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

The original version again:
X = 0.99999.....
10X = 9.999999....
10X - X = 9 <<<:replaces 9X with a real number... logical inconsistency (loss of context)
9X = 9
X = 1 <<: irrational number now converted to real

Therefore X = 1 = 0.9999.... <<: 1 is now an irrational number
QQ, this really doesn't make a lot of sense. Algebra doesn't work the way you seem to think.

So let's try it with only arithmetic, no algebra.

Do you agree with these identities? Each one stands alone.

10 x 0.999... = 9.999...

10 x 0.999... - 0.999... = 9 x 0.999...

9.999... - 0.999... = 9

9 x 0.999... / 9 = 0.999...

9 / 9 = 1

QQ, you don't seem to understand how algebra works.

So far so good.
Next step:
x = 7/9

You've divided the let hand side by 9x, and the right hand side by 9. Are you thinking that x=1, instead of x=0.777...?
nah stupid mistake using a stupid example.. sorry
9x=7
therefore
1x = 7/9
1x= 0.777777778

QQ, this really doesn't make a lot of sense. Algebra doesn't work the way you seem to think.
from what I see you are obviously quite correct...

So let's try it with only arithmetic, no algebra.

Do you agree with these identities? Each one stands alone.

10 x 0.999... = 9.999...

10 x 0.999... - 0.999... = 9 x 0.999...

9.999... - 0.999... = 9

9 x 0.999... / 9 = 0.999...

9 / 9 = 1
sure... no problemo
just that it appears to me you are mixing the context if algebra x is used.

of course there are 9 of 0.999....
but this does not make 1= 0.999...
it only mean there are only 1 of 0.999...

To me it is the difference in the use of the equal sign..
"of" vs "equal" that allows the method to "fudge" it's way to allowing 1 to equal 0.999...
hence the quantity vs quality comment I made earlier

try to find 1= 0.999 in the same way with out using the algebraic X
I will be most surprised if you can...

The equality of 0.999... and 1 is closely related to the absence of nonzero infinitesimals in the real number system, the most commonly used system in mathematical analysis. Some alternative number systems, such as the hyperreals, do contain nonzero infinitesimals. In most such number systems, the standard interpretation of the expression 0.999... makes it equal to 1, but in some of these number systems, the symbol "0.999..." admits other interpretations that contain infinitely many 9s while falling infinitesimally short of 1.

The equality 0.999... = 1 has long been accepted by mathematicians and is part of general mathematical education. Nonetheless, some students find it sufficiently counterintuitive that they question or reject it, commonly enough that the difficulty of convincing them of the validity of this identity has been the subject of numerous studies in mathematics education.
wiki: http://en.wikipedia.org/wiki/0.999...

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By abandoning the reals and ability to do arithmetic with them such as $$\left( \sqrt{2} \right)^2 =2$$, you can have a consistent theory of positive rational numbers and quasi-decimal representations and quasi-infinitesimals if you limit yourself to one base.

Rather than the standard definition of $${ 0.a_1 a_2 a_3 a_4 a_5 .... }_b = \lim_{n\to \infty} \sum_{k=1}^n \frac{a_k}{b^k}$$, this model of the rational numbers adopts $${ 0.a_1 a_2 a_3 a_4 a_5 .... }_b = \sum_{k=1}^{\tilde{\omega}} \frac{a_k}{b^k}$$ where $$\tilde{\omega}$$ is defined as an integer large enough to be divisible by all relevant figures.

Thus we have the equality $$\frac{p}{q} = \lim_{n\to \infty} \sum_{k=1}^n \frac{a_k}{b^k} = \sum_{k=1}^{\tilde{\omega}} \frac{a_k}{b^k} + \frac{ \sum_{k=\tilde{\omega}+1}^{\tilde{\omega}+\ell} \frac{a_k}{b^k} }{\sum_{k=\tilde{\omega}+1}^{\tilde{\omega}+\ell} \frac{b-1}{b^k} } \epsilon_{b,\tilde{\omega}}$$ where $$\ell$$ is the repetition length of the rational number.

So we have
$$\begin{eqnarray} 1 & = & {0.111...}_2 + \epsilon_{2,\tilde{\omega}} & = & {0.222...}_3 + \epsilon_{3,\tilde{\omega}} & = & {0.444...}_5 + \epsilon_{5,\tilde{\omega}} & = & {0.666...}_7 + \epsilon_{7,\tilde{\omega}} & = & {0.999...}_{10} + \epsilon_{10,\tilde{\omega}} \frac{1}{2} & = & {0.1}_2 & = & {0.111...}_3 + \frac{1}{2} \epsilon_{3,\tilde{\omega}} & = & {0.222...}_5 + \frac{1}{2} \epsilon_{5,\tilde{\omega}} & = & {0.333...}_7 + \frac{1}{2}\epsilon_{7,\tilde{\omega}} & = & {0.5}_{10} \frac{1}{3} & = & {0.010101...}_2 + \frac{1}{3} \epsilon_{2,\tilde{\omega}} & = & {0.1}_3 & = & {0.131313...}_5 + \frac{1}{3} \epsilon_{5,\tilde{\omega}} & = & {0.222...}_7 + \frac{1}{3} \epsilon_{7,\tilde{\omega}} & = & {0.333...}_{10} + \frac{1}{3} \epsilon_{10,\tilde{\omega}} \frac{1}{5} & = & {0.001100110011...}_2 + \frac{1}{5} \epsilon_{2,\tilde{\omega}} & = & {0.012101210121...}_3 + \frac{1}{5} \epsilon_{3,\tilde{\omega}} & = & {0.1}_5 & = & {0.125412541254...}_7 + \frac{1}{5} \epsilon_{7,\tilde{\omega}} & = & {0.2}_{10} \frac{1}{7} & = & {0.001001...}_2 + \frac{1}{7} \epsilon_{2,\tilde{\omega}} & = & {0.010212...}_3 + \frac{1}{7} \epsilon_{3,\tilde{\omega}} & = & {0.032412...}_5 + \frac{1}{7} \epsilon_{5,\tilde{\omega}} & = & {0.1}_7 & = & {0.142857...}_{10} + \frac{1}{7} \epsilon_{10,\tilde{\omega}} \frac{1}{210} & = & \frac{53}{105} - \frac{1}{2} & = & \frac{47}{70} - \frac{2}{3} & = & \frac{17}{42} - \frac{2}{5} & = & \frac{13}{30} - \frac{3}{7} & = & \frac{19}{21} - \frac{9}{10} \frac{1}{210} & = & {0.000000010011100000010011...}_2 + \frac{53}{105} \epsilon_{2,\tilde{\omega}} & = & {0.000010110201200010110201...}_3 + \frac{47}{70} \epsilon_{3,\tilde{\omega}} & = & {0.000244200244200244...}_5 + \frac{17}{42} \epsilon_{5,\tilde{\omega}} & = & {0.0014301430143014...}_7 + \frac{13}{30} \epsilon_{7,\tilde{\omega}} & = & {0.004761904761904761...}_{10} + \frac{19}{21} \epsilon_{10,\tilde{\omega}} \end{eqnarray}$$

But since $$\tilde{\omega}$$ isn't infinite (it's just very large compared to any number we choose to think of), to be consistent you need particular rules to handle addition and multiplication. In particular since $$\tilde{\omega}$$ isn't infinite there is a "last digit" which eats carries from the quasi-infinitesimal.

Example:

$$\frac{3}{210} = 3 \times \frac{1}{210} = 3 \times ( \frac{19}{21} - \frac{9}{10} ) = \frac{57}{21} - \frac{27}{10} = \frac{15}{21} - \frac{7}{10} = \frac{5}{7} - \frac{7}{10} \frac{3}{210} = 3 \times \left( {0.004761904761904761...}_{10} + \frac{19}{21} \epsilon_{10,\tilde{\omega}} \right) = 3 \times {0.004761904761904761...}_{10} + 2 \epsilon_{10,\tilde{\omega}} + \frac{15}{21} \epsilon_{10,\tilde{\omega}} = {0.014285714285...}_{10} + \frac{15}{21} \epsilon_{10,\tilde{\omega}}$$

But, this does not give you a math system more powerful than the rational numbers. It does not have any of the benefits of the real numbers and irrational numbers don't have a representation. It's just a wasteful way to think about the rational numbers unless you use logic and mathematical rigor to bridge the gap between numbers too large to think of and the infinite.

That is the benefit of the real numbers. They obey the axioms common to other number systems but with the concept of continuity so that numbers defined as limits or boundaries of sets actually exist.

 $$0 \neq 1$$ $$0 \in \mathbb{X}$$ $$1 \in \mathbb{X}$$ $$p + q \in \mathbb{X}$$ $$p \times q \in \mathbb{X}$$ $$p + 0 = p$$ $$p \times 1 = p$$ $$p + q = q + p$$ $$p \times q = q \times p$$ $$(p + q) + r = p + ( q + r )$$ $$(p \times q) \times r = p \times ( q \times r )$$ $$\exists s \in \mathbb{X} \quad p + s = 0$$ $$p \neq 0 \quad \Rightarrow \quad \exists s \in \mathbb{X} \quad p \times s = 1$$ $$p \times ( q + r ) = ( p \times q ) + ( p \times r )$$ $$p \lt q \quad \Leftrightarrow \quad p \neq q \; \textrm{and} \; p \not \gt q$$ $$p \lt q \; \textrm{and} \; q \lt r \quad \Rightarrow \quad q \lt r$$ $$p \lt q \quad \Rightarrow \quad r + p \lt r + q$$ $$0 \lt p \; \textrm{and} \; 0 \lt q \quad \Rightarrow \quad 0 \lt p \times q$$

It doesn't matter if $$\mathbb{X}$$ stands for the rationals or the reals, these rules apply to them all. This is the burden of claiming one has a system of infinitesimal numbers -- one has to play by the same rules as any other number. The quasi-infinitesimals above play by these rules and turn out to be just rational numbers by a different name.

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hee hee, doing some research... I had no idea that this issue was so uhm controversial....[chuckle]
from wiki:

Q: How many mathematicians does it take to screw in a lightbulb?

A: 0.999999....

Pete said:
Do you agree with these identities? Each one stands alone.
10 x 0.999... = 9.999...
10 x 0.999... - 0.999... = 9 x 0.999...
9.999... - 0.999... = 9
9 x 0.999... / 9 = 0.999...
9 / 9 = 1
sure... no problemo
...
try to find 1= 0.999 in the same way with out using the algebraic X
I will be most surprised if you can...
10 x 0.999... = 9.999...
10 x 0.999... - 0.999... = 9.999... - 0.999...
9 x 0.999... = 9
0.999... = 1

Each step uses only the agreed identities.

10 x 0.999... = 9.999...
10 x 0.999... - 0.999... = 9.999... - 0.999...
9 x 0.999... = 9
0.999... = 1

Each step uses only the agreed identities.

close IMO but it could read:
10 x 0.999... = 9.999...
9.999... - 0.999... = 9
9 x 0.999 = 8.999...

oops!
we can't presume 0.999...= 1 until proved

we can't presume 0.999...= 1 until proved

We don't.
The only things we presume are what we already agreed on.

We don't.
The only things we presume are what we already agreed on.
well 9 x 0.999... = 8.999.... not 9.
how do you then derive 1= 0.999...?

10 x 0.999... = 9.999...
10 x 0.999... - 0.999... = 9.999... - 0.999...
9 x 0.999... = 9
0.999... = 1

Each step uses only the agreed identities.

well 9 x 0.999... = 8.999.... not 9.
8.999... = 9

how do you then derive 1= 0.999...?
Like I posted:

10 x 0.999... = 9.999...
10 x 0.999... - 0.999... = 9.999... - 0.999...
9 x 0.999... = 9
0.999... = 1

Every step uses only identities we already agreed on.

8.999... = 9

how so?
As I said normally you don't use a proof until it is proven and can't presume the rounding of 0.999.... to a whole number

Like I posted:

10 x 0.999... = 9.999...
10 x 0.999... - 0.999... = 9.999... - 0.999...
9 x 0.999... = 9
0.999... = 1

Every step uses only identities we already agreed on.

According to my calculator, disregarding the rounding effect:

9x 0.999... = 8.999.... NOT 9

According to my calculator, disregarding the rounding effect:

9x 0.999... = 8.999.... NOT 9
That step is deduced from the previous step and the agreed identities.
Let me spell it out.

We agree on these identities:
• 10 x 0.999... = 9.999...
• 10 x 0.999... - 0.999... = 9 x 0.999...
• 9.999... - 0.999... = 9
• 9 x 0.999... / 9 = 0.999...
• 9 / 9 = 1
Start from identity a:
10 x 0.999... = 9.999...

Subtract 0.999... from each side:
10 x 0.999... - 0.999... = 9.999... - 0.999...

Note that the left hand side of this equation matches one side of identity b. This means we can switch it for the other side of that identity, while keeping the right hand side unchanged:
9 x 0.999... = 9.999... - 0.999...

We do the same with the right hand side using identity c:
9 x 0.999... = 9

ok I think I see what is happening ...

take two identities that are treated entirely separately [no rounding allowed]
1]

9.999... - 0.999... = 9

2]

9 x 0.999... = 8.999....

I believe both equations are correct if no presumptions are made

or

consider regarding the OP
in identity one rounding is not necessary [infinitesimal not needed to = 9]
in identity two rounding is necessary [inclusion of an infinitesimal is needed to =9]

essentially what I see is this:
9 x 0.999....+(1/infinity) = 9.999... - 0.999...

ok I think I see what is happening ...
I'm not sure you do, because you're not addressing the logic involved in the step of the proof in question.

Do you agree with this equation:
10 x 0.999... - 0.999... = 9.999... - 0.999...

Do you agree with this equation:
10 x 0.999... - 0.999... = 9 x 0.999...

Do you agree with this equation:
9.999... - 0.999... = 9

If you agree with those three equations, then you should logically agree with this one:
9 x 0.999... = 9

take two identities that are treated entirely separately [no rounding allowed]
1]

9.999... - 0.999... = 9

2]

9 x 0.999... = 8.999....

which identity is correct?
Both are correct. The second is not relevant to the proof.

I'm not sure you do, because you're not addressing the logic involved in the step of the proof in question.

Do you agree with this equation:
10 x 0.999... - 0.999... = 9.999... - 0.999...

Do you agree with this equation:
10 x 0.999... - 0.999... = 9 x 0.999...

Do you agree with this equation:
9.999... - 0.999... = 9

If you agree with those three equations, then you should logically agree with this one:
9 x 0.999... = 9

Both are correct. The second is not relevant to the proof.
see edited post above... sorry..

see edited post above... sorry..
Do you agree with those three equations, do you see that they logically imply the fourth?

Do you agree with this equation:
10 x 0.999... - 0.999... = 9 x 0.999...
no...
I see:
10 x 0.999... - 0.999... = 9 x 0.999...+ (1/infinity)
as being correct...

To be specific:
I'm not sure you do, because you're not addressing the logic involved in the step of the proof in question.

Do you agree with this equation:
10 x 0.999... - 0.999... = 9.999... - 0.999... <<:agrees

Do you agree with this equation:
10 x 0.999... - 0.999... = 9 x 0.999... <<disagrees should be: 10 x 0.999... - 0.999... = 9 x 0.999...+ (1/infinity)

Do you agree with this equation:
9.999... - 0.999... = 9 <<agrees

If you agree with those three equations, then you should logically agree with this one:
9 x 0.999... = 9 << disagrees see above: 9 x 0.999...+(1/infinity) =9

Both are correct. The second is not relevant to the proof.<<disagrees ..very relevant to the proof

Pete said:
Do you agree with this equation:
10 x 0.999... - 0.999... = 9 x 0.999...
no...
You should, it's very simple maths and you agreed to it before (post 83).
10 x 0.999... - 0.999... = 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... - 0.999...
= 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999... + 0.999...
= 9 x 0.999...

But, we can make this even simpler.
Try this:
0.999... + 0.999... = 2 x 0.999... = 1.999... = 1 + 0.999...