# A shorter proof to 0.999 = 1

rachmaninoff::
Hey thanks for pointing out my error, ill use $\overline{x}$ Instead of $x infty$ from now on ;) Atleast you understand my point and I wasnt called a wacko :lol:

nice proof bao_ho, btw.

Zurtex
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eNathan said:
rachmaninoff::
Hey thanks for pointing out my error, ill use $\overline{x}$ Instead of $x infty$ from now on ;) Atleast you understand my point and I wasnt called a wacko :lol:

nice proof bao_ho, btw.
Yeah, kind of my point, with all the infinities floating round I still can't really make sense what you are saying.

EnumaElish
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bao_ho said:
1 / 9 = 0.1111recuring
0.9999recuring / 9 = 0.1111recuring

then 1 = 0.9999recuring
Your 1st step assumes your conclusion, when you replace 1 with "0.9999recuring".
eNathan said:
Sorry I forgot the latex code for <> or !=, what is it again \ne

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Zurtex
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EnumaElish said:
Your 1st step assumes your conclusion, when you replace 1 with "0.9999recuring".
It's not a rigorous proof on many levels but it doesn't do that. It is showing that you get the same answer when you divide both of them by 9, hence they must be the same number.

EnumaElish
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Okay, I got you.

It is wrong to say that 1 / 3 = 0.3...
Despite math ascertion to the otherwise it should be:
1 / 3 = 0.3... r 0.0...1
You try doing it on paper and see if you don't have to keep carrying the one.

arildno
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Dearly Missed
It is wrong to say that 1 / 3 = 0.3...
Despite math ascertion to the otherwise it should be:
1 / 3 = 0.3... r 0.0...1
You try doing it on paper and see if you don't have to keep carrying the one.
Complete nonsense. Learn some maths before posting.

JasonRox
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Gold Member
I concur with arildno.

When is the Pi=3 thread going to come back up?

HallsofIvy
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The greatest number of all real numbers. Obviously, you won't be able to find what number actually infinity is.
----
@eNathan:
How come 1 - 0.99... = a row of 0s? Where is the little number 1.
I agree 0.99... = 1, but your proof does not make much sense to me.
Viet Dao,
If there WERE a "greatest number of all real numbers" many important theorems in mathmatics would be untrue. Perhaps the most important of them would be the "Archimedean Property of Positive Integers": If x is any real number, there exist a positive integer, n, with n> x".

Fortunately for us all, there is NO "greatest of all real numbers" and "infinity", however you define it, is NOT a real number.

Integral
Staff Emeritus
Gold Member
It is wrong to say that 1 / 3 = 0.3...
Despite math ascertion to the otherwise it should be:
1 / 3 = 0.3... r 0.0...1
You try doing it on paper and see if you don't have to keep carrying the one.
Fortunatly we do not have to do it on paper.

Now, I have to point out that your notation is faulty or at least it does not work the way you want it to. You seem to want to claim that .00...1 where the ellipsis represents a infinite number of zeors has meaning. In a well formed decimal number every digit must have a place value, as soon as you place the 1 you have terminated the string of zeros, thus the difference really represented is 1-.999...9 a finite length string of 9s. Of course this difference is greater then 0. Since .999... represents an endless string of 9's you can never reach the end so the 1 never appears.

Contrary to your apparent claim, infinity is NOT just a large real number.

Hi Integral

The only reason we are disallowed from using 0.0...1 notation - even though it is easy to work out what is meant from the notation - is because this would destroy every proof that 1 = 0.9...
There is no other reason to disallow this notation.

But let's look at 1 / 3 seeing this is one of the simpler proofs used (although all proofs no matter how cool they look ultimately are just re-representations of this simplest proof).

Let's work it out:
1. 3 ) 1 (can't do)
2. 3 ) 1.0 = 0.3 r 0.1 (as 0.3 x 3 = 0.9)
3. 3 ) 1.00 = 0.33 r 0.01 (as 0.33 x 3 = 0.99)
4. 3 ) 1.000 = 0.333 r 0.001 (as 0.333 x 3 = 0.999)
5. 3 ) 1.0000 = 0.3333 r 0.0001 (as 0.3333 x 3 = 0.9999)
and we could attempt to keep going for ever.

And in all respects each of these answers is correct on its own.
ie 1/3 = 0.3 r 0.1 = 0.33 r 0.01 = 0.333 r 0.001 etc.

The one thing that we can clearly see from the above is that we always have a remainder. The disappointing thing is that the maintainers of the 1 = 0.9... "proof" want us to simply let that remainder drop off the edge of the universe and to just forget about it.

I know I will never be able to convince you but I do have a question that perhaps you can help me with:

Why is it so important to mathematics that 1 = 0.9...?
What is at stake?

Thanks

Integral
Staff Emeritus
Gold Member
Hi Integral

The only reason we are disallowed from using 0.0...1 notation - even though it is easy to work out what is meant from the notation - is because this would destroy every proof that 1 = 0.9...
There is no other reason to disallow this notation.

But let's look at 1 / 3 seeing this is one of the simpler proofs used (although all proofs no matter how cool they look ultimately are just re-representations of this simplest proof).

Let's work it out:
1. 3 ) 1 (can't do)
2. 3 ) 1.0 = 0.3 r 0.1 (as 0.3 x 3 = 0.9)
3. 3 ) 1.00 = 0.33 r 0.01 (as 0.33 x 3 = 0.99)
4. 3 ) 1.000 = 0.333 r 0.001 (as 0.333 x 3 = 0.999)
5. 3 ) 1.0000 = 0.3333 r 0.0001 (as 0.3333 x 3 = 0.9999)
and we could attempt to keep going for ever.

And in all respects each of these answers is correct on its own.
ie 1/3 = 0.3 r 0.1 = 0.33 r 0.01 = 0.333 r 0.001 etc.

The one thing that we can clearly see from the above is that we always have a remainder. The disappointing thing is that the maintainers of the 1 = 0.9... "proof" want us to simply let that remainder drop off the edge of the universe and to just forget about it.

I know I will never be able to convince you but I do have a question that perhaps you can help me with:

Why is it so important to mathematics that 1 = 0.9...?
What is at stake?

Thanks
Once again infinity is NOT a very large real number. That is the way you treat it and that is why you are being fooled into thinking you are doing math.

Hi Integral

The only reason we are disallowed from using 0.0...1 notation - even though it is easy to work out what is meant from the notation - is because this would destroy every proof that 1 = 0.9...
There is no other reason to disallow this notation.
Please state a well defined definition of what "0.0...1" means and show that it is a real number. If you do this im fairly certain everyone will let you use it in this discussion.

Why is it so important to mathematics that 1 = 0.9...?
What is at stake?
I believe completeness is at stake.

Hi Integral

Is the following math correct or wrong?

1 / 3
= 0.3 r 0.1
= 0.33 r 0.01
= 0.333 r 0.001

Hi Integral

Is the following math correct or wrong?

1 / 3
= 0.3 r 0.1
= 0.33 r 0.01
= 0.333 r 0.001
It is correct, but behavior after a finite number of steps does not dictate the behavior at infinity, certainly not in the way you are treating it.

arildno
Homework Helper
Gold Member
Dearly Missed
$$\frac{1}{3}=0.3+\frac{1}{30}=0.33+\frac{1}{300}=0.333+\frac{1}{3000}=0.3333+\frac{1}{30000}$$
and so on.
Seems you didn't know what division and remainders was after all.

Hurkyl
Staff Emeritus
Gold Member
Why is it so important to mathematics that 1 = 0.9...?
It's not. It's a rather uninteresting consequence of

(1) the axioms of the real numbers
(2) the definition of decimal representations of real numbers

I touched this topic before, here it goes again for you: let's solve X/10 + X/100 + X/100 + ... to some N/M. we will plug X=9 in then and see If N = M. so. above series sum is X * (1/10 + 1/100 + 1/1000 + ...)

expression in () is infinite geometric progression with a = r = 0.1, hence the sum is 0.1/(1-0.1)

which yields N = K * 0.1 * X, M = K * (1 - 0.1)

for X = 9, N = K * 0.9, M = K * 0.9

so N = M

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lot of unrelated chatter which doesn't really answer the real question...ie 1/9 = 0.11111... so is 1 = 0.9999999.....?
Now as there are infinite numbers between 0 & 1, and suppose an insect is moving on a table...it starts at point zero and starts moving at a constant speed...so now it covers 0, 0.00000000...1, this last '1' no one knows when it comes...!! now you come back after one hour and see where the insect is..definitely it would have moved,say 10cm. Now how did the insect cross infinite numbers and reach '1' cm and so on till '10' cm????

HallsofIvy
Homework Helper
Hi Integral

The only reason we are disallowed from using 0.0...1 notation - even though it is easy to work out what is meant from the notation - is because this would destroy every proof that 1 = 0.9...
There is no other reason to disallow this notation.

But let's look at 1 / 3 seeing this is one of the simpler proofs used (although all proofs no matter how cool they look ultimately are just re-representations of this simplest proof).

Let's work it out:
1. 3 ) 1 (can't do)
2. 3 ) 1.0 = 0.3 r 0.1 (as 0.3 x 3 = 0.9)
3. 3 ) 1.00 = 0.33 r 0.01 (as 0.33 x 3 = 0.99)
4. 3 ) 1.000 = 0.333 r 0.001 (as 0.333 x 3 = 0.999)
5. 3 ) 1.0000 = 0.3333 r 0.0001 (as 0.3333 x 3 = 0.9999)
and we could attempt to keep going for ever.
And each of those can be written
2. 0.3+ .1/3
3. 0.33+ .01/3
4. 0.333+ .001/3
5. 0.3333+ .0001/3
Now what is the limit of that fraction as the number of 0s goes to infinity?

And in all respects each of these answers is correct on its own.
ie 1/3 = 0.3 r 0.1 = 0.33 r 0.01 = 0.333 r 0.001 etc.
Do you understand what each of decimal notation MEANS? The basic decimal notation: 0.a1a2a3... MEANS the limit of the sequence 0.a1, 0.a1a2, a1a2a3, ... (the limit of the sequence, not the sequence itself. If you disagree with that, you disagree with the basic definition of a 'base 10 numeration system'!). We don't need to "attempt to keep going for ever", anyone who has taken enough "precalculus" to have seen geometric series should know exactly what the result is.
A geometric series: the sum of a finite number of terms of the form a+ ar+ ar2+ ...+ arn can be shown to be equal to
$$\frac{a- r^{n+1}}{1- r}$$
and, in the case that -1< r< 1, the sequence, as n goes to infinity, has limit $$\frac{a}{1-r}$$

While very few decimal sequences correspond to geometric series, the ones here do. 0.333... is DEFINED as the limit of the sequence .3, .33= .3+ .03, .333= .3+.03+.003, .3333= .3+ .03+ .003+ .0003, ... , a geometric series with a= 0.3 and r= .1< 1. The limit of that sequence is
$$\frac{0.3}{1- .1}= \frac{0.3}{.9}= \frac{1}{3}$$
Similarly, 0.999.. is DEFINED as the limit of the sequence .2, .99= .9+ .09, .999= .9+ .09+ .009, .9999= .9+ .09+ .009+ .0009, ..., a geometric series with a= 0.9 and r= .1< 1. The limit of that sequence is
$$\frac{0.9}{1- .1}= \frac{0.9}{0.9}= 1$$

The one thing that we can clearly see from the above is that we always have a remainder. The disappointing thing is that the maintainers of the 1 = 0.9... "proof" want us to simply let that remainder drop off the edge of the universe and to just forget about it.[\quote]
The remainder "drops of the edge" only if you do not understand that the number represented by a decimal sequence is the limit of that sequence and not the sequence itself!

I know I will never be able to convince you but I do have a question that perhaps you can help me with:

Why is it so important to mathematics that 1 = 0.9...?
What is at stake?

Thanks
It is as important as 1+ 1= 2 or any other equation! There is nothing "at stake" except for the people who make such mistakes. Unfortunately, mathematicians have this compulsion to keep trying to correct mistakes about mathematics!

Hurkyl
Staff Emeritus