# 1 is 0.9999999999999............

I remember doing a "proof" for this at school - but my understanding may be woeful in the grand scheme of things.
Anyhoo...

X = 0.99999.....
10X = 9.999999....
10X - X = 9
9X = 9
X = 1

Therefore X = 1 = 0.9999....

QED

The key, I guess, is in accepting that 0.999.... multiplied by 10 is 9.999....
And at the level of maths I was told this "proof", we had no reason to dispute the idea.

You are under the assumption that 10 = 9.99999........ and prove 1 = .99999...........

Show by induction, where this hold true.

Are you saying that 1 divided by 3 is .333...? If so, I am saying that ".333..." is a notation that represents the result of 1 divided by 3. Let me clarify that ".333...", i.e. a short sequence of threes followed by "..." is a short hand notation that is defined to equal 1/3, while 1 divided by 3 actually is a decimal point followed by an infinite sequence of threes.

The Euclidean algorithm can only produce finite sequences. Look at my recursive argument above.

In other words, at what part of the loop will the division 1/3 terminate? The answer is never.

So, 1/3 does not produce an infinite sequence of digits in spite of ancient folklore.

Really, at which n is it the same number when for all n it is a different number.

Show me a valid argument by induction where your series is only 1 specific number.

It's not done with induction , chinglu, it is done with limits:

$$\frac{1}{10}+\frac{1}{10^2}+.....+\frac{1}{10^n}= \frac{1-1/10^{n+1}}{1-1/10}-1$$

Can you calculate the limit of the above for $$n->\infty$$ all by yourself?

The Euclidean algorithm can only produce finite sequences.
I don't understand. Would it change things if we said that the result of dividing 1 by 3 gives us a POTENTIALLY infinite sequence?
Look at my recursive argument above.
I didn't get it, where is your recursive arguement? Which post?
In other words, at what part of the loop will the division 1/3 terminate? The answer is never.
Good, I understand that, and so by saying that the sequence is potentially infinite conveys that fact, doesn't it?
So, 1/3 does not produce an infinite sequence of digits in spite of ancient folklore.
Hmm, ancient folklore? Doesn't the concept of .999... being a convention or notation representing the value of 1 resolve the issue? Why not?

X = 0.99999.....
10X = 9.999999....
10X - X = 9
9X = 9
X = 1

If you were to subtract x from one side, you would have to do the same with the other.

10x = 9.99999...
10x - x = 9.99999... - x
Right, that's what he did. There are a couple of trivial steps that weren't explicitly spelled out, but it's easy to fill them in:

X = 0.99999.....
10X = 10 x 0.99999.....
10X = 9.999999....
10X - X = 9.99999... - X
10X - X = 9.99999... - 0.99999...
10X - X = 9
9X = 9
9X / 9 = 9 / 9
X = 1

You are under the assumption that 10 = 9.99999...
No, that's not part of the chain of reasoning.

I remember doing a "proof" for this at school - but my understanding may be woeful in the grand scheme of things.
Anyhoo...

X = 0.99999.....
10X = 9.999999....
10X - X = 9
9X = 9
X = 1

Therefore X = 1 = 0.9999....
nope: 1 = a value of 0.9999.... not a quantity of 0.99999
we have 1 OF 0.999999 this is not the same as 1 equals 0.9999

QED

The key, I guess, is in accepting that 0.999.... multiplied by 10 is 9.999....
And at the level of maths I was told this "proof", we had no reason to dispute the idea.
I believe that the teachers were playing with context... an old trick..

X = 0.99999..... : value of one "x" only. Context:"We have one object with the value of 0.9999..."
10X = 9.999999....: quantity of x and subsequent value of x. Context: "we still have only 10 OF object 0.999....."
10X - X = 9 :quantity of x - the value of x Contextual : mixing
9X = 9 : quantity of value only
X = 1 :quantity of value only

Therefore X = 1 = 0.9999....

Nope: 1 = a value of 0.9999.... not a quantity of 0.99999
we have 1 OF 0.999999 this is not the same as 1 equals 0.9999

the logical fallacy of mixing context seems to be happening here...
On the surface it makes sense but if you look at it deeper it is logically unsound IMO
The equal sign is being used in two ways.

Example:
say
1x= 1a
10x = 10a
10x - 1a = 9x
9x = 9a
1x= 1a
Proves nothing other than multiplying "1a" gives you multiples of "1a"..
on the other hand... if we were to write:
0.999...= 0.999...
and so on
do you see the logical context problem?

The notations have been defined. .333... is defined as equal to 1/3, and .999... is defined as equal to 1.
No.
0.333... is intuitively defined as a decimal point followed by an infinite sequence of threes.
More formally, it could be defined as:
$$\lim_{n \to \infty} \sum_{i=1}^{n} 3\times 10^{-i}$$

Similarly for 0.999...

Right, that's what he did. There are a couple of trivial steps that weren't explicitly spelled out, but it's easy to fill them in:

X = 0.99999.....
10X = 10 x 0.99999.....
10X = 9.999999....
10X - X = 9.99999... - X
10X - X = 9.99999... - 0.99999...
10X - X = 9
9X = 9
9X / 9 = 9 / 9
X = 1
'tis only mixing context... that allows such a result...IMO
x = 0.999.... is the same as saying
a ball (x) is labelled 0.999...

1x = 0.999...

10x = 10x [10 balls of 0.999...]
"I have a million balls each with a value of 0.999....
I subtract 10 balls ..how many balls do I have remaining ? Answer: I have 999,990 balls
rather than whats the value I end up with.
type contextual mix up

No.
0.333... is intuitively defined as a decimal point followed by an infinite sequence of threes.
More formally, it could be defined as:
$$\lim_{n \to \infty} \sum_{i=1}^{n} 3\times 10^{-i}$$

Similarly for 0.999...
Ok, then I will stop calling them notations and short cuts. Thanks.

Right, that's what he did. There are a couple of trivial steps that weren't explicitly spelled out, but it's easy to fill them in:

X = 0.99999.....
10X = 10 x 0.99999.....
10X = 9.999999....
10X - X = 9.99999... - X
10X - X = 9.99999... - 0.99999...

10X - X = 9
9X = 9
9X / 9 = 9 / 9
X = 1

That would imply that x=0.9999..., so you might as well convert ALL the xs into 0.9999... and say:

10x - x = 9.9999... - x
10(0.9999...) - 0.9999... = 9.9999... - 0.99999...
9.9999... - 0.9999... = 9.9999... - 0.9999...
9 = 9

...
Is everyone having trouble with simple math here?

That would imply that X=0.9999...
That's what we started from. The proof starts by setting X = 0.999...

so you might as well convert ALL the xs into 0.9999...
That works, too. You get this:
10 x 0.999... = 9.999...
10 x 0.999... - 0.999... = 9.999... - 0.999...
9 x 0.999... = 9
0.999... = 1

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If you cut a cake into thirds, the mathematical representation for each third is 0.3 recurring.
Add those recurring amounts together, and you get 0.9 recurring.
But add the real pieces of cake together and you get 1 whole cake.
So, 0.9 recurring is equal to 1.

If you cut a cake into thirds, the mathematical representation for each third is 0.3 recurring.
Add those recurring amounts together, and you get 0.9 recurring.
But add the real pieces of cake together and you get 1 whole cake.
So, 0.9 recurring is equal to 1.
so you are suggesting that 1/infinity = 0 ?

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No, but this.
Taking the example of the cake, the recurring number 0.3333.... is a representation of one third in numbers.
It goes on forever. The cake does not.
If you think that you can't add the pieces of the cake together to make a whole, you are repeating Zeno's paradox.

If however, the recurring number does not represent one third, but only itself, then it need not be equal to one third.

If you take an arrow and continually move it half the distance between itself and a target, you will never reach the target.
It is only when all those distances added together represent a unity, that the arrow reaches the target.

No, but this.
Taking the example of the cake, the recurring number 0.3333.... is a representation of one third in numbers.
It goes on forever. The cake does not.
If you think that you can't add the pieces of the cake together to make a whole, you are repeating Zeno's paradox.

If however, the recurring number does not represent one third, but only itself, then it need not be equal to one third.

If you take an arrow and continually move it half the distance between itself and a target, you will never reach the target.
It is only when all those distances added together represent a unity, that the arrow reaches the target.
but if the arrow reaches the target then there is no paradox yes?

3.333... +3.333... + 3.3333 still doesn't add up to 1 cake, only 0.99999999...of 1 cake.
Unless you want to resolve the paradox and fudge the cake so to speak.
Using a base ten it is impossible to cut the cake exactly, initially, into three equal pieces with out calling on 1/infinity.

1/infinity can never equal zero, not rationally any way....

If the hare catches the tortoise the game is over.... the hare must never catch the tortoise...
and this game we are playing with 1 = 0.99999... is simply trying to say that hare can catch the tortoise with out blowing the game. IMO
in the example:
to say that
x=0.99999...
we are really saying we have one set of 0.99999... [therfore one object] and not that x= 0.9999...
which is saying that 2x= 2 objects
The context of the algebraic symbols is changed from "quantity of" to "value of" and that as far as I can tell creates the "fudge on the cake"..so to speak.

Example to help explain my thoughts:
say
1x= 1a
10x = 10a
10x - 1a = 9x
9x = 9a
1x= 1a
Proves nothing other than multiplying "1a" gives you multiples of "1a"..
on the other hand... if we were to write:
0.999...= 0.999...
and so on
do you see the logical context problem?

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1 of x = 1 of (0.999...)
10 of x = 10 of (0.999...)
10 of x - 1 of (0.999...) = 9 of 1(0.999...)
9 of 1(0.999...) = 9 of 1(0.999...)
1 of 1(0.999...) = 1 of 1(0.999...)

context is maintained as consistent
I accept that I might be totally off world with this, but I don't think so...

"3.333... +3.333... + 3.3333 still doesn't add up to 1 cake, only 0.99999999...of 1 cake."

You mean 0.3333.......
Yes, but when the recurring numbers each represent one third, they do make a whole cake.
Say we are talking about a real cake, which you have really divided into three parts.
It is not less than a whole cake when you put it back together just because fractions have a difficult time representing the parts.
It does not somehow become 0.9999..... of a cake.

0.3333...... is the fraction we use to represent one third. It is always slightly inaccurate for this purpose,
no matter how many decimal places you take it to.
0.3 would be very inaccurate, 0.33 less inaccurate, and so on.
If you have an infinite number of 3's then the inaccuracy is infinitely small, but it is still inaccurate.

It is also a number in its own right, not representing one third, and as such it is perfectly accurate, and does not quite equal one third.

"3.333... +3.333... + 3.3333 still doesn't add up to 1 cake, only 0.99999999...of 1 cake."

You mean 0.3333.......
Yes, but when the recurring numbers each represent one third, they do make a whole cake.
Say we are talking about a real cake, which you have really divided into three parts.
0.3333...... is the fraction representing one third. It is always slightly inaccurate, no matter how many places you take it to.

It is also a number in its own right, and as such it is perfectly accurate, and does not quite equal one third.
yeah big oopsy... sorry...

1/3 only equals 0.333.... in base 10, because of the existence of an infinitesimal.
To exclude the infinitesimal later would lead to a false result.

"Trying to rationalize the irrational may be an issue in this case"
so 0.333.. + 0.333... + 0.333... = 1 - (1/infinity)

any way I really am over my head I think as I am sure there are proofs I am not able to know nor aware of.

I think you've got it.
It shows the limitation of that method of representing the quantity,
which doesn't occur when you use fractions.
Once you get past a few hundred zeroes, it would be a strict man who wouldn't let you round it up anyway.