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

Discussion in 'Alternative Theories' started by chinglu, Oct 27, 2013.

1. ### PeteIt's not rocket surgeryRegistered Senior Member

Messages:
10,167
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

3. ### Quantum QuackLife's a tease...Valued Senior Member

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23,328
nah stupid mistake using a stupid example.. sorry
9x=7
therefore
1x = 7/9
1x= 0.777777778

5. ### Quantum QuackLife's a tease...Valued Senior Member

Messages:
23,328
from what I see you are obviously quite correct...

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...

wiki: http://en.wikipedia.org/wiki/0.999...

Last edited: Oct 30, 2013

7. ### rpennerFully WiredValued Senior Member

Messages:
4,833
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.

[table="width: 500"]
[tr]
[td]$0 \neq 1$[/td]
[/tr]
[tr]
[td]$0 \in \mathbb{X}$[/td]
[td]$1 \in \mathbb{X}$[/td]
[/tr]
[tr]
[td]$p + q \in \mathbb{X}$[/td]
[td]$p \times q \in \mathbb{X}$[/td]
[/tr]
[tr]
[td]$p + 0 = p$[/td]
[td]$p \times 1 = p$[/td]
[/tr]
[tr]
[td]$p + q = q + p$[/td]
[td]$p \times q = q \times p$[/td]
[/tr]
[tr]
[td]$(p + q) + r = p + ( q + r )$[/td]
[td]$(p \times q) \times r = p \times ( q \times r )$[/td]
[/tr]
[tr]
[td]$\exists s \in \mathbb{X} \quad p + s = 0$[/td]
[td]$p \neq 0 \quad \Rightarrow \quad \exists s \in \mathbb{X} \quad p \times s = 1$[/td]
[/tr]
[tr]
[td]$p \times ( q + r ) = ( p \times q ) + ( p \times r )$[/td]
[/tr]
[tr]
[td]$p \lt q \quad \Leftrightarrow \quad p \neq q \; \textrm{and} \; p \not \gt q$[/td]
[/tr]
[tr]
[td]$p \lt q \; \textrm{and} \; q \lt r \quad \Rightarrow \quad q \lt r$[/td]
[/tr]
[tr]
[td]$p \lt q \quad \Rightarrow \quad r + p \lt r + q$[/td]
[td]$0 \lt p \; \textrm{and} \; 0 \lt q \quad \Rightarrow \quad 0 \lt p \times q$[/td]
[/tr]
[/table]

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.

Last edited: Oct 30, 2013
8. ### Quantum QuackLife's a tease...Valued Senior Member

Messages:
23,328
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....

9. ### PeteIt's not rocket surgeryRegistered Senior Member

Messages:
10,167
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. ### Quantum QuackLife's a tease...Valued Senior Member

Messages:
23,328
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

11. ### PeteIt's not rocket surgeryRegistered Senior Member

Messages:
10,167
Which step do you disagree with?

But it doesn't.

Why oops?

12. ### PeteIt's not rocket surgeryRegistered Senior Member

Messages:
10,167
We don't.
The only things we presume are what we already agreed on.

13. ### Quantum QuackLife's a tease...Valued Senior Member

Messages:
23,328
well 9 x 0.999... = 8.999.... not 9.
how do you then derive 1= 0.999...?

14. ### PeteIt's not rocket surgeryRegistered Senior Member

Messages:
10,167
8.999... = 9

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.

15. ### Quantum QuackLife's a tease...Valued Senior Member

Messages:
23,328

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

According to my calculator, disregarding the rounding effect:

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

16. ### PeteIt's not rocket surgeryRegistered Senior Member

Messages:
10,167
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

17. ### Quantum QuackLife's a tease...Valued Senior Member

Messages:
23,328
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...

18. ### PeteIt's not rocket surgeryRegistered Senior Member

Messages:
10,167
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.

19. ### Quantum QuackLife's a tease...Valued Senior Member

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23,328
see edited post above... sorry..

20. ### PeteIt's not rocket surgeryRegistered Senior Member

Messages:
10,167
Do you agree with those three equations, do you see that they logically imply the fourth?

21. ### Quantum QuackLife's a tease...Valued Senior Member

Messages:
23,328
no...
I see:
10 x 0.999... - 0.999... = 9 x 0.999...+ (1/infinity)
as being correct...

22. ### Quantum QuackLife's a tease...Valued Senior Member

Messages:
23,328
To be specific:

23. ### PeteIt's not rocket surgeryRegistered Senior Member

Messages:
10,167
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...