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Why does 0.999... = 1?

Algebraically, it makes sense somewhat, but in theoretical terms, where the number line is infinite, then an infinite amount of numbers assumes an infinite amount of space, with each number having it's own allocated slot. Therefore, no matter how large 0.999... becomes, it cannot suddenly assume a different unconnected number. Of course, in reality, nothing is infinite and only occupies a finite space, so it's not that bad.

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Someone didn't do GCSE maths
They're the same number; it's not a case of 'assuming a different unconnected number.' The only difference between the two is how we write them.
(edited 4 years ago)
0.999...= 1/3 x 3
0.999.../3= 1/3
(0.999.../3) x 3 =1

That's the proof but I am not sure what the underlying theory is.
(edited 6 years ago)
Let n = 0.999...
So, 10n = 9.999...
10n - n = 9n, meaning 9.999... - 0.999... = 9
9n = 9
n = 1
(edited 6 years ago)
Reply 5
0.111... = 1/9
0.2222... = 2/9
0.333... = 3/9
0.444... = 4/9
0.999... = 9/9
and 9/9 = 1

it's a geometric series and we can find it's value using formula Sn=a1(1−rn)1−r,r≠1
Reply 6
This should probably be in maths :smile:
Original post by Notnek
This should probably be in maths :smile:


Done :smile: thank you!
Reply 8
No one is giving me an answer, just algebraic solutions.
What's the difference between 11 and 0.9˙0.\dot{9}?

If there's infinite space for an infinite number of numbers, how much space does each get? Is it well-defined?
(edited 4 years ago)
Original post by RogerOxon
What's the difference between 11 and 0.9˙0.\dot{9}?

If there's infinite space for an infinite number of numbers, how much space does each get? Is it well-defined?


If there are an infinite amount of numbers, then the number line they can be placed on must be infinite, meaning every number has its own space to preoccupy on this number line, which is why two cannot one space.
The point is that what you think is '0.999...' has no meaning. I'm guessing you think it's '0.999... where there are infinitely many 9s' but such a thing has no meaning in itself. That's like considering the number '111... where there are infinitely many 1s'. There is no ambiguity in maths.

The idea is that what we call '0.999...' is actually a limit (which has a precise mathematical meaning and does not use some vague concept of 'infinity' in its definition) of a sequence, namely the sequence 0.9, 0.99, 0.999, ... , all of which are well defined numbers. It happens that this limit is 1, and this is quite intuitive anyway.

Certainly '0.999...=1' has no meaning in itself, without preliminary definitions (as mentioned above).
I must emphasise that the argument that the 'difference between the two numbers is infinitely small' is not in any way mathematics. The same goes for all the other intuitive explanations. They're by no means proofs.
It has got nothing to do with rounding. They are different representations of the same number.

If you would like a more analytical proof, using slightly more rigorous mathematics, then I'll sketch one here. As there is an infinite number of "9"s, the number is equal to the infinite series below.

0.90.9+0.09+0.0099(0.1+0.01+0.001)9k=1110k\displaystyle 0.\overline{9} \equiv 0.9 + 0.09 + 0.009 \ldots \equiv 9(0.1 + 0.01 + 0.001 \ldots) \equiv 9\sum_{k=1}^{\infty} \frac{1}{10^k}

As 110<1\displaystyle \left|\frac{1}{10}\right| < 1, we know that the sum of the series will converge. In very very loose Layman's terms, this means that the difference between terms and a given limit LL, ie. Lxn|L-x_n| becomes arbitrarily small through increasing n. You may think of this as the value added on to the sum being increasingly small to the point at which the difference is almost incomprehensible, meaning that the sum will approach a specific value. You may notice that for r>1|r|>1, the series will diverge as there exists no LL such that Lxn|L - x_n| becomes arbitrarily small through increasing nn, meaning that the sum will diverge. Using quite elementary techniques, we can work out this "specific value", as the given series is a geometric series, S, knowing the sum of a geometric series as S=a11r\displaystyle S = \frac{a_1}{1-r}, where rr is the common ratio, in this case 0.1

9k=1110k=(910)1110=910910=1\displaystyle 9\sum_{k=1}^{\infty} \frac{1}{10^k} = \frac{(\frac{9}{10})}{1-\frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1.

Euler actually used this method to prove that 10=9.910 = 9.\overline{9}.
(edited 6 years ago)
If there are an infinite amount of numbers, then the number line they can be placed on must be infinite, meaning every number has its own space to preoccupy on this number line, which is why two cannot one space.

I'm not a mathematician, but I thought that infinity divided by infinity wasn't well-defined.

You are assuming that 11 and 0.9˙0.\dot{9} are different numbers. We could do with a mathematician to answer that, but, as the difference is 0, I would say that they're the same number.
(edited 5 years ago)
No one is giving me an answer, just algebraic solutions.


You can't ask a mathematical question and expect a non mathematical answer.
(edited 5 years ago)
Original post by Sternumator
You can't ask a mathematical question and expect a non mathematical answer.


I want a mathematical explanation, not solution.
Original post by _gcx
It has got nothing to do with rounding. They are different representations of the same number.

If you would like a more analytical proof, using slightly more rigorous mathematics, then I'll sketch one here. As there is an infinite number of "9"s, the number is equal to the infinite series below.

0.90.9+0.09+0.0099(0.1+0.01+0.001)9n=1110k\displaystyle 0.\overline{9} \equiv 0.9 + 0.09 + 0.009 \ldots \equiv 9(0.1 + 0.01 + 0.001 \ldots) \equiv 9\sum_{n=1}^{\infty} \frac{1}{10^k}

As 110<1\displaystyle \left|\frac{1}{10}\right| < 1, we know that the sum of the series will converge. In very very loose Layman's terms, this means that the difference between terms and a given limit LL, ie. Lxn|L-x_n| becomes arbitrarily small through increasing n. You may think of this as the value added on to the sum being increasingly small to the point at which the difference is almost incomprehensible. You may notice that for r>1|r|>1, the series will diverge as there exists no LL such that Lxn|L - x_n| becomes arbitrarily small through increasing nn, meaning that the sum will diverge. Using quite elementary techniques, we can work out this "specific value", as the given series is a geometric series, L, knowing the sum of a geometric series as L=a11r\displaystyle L = \frac{a_1}{1-r}, where rr is the common ratio, in this case 0.1

9n=1110k=(910)1110=910910=1\displaystyle 9\sum_{n=1}^{\infty} \frac{1}{10^k} = \frac{(\frac{9}{10})}{1-\frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1.

Euler actually used this method to prove that 10=9.910 = 9.\overline{9}.


This is the summation to infinity, which is impossible. So theoretically yes it does = 1 but in reality, it never can.

Original post by RogerOxon
I'm not a mathematician, but I thought that infinity divided by infinity wasn't well-defined.

You are assuming that 11 and 0.9˙0.\dot{9} are different numbers. We could do with a mathematician to answer that, but, as the difference is 0, I would say that they're the same number.


Anything involving infinity isn't well defined. When you introduce infinity you move from absolute answers to theoretical absolutes.
Reply 17
I want a mathematical explanation, not solution.


I don't think you know what these words mean.
(edited 5 years ago)
Original post by Pirko
I don't think you know what these words mean.


Ok.
Yo wot m8...
(edited 6 years ago)

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