X = .99999999999... (.999999999 repeating ....)

10 X = 9.99999999999...

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

Then subtract the 2 equations

10 X = 9.9999999999999...
- X = .9999999999999...
===================
9 X = 9
X = 1

THUS: .99999999999... = 1

++++++++++++++++++++++++++

Ok, the proof makes sense but the big picture doesn't.
.99999999999999... = 1 ? Seems weird. Makes sense from the proof but otherwise it just doesn't seem right.

Any other explanations for this out there that may be easier to see/from a different angle?

Technically, .99999..... is not an infinite number of 9's! .9999..... is an infinite sequence of 9's. The sequence is the obvious one; the first digit is 9, the second digit is 9, the third digit is 9, et cetera.

So what can we possibly mean when we say something like:

1/3 = 0.33333...

?

Well, what that means is if we take the sequence of partial decimals:

1st: 0.3
2nd: 0.33
3rd: 0.333
4th: 0.3333
et cetera

then this sequence of partial decimals converges to 1/3, which means essentially that the difference between 1/3 and the n-th partial decimal decreases to zero as n grows.

Now, for 0.99999... = 1 it should be clear! The difference between 1 and the decimal containing n 9's decreases to zero as n grows.

Hurkyl

But .99999999999999999999... and 1 are clearly not the same!

There is no number between these 2, but still, they are two different numbers! So how can we call them the same?

The proof works but the big picture doesn't make sense (for me at least).

When you write it as 0.9999999... it's easy to think that it will always be a tiny bit smaller than one, but it is more correct to think of it as the limit of 0.9999... as the number of decimal places increases to infinity. When the number of decimal places increases to infinity, the difference between 0.99999... and 1 decreases to zero.

1, 1/2, 1/3, 1/4, 1/5, 1/6, ...

and

1, 1/2, 1/4, 1/8, 1/16, 1/32, ...

Clearly both converge to zero, yet the sequences are different!

Mathematically, there's no reason to think that a decimal number has a unique representation... but I know that's unsatisfactory to the layperson, so think of it this way:

A decimal isn't the actual number, but a representation of that number. We know from common experience that the same number can have different representations; for instance the famous phrase: "six of one and a half-dozen of the other". Once you accept that .99999... isn't the actual number but just a way to write that number, it doesn't seem so paradoxical for .9999... = 1 to be true.

Hurkyl

The confusion here is how you look at it.

0.999999999999..... is not, rigorously speaking, a number.
Technically no matter how many nines we place after the decimal we will never get the number 1.

Consider the number 10^-n .
Now let n tend to infinity. Clearly 10^-n then tends to zero.

Now if 0.9_n is a decimal point with n nines after it then

0.9_n + 10^-n = 1

Now as n tends to infinity the equation still holds right?
And also 10^-n will tend to zero right.
So 0.9_n will tend to one.

But let n "equal" infinity
Then 10^-n = 1/
but 1/ is not defined.
so

0.9 + 10^- = 1
does not make sence. the second term on the left hand side is not a number, so we cannot place it in an equation.

Well I hope that makes it clear that 0.9999999.... is not a number, or maybe it was just confusing. Sorry. So if its not a number then
0.999999999.... = 1
Does not make sence.

We are better off saying

lim_n->0.9_n = 1

This at least make sence.

When dealing with infinities it's always better to talk in limits rather than equalities.

quote:

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Originally posted by ObsessiveMathsFreak:
Actually 1 + 1/2 + 1/3 + 1/4 + 1/5 +1/6 + ......
doesn't converge. It's a divergent sequence. Strange but true.

---

Correct. I think that Hurkyl was talking about the sequence though, not the series.

With what frequency does this argument occur.

This and the other recurrent one about division by zero.

For every error tolerance e, there's a point in the sequence where all future terms are within e of zero.

For example, with the sequence of harmonic numbers:

1, 1/2, 1/3, 1/4, ...

If we choose the error tolerance 0.01, then every number in the sequence after 1/100 is less than 0.01 from 0. If we choose 0.001, then every number after 1/1000 is less than 0.001 from 0.

If we choose an arbitrary error tolerance a / b, then every term in the sequence after 1/b is closer to 0 than this error tolerance. (any irrational error tolerance can be replaced with a stricter rational tolerance)

That's the nitty gritty proof that this sequence decreases to 0.

hurkyl

.9 + .09 + .009 + .0009 + ... + 9/10ⁿ + ... is an infinite series.

.99999999... , as Hurkyl and others have said, is not a number, just a form of notation meaning _n=19/10ⁿ
which is the sum of the above series.

Given a sequence {a_n} = a₁, a₂, a₃, ... , a_n
you can define a series a_n = a₁ + a₂ + a₃ + ... + a_n

Then, you can define a sequence of partial sums {s_n} = s₁, s₂, s₃, ... , s_n
where
s₁ = a₁
s₂ = a₁ + a₂
s₃ = a₁ + a₂ + a₃
s₄ = a₁ + a₂ + a₃ + a₄
and so on...

and then, the sum of the infinite series a_n is defined as the limit of the sequence of partial sums, lim_n-> {s_n} (if that limit exists as a number).

quote:

---

Originally posted by vanwinkel:
Since we're being "picky":
.9, .09, .009, .0009, ... 9/10ⁿ, ... is an infinite sequence.

.9 + .09 + .009 + .0009 + ... + 9/10ⁿ + ... is an infinite series.

.99999999... , as Hurkyl and others have said, is not a number, just a form of notation meaning _n=19/10ⁿ
which is the sum of the above series.

Given a sequence {a_n} = a₁, a₂, a₃, ... , a_n
you can define a series a_n = a₁ + a₂ + a₃ + ... + a_n

Then, you can define a sequence of partial sums {s_n} = s₁, s₂, s₃, ... , s_n
where
s₁ = a₁
s₂ = a₁ + a₂
s₃ = a₁ + a₂ + a₃
s₄ = a₁ + a₂ + a₃ + a₄
and so on...

and then, the sum of the infinite series a_n is defined as the limit of the sequence of partial sums, lim_n-> {s_n} (if that limit exists as a number).

---

Going a bit further the sum of the series which you have shown above can be shown to be 1/(1-9/10) if you take the sum from 0 to instead. This is equal to 1/.1 which equals 10. So if you subtract off the first term (9/10⁰ or 9) you will get 1, which shows that .999999..... = 1.

Just wanted to add something that could not ne a number for the same reason as 0.999.... ,or does this not make any sense out of convergence? So would an exponential graph converge also, as it seems to become infinitely close to a number?

quote:

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Originally posted by thatoneguy:
I think the misunderstanding is because 0.9999999999... is not a number, it is an infinite sequqence.

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Hmmm

.999999999999... cannot be represented by a fraction, correct?

i.e.

1/9 = .11111111...
2/9 = .22222222...
4/9 = .44444444...
5/9 = .55555555...
7/9 = .77777777...
8/9 = .88888888...

don't think .999999999999... can be a fraction.. can it?
because if it could, then i guess we could call it a number (a/b where a and b a are integers)

Many like to induce from your line of examples that 0.999... = 9/9 which is also correct.

That brings up another relevant example. 1 = 1 / 1 = 9/9. They're all the same value! 1/1 and 9/9 are different fractions, but they have the same value; they represent the same number, analogously to how 0.9999... and 1 represent the same number.

Hurkyl

1/9 = .11111111...
2/9 = .22222222...
3/9 = .33333333...
4/9 = .44444444...
5/9 = .55555555...
6/9 = .66666666...
7/9 = .77777777...
8/9 = .88888888...
so take a look at what comes next
9/9 = .99999999... = 1

That's kinda a cool way of looking at it I think. I don't know if this would count as a mathematically 'rigorous' proof, but it makes good intuitive sense.

So .999999.... = 1 because in fraction form, its 9/9! That is so cool. I still don't get one thing though. When written as a decimal, its not 1 (to my understanding) but really really close, but not one.

quote:

---

Originally posted by brum:
10 X = 9.9999999999999...
- X = .9999999999999...
===================
9 X = 9
X = 1

THUS: .99999999999... = 1

---

the problem here is one of equivalance. the .999 of 0.999 is NOT the same as the .999 of 9.999. the reason is that the .999 of 9.999 is 10 times larger than the .999 of 0.999 (as proven by your equation 10 x 0.999). your math is wrong because you have truncated the .999 and essentially rounded the .999 of 0.999 up to the same .999 as 9.999.

quote:

---

Originally posted by climbhi:
Dude you almost gave another proof for this .99999999.... = 1, but you stopped just short of realizing it. Take a look at what you gave:

1/9 = .11111111...
2/9 = .22222222...
3/9 = .33333333...
4/9 = .44444444...
5/9 = .55555555...
6/9 = .66666666...
7/9 = .77777777...
8/9 = .88888888...
so take a look at what comes next
9/9 = .99999999... = 1

That's kinda a cool way of looking at it I think. I don't know if this would count as a mathematically 'rigorous' proof, but it makes good intuitive sense.

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just because 7/9 = .7777 and 8/9 = .8888 doesn't mean that 9/9 = .99999 - this is a problem of precision. its the same as saying that 1/3 = .3333 2/3 = .6666 and 3/3 = .9999 - which isn't true.

9/9 = 1
3/3 = 1

3 * .3333... = 1
9 * .1111... = 1

you can't forget the "..."

quote:

---

Originally posted by underworld:
the problem here is one of equivalance. the .999 of 0.999 is NOT the same as the .999 of 9.999. the reason is that the .999 of 9.999 is 10 times larger than the .999 of 0.999 (as proven by your equation 10 x 0.999). your math is wrong because you have truncated the .999 and essentially rounded the .999 of 0.999 up to the same .999 as 9.999.

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just to clarify my own post:

imagine instead of 0.9999 you have 0.900999...

now you have x = 0.900999... and 10x =9.00999...

notice that when you multiply by 10 you "left-shift" the 9s.

so 10X = 9.00999...
X = .900999....
--------------
9X = 8.108999....
X = .90999.....

the same is true for .9999 - you are simply discounting the left-shift as insignificant and truncating or rounding it out of the equation.

quote:

---

Originally posted by climbhi:
Uhh, If you and I are talking about the same exponential graph (like e^x) then the 'number' its 'converging' to is , which means actually that it is not convergent.

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