never seen this book), the construction uses approximations (or, if one prefers -- limits).
 

Rainsford

Lifer
Apr 25, 2001
17,515
0
0
I would say yes because that proof makes sense. However, could someone explain something to me, I still do not understand this. Hopefully I can make this clear enough.

.9 is 9/10, right? So it could be considered as you moving 9/10 of the distance between two points (say, 0 and 1), right?
Now .09 is 9/100 which could also be thought of as 9/10 * 1/10. When you add .09 to .9 this could be thought of as moving 9/10 of the remaining 1/10 of the distance, right?
So for each 9 you add to the end, you can think of it as moving 9/10 of the remaining distance. So the distance between you and the second point (1) gets smaller and smaller with each 9 added on to the end. However, it does not seem like you would ever actually reach the second point because you are only moving a fraction of the remaining distance each time. No matter how small, that distance would still exist, right? Or am I missing something?
 

Haircut

Platinum Member
Apr 23, 2000
2,248
0
0
Originally posted by: Rainsford
I would say yes because that proof makes sense. However, could someone explain something to me, I still do not understand this. Hopefully I can make this clear enough.

.9 is 9/10, right? So it could be considered as you moving 9/10 of the distance between two points (say, 0 and 1), right?
Now .09 is 9/100 which could also be thought of as 9/10 * 1/10. When you add .09 to .9 this could be thought of as moving 9/10 of the remaining 1/10 of the distance, right?
So for each 9 you add to the end, you can think of it as moving 9/10 of the remaining distance. So the distance between you and the second point (1) gets smaller and smaller with each 9 added on to the end. However, it does not seem like you would ever actually reach the second point because you are only moving a fraction of the remaining distance each time. No matter how small, that distance would still exist, right? Or am I missing something?
You are correct with your thinking here.
If we have a line where we start at 0.9 and for each unit of length we move along that line we add another 9 to the end of the previous number, hence at point 2 we have 0.99, point 3 0.999 and so on.
We can keep going on this line for as long as we like and we will never reach 1, there will always be some non-zero number between the number at your current position on the line and 1.
Similarly we will never reach 0.999..., it isn't that it is at the end of the line or that it is a long way down the line, 0.999... does not exist as a point on this line at all in the same way that infinity does not exist on the real line.
This is quite a difficult concept to grasp and it is obvious that some of the posters here who have tried to dispute that 0.999... = 1 don't fully get this concept either.

 

Yomicron

Golden Member
Mar 5, 2002
1,735
1
81
Originally posted by: Rainsford
I would say yes because that proof makes sense. However, could someone explain something to me, I still do not understand this. Hopefully I can make this clear enough.

.9 is 9/10, right? So it could be considered as you moving 9/10 of the distance between two points (say, 0 and 1), right?
Now .09 is 9/100 which could also be thought of as 9/10 * 1/10. When you add .09 to .9 this could be thought of as moving 9/10 of the remaining 1/10 of the distance, right?
So for each 9 you add to the end, you can think of it as moving 9/10 of the remaining distance. So the distance between you and the second point (1) gets smaller and smaller with each 9 added on to the end. However, it does not seem like you would ever actually reach the second point because you are only moving a fraction of the remaining distance each time. No matter how small, that distance would still exist, right? Or am I missing something?
lets say you have n number of 9's, as n approaches infinity the value approaches 1, but we are conserned with when n equals infintiy.

if you were to just keep adding 9s you would never reach 1, because you are looking at a finite number of 9s
 

josphII

Banned
Nov 24, 2001
1,490
0
0
Originally posted by: Woodchuck2000
incorrect. as stated earlier a single number is in fact, a single number and nothing more. single numbers, such as 0.999..., tend toward nothing. only series of numbers, or sums, can trend toward something, or have a limit.
Any how do you intend to define .99 reccuring without using some kind of series? As the floating point precision with which you define the number increases, it tends towards one.

As far as your 'proof' goes, I'd like to start by challenging
Proof: 0.9999... = Sum 9/10^n
(n=1 -> Infinity)
which makes no sense whatsoever.

followed by
= lim sum 9/10^n
(m -> Infinity) (n=1 -> m)
Where you've introduced an undefined variable m and done mysterious things with it.
I'll challenge the rest once you've explained the first premises.

BTW You're going to have to explain your notation a little as it's mathematical in no sense of the word.
Does m -> infinity mean 'as m tends towards infinity' or are you defining a range?

omfg your a moron. that proof is stated in about 1000 math text books. your not arguing with me on this one but the ENTIRE math community. and if you cant understand the notation, then you arent even qualified to participate in this discussion
 

josphII

Banned
Nov 24, 2001
1,490
0
0
Originally posted by: DoNotDisturb
Originally posted by: bleeb
They are not the same because you can NEVER reach infinity. And the only way for 0.9999... to converge to 1 is if you reach infinity, but there is no way to reach infinity. THERFORE 0.9999.... != 1. (hhaha i love it)

that is true. anything that is considered "infinite" is theoretical, the LIMIT approaches 1 as n -> infinity, HOWEVER, there is theoretically no "end point", thus you cannot say it's exactly equal to 1. it gets closer and closer to 1, but will never equal it because infinity is not attainable. if you don't know why its theoretical, read on infinity. if you look at the proof in that dr.math site, n -> infinity.

god damn ignorant fuks like you irritate me. let me walk you through the proof

the dr math PROOF isnt arbitrarily taking the limit of something. the quantity {9/10^n (for n=1 to inf)} EQUALS the limit of 9/10^n, for n=1 to m, and as m approaches inf.

the proof is not saying the limit of 0.99999.... is something, the proof says 0.999.... equals the limit of something
 

josphII

Banned
Nov 24, 2001
1,490
0
0
Originally posted by: Woodchuck2000
As proofs go, it's badly phrased in my opinion...
As i stated earlier, I've seen a much better proof of the proposition, I was just playing devil's advocate because josphII was bullsh*tting...

how was i bullshting????

i stated the proof, period. what you want me to do, hold your hand and walk you through it?
 

josphII

Banned
Nov 24, 2001
1,490
0
0
Originally posted by: TuxDave
Originally posted by: FrustratedUser
Originally posted by: TuxDave
Originally posted by: McPhreak
Originally posted by: TuxDave
That proof looks good to me. I believe 0.9999... = 1. For those who don't believe still, show me 1-0.999... doesn't equal 0.

And if you say it equals 0.000000.....001... there's an infinite number of zeros, so that '1' never really exists, so technically 1-0.999... = 0

Does that mean 0.99999.....98 = 0.999999999.....99?

Ahh.. but there's a freaky difference. I say numbers like 0.9999....98 doesn't exist because we define there to be an infinite number of nines... and here with 0.9999...98 we say, there's a last nine, but there isn't.. there's infinite!

Well, that means 0.999999.... is infinitely smaller than 1.

Hence my latter question, is there such thing as a number that's infinitely smaller than 1. Because that implies that there exists a largest number smaller than 1. And I say no... such a number does not exist...

<edit>
just as there is no smallest number greater than 1. So since nothing can go between 0.999... and 1, AAND we say that there is no such largest number smaller than 1, we must conclude that they are equal.

sure you can say that 0.9999..... is infinetely smaller than 1, but 1/inf (infinetely small) equals 0
 

TuxDave

Lifer
Oct 8, 2002
10,571
3
71
<edit> oh.. new post.. one sec... sure you can say that 0.9999..... is infinetely smaller than 1, but 1/inf (infinetely small) equals 0

Agreed, 1/inf EQUALS 0... so if the difference of 0.9999... and 1 is zero... then aren't they equal. If not, well.. then I'm out of arguments... heh
 

josphII

Banned
Nov 24, 2001
1,490
0
0
Originally posted by: FrustratedUser
Originally posted by: halik
.9999 infinatelly repeating is EXACTLY equal to 1...sheesh people

I must be stupid but an infinitely long string of 0.999999.... still need to be added to 0.0000......001 to be = to 1.

there is no such thing as 0.000.....0001 !!!

look at the proof that has been posted for a complete understanding
 

josphII

Banned
Nov 24, 2001
1,490
0
0
Originally posted by: kenleung
it sounds like an asymptotic (sp?) number to me... well for a function that is but i can find the similarities

graph f(x)=1/(x-1). x approaches 1 but it never reaches 1

apply the same logic to this question, .999999.... isn't 1

0.999... isnt a function thus you can not apply your logic, 0.9999... is a number, if graphed it would be a straight line on 1
 

josphII

Banned
Nov 24, 2001
1,490
0
0
Originally posted by: SilentRunning
My point about e=mc^2 is that that is not an actual equation, It is an approximation of the actual equation.

And you seem to be missing the point

1/3 = 0.3333........ + 10^(-n) as n approaches infinity


to say that as n approaches infinity 10^(-n) = zero is wrong, it approaches zero.

Nothing you can say will change that fact. So 0.33333...... is an approximation of 1/3

your wrong. for the same reasons that 0.9999.... = 1, 0.3333.... equals 1/3. 0.333333..... is not an approxamation for 1/3, it IS 1/3

the proof is similar to the proof for 0.999... = 1
 

RSMemphis

Golden Member
Oct 6, 2001
1,521
0
0
Heh, heh... It really helps to have taken calculus. I always thought "no", until I did the proof myself.

Proof I liked even more: In the space of the real numbers, you can project all the numbers from negative to positive infinity to the span of negative to positive one (or any other two numbers), because between every rational number, there is an infinite number of real numbers.

Talk about headache - but once understood, the coolest thing evar.
 

spidey07

No Lifer
Aug 4, 2000
65,469
5
76
Yeah,

Funny how you can perform so many proofs using series (supports the fact the the two are the same number), then how the definition of the real number system supports it, and so on.

The only real argument is some folks just don't get it.
 

Haircut

Platinum Member
Apr 23, 2000
2,248
0
0
Originally posted by: spidey07
Yeah,

Funny how you can perform so many proofs using series (supports the fact the the two are the same number), then how the definition of the real number system supports it, and so on.

The only real argument is some folks just don't get it.
True, I'm amazed that 117 people are confident that it is wrong.
 

erikiksaz

Diamond Member
Nov 3, 1999
5,486
0
76
I'm not sure if this applies fully, but as a store clerk, would you accept 99 cents if the item being bought is worth a dollar?
 

Rainsford

Lifer
Apr 25, 2001
17,515
0
0
Originally posted by: Yomicron
Originally posted by: Rainsford
I would say yes because that proof makes sense. However, could someone explain something to me, I still do not understand this. Hopefully I can make this clear enough.

.9 is 9/10, right? So it could be considered as you moving 9/10 of the distance between two points (say, 0 and 1), right?
Now .09 is 9/100 which could also be thought of as 9/10 * 1/10. When you add .09 to .9 this could be thought of as moving 9/10 of the remaining 1/10 of the distance, right?
So for each 9 you add to the end, you can think of it as moving 9/10 of the remaining distance. So the distance between you and the second point (1) gets smaller and smaller with each 9 added on to the end. However, it does not seem like you would ever actually reach the second point because you are only moving a fraction of the remaining distance each time. No matter how small, that distance would still exist, right? Or am I missing something?
lets say you have n number of 9's, as n approaches infinity the value approaches 1, but we are conserned with when n equals infintiy.

if you were to just keep adding 9s you would never reach 1, because you are looking at a finite number of 9s

Ah, so we are thinking about a situation that could never be acheived with just adding 9's to the end of the number. Now I think I get it.
 

MadRat

Lifer
Oct 14, 1999
11,965
278
126
Originally posted by: silverpig
Originally posted by: MadRat
Hence, objective values cannot be used to define subjective ones.

What, pray tell, is an objective value, and what is a subjective value?

Null and Infinity do not exist in the natural universe, they are subjective definitions to represent fantasy.

An objective value can be expressed in a concrete sense.

While .333... is approximately 1/3 it is not equal to the latter. We use approximations based on objective values to solve problems close enough to an answer than it acceptable and within significant tolerances, but never do they truly represent the subjective value within the original equation. Substitution of the objective value for the subjective value is simply the key to approximation.
 
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