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

Notation Guide
$$ \left[ \psi \right] $$ is 1 if $$\psi$$ is true and 0 otherwise.
$$ \psi \wedge \phi $$ is true only if both $$\psi$$ and $$\phi$$ are true.
$$ x | y $$ is true only if $$\frac{y}{x}$$ is an integer.
$$ \lfloor x \rfloor$$ is the largest integer equal or smaller than x.

Shared Assumptions

$$p \in \mathbb{N}_0, q \in \mathbb{N}, b \in \mathbb{N}, b > 1$$

Division algorithm 1 for positive rationals
Let $$a_0 = \lfloor p / q \rfloor, \quad s_0 = p - q a_0 , \quad z_0 = a_0$$ then it follows that $$a_0 \in \mathbb{N}_0, \; s_0 \in \mathbb{N}_0 , \; s_0 < q, \; z_0 = \sum_{k=0}^{0} a_k b^{-k}, \; b^0 p = s_0 + b^0 q z_0 $$
Let $$a_{n+1} = \lfloor b s_n / q \rfloor, \quad s_{n+1} = b s_n - q a_{n+1} , \quad z_{n+1} = z_n + b^{-n-1} a_{n+1}$$ then it follows that $$a_{n+1} \in \mathbb{N}_0, \; s_{n+1} \in \mathbb{N}_0 , \; s_{n+1} < q, \; a_{n+1} < b, \; z_{n+1} = \sum_{k=0}^{n+1} a_k b^{-k}, \; b^{n+1} p = s_{n+1} + b^{n+1} q z_{n+1} $$

Division algorithm 2 for positive rationals
Let $$s'_0 = ( p \; \textrm{mod} \; q ) + q \left[ q|p \; \wedge \; p \neq 0 \right] , \quad a'_0 = \frac{ p - s'_0 }{q}, z'_0 = a'_0$$ then it follows that $$a'_0 \in N_0, \; s'_0 \in N_0 \; s'_0 \leq q, \; z'_0 = \sum_{k=0}^{0} b^{-k} a'_k , \; b^0 p = s'_0 + b^0 q z'_0 $$
Let $$s'_{n+1} = \left( (b s'_n) \; \textrm{mod} \; q \right) + q \left[ q|(b s'_n) \; \wedge \; s'_n \neq 0 \right] , \quad a'_{n+1} = \frac{ b s'_n - s'_{n+1} }{q}$$ then it follows that $$a'_0 \in \mathbb{N}_0, \; s'_0 \in \mathbb{N}_0 \; s'_{n+1} \leq q, \; a'_{n+1} < b, \; z'_{n+1} = \sum_{k=0}^{n+1} b^{-k} a'_k , \; b^{n+1} p = s'_{n+1} + b^{n+1} q z'_{n+1}$$

Induction
$$\forall n \in \mathbb{N}_0 \quad a_n, a'_n, s_n, s'_n \in \mathbb{N}_0 \; \wedge \; s_n < q \; \wedge \; s'_n \leq q \; \wedge \; z_n = \sum_{k=0}^{n} a_k b^{-k} \; \wedge \; z'_n = \sum_{k=0}^{n} a'_k b^{-k} \; \wedge \; b^n p = s_n + b^n q z_n \; \wedge \; b^n p = s'_n + b^n q z'_n
\forall n \in \mathbb{N} \quad a_n < b \; \wedge \; a'_n < b
p < b q \Rightarrow \forall n \in \mathbb{N}_0 \quad a_n < b \; \wedge \; a'_n < b$$

Thus
$$\forall n \in \mathbb{N}_0 \quad \frac{p}{q} \; - \; z_n \; = \; \frac{s_n}{q b^n} \\ \forall n \in \mathbb{N}_0 \quad \frac{p}{q} \; - \; z'_n \; = \; \frac{s'_n}{q b^n} \\ \forall n \in \mathbb{N}_0 \quad 0 \; \leq \; \frac{p}{q} \; - \; z_n \; < \frac{1}{b^n} \quad \wedge \quad 0 \; \leq \; \frac{p}{q} \; - \; z'_n \; \leq \frac{1}{b^n}$$

Limits
Since we define $$\sum_{k=0}^{\infty}$$ as $$ \lim_{n\to\infty}\sum_{k=0}^{n}$$ we have $$\sum_{k=0}^{\infty} a_k b^{-k} = \lim_{n\to\infty} z_n $$ and likewise for $$z'_n$$.

$$ \lim_{n\to\infty} z_n = \lim_{n\to\infty} z'_n = \frac{p}{q}$$
Thus both division algorithms deliver a valid sequence of digits that sum to the desired rational number.

Repetition
By the pigeonhole principle it follows that
$$\exists \ell \in \left{ 1, ..., q \right} \quad n > q \quad \Rightarrow \quad s_{n+\ell} = s_{n}
\exists \ell \in \left{ 1, ..., q \right} \quad n > q \quad \Rightarrow \quad s'_{n+\ell} = s'_{n}$$
Thus all rational numbers lead to repetition of in s or s' which means that a and a' (the digits of the expansion) also repeat with period $$\ell$$.
$$\exists \ell \in \left{ 1, ..., q \right} \quad n > q \quad \Rightarrow \quad a_{n+\ell} = a_{n}
\exists \ell \in \left{ 1, ..., q \right} \quad n > q \quad \Rightarrow \quad a'_{n+\ell} = a'_{n}$$
Thus all rational numbers have only repeating expansions when represented as digits although the digits that repeat may be all zeros.

Example 1
Let $$p = 1, q = b$$
Then $$a_0 = 0, a_1 = 1, a_2 = 0, a_3 = 0, ..., s_0 = 1, s_1 = 0, s_2 = 0, s_3 = 0, ... , z_0 = 0, z_1 = 1/b, z_2 = 1/b, z_3 = 1/b, ...$$
and $$a'_0 = 0, a'_1 = 0, a'_2 = b-1, a'_3 = b-1, ..., s'_0 = 1, s'_1 = b, s'_2 = b, s'_3 = b, ... , z'_0 = 0, z'_1 = b^{-1} - b^{-1}, z'_2 = b^{-1} - b^{-2}, z'_3 = b^{-1} - b^{-3}, ...$$
And $$ \lim_{n\to\infty} z_n = \lim_{n\to\infty} z'_n = \frac{1}{b}$$

So 0.1 = 0.0999... = 1/10.

Example 2
Let $$p = 1, q = b-1$$
Then $$a_0 = 0, a_1 = 1, a_2 = 1, a_3 = 1, ..., s_0 = 1, s_1 = 1, s_2 = 1, s_3 = 1, ... , z_0 = 0, z_1 = \frac{b^1 -1}{(b-1)b^1}, z_2 = \frac{b^2 -1}{(b-1)b^2}, z_3 = \frac{b^3 -1}{(b-1)b^3}, ...$$
and $$a'_0 = 0, a'_1 = 1, a'_2 = 1, a'_3 = 1, ..., s'_0 = 1, s'_1 = 1, s'_2 = 1, s'_3 = 1, ... , z'_0 = 0, z'_1 = \frac{b^1 -1}{(b-1)b^1}, z'_2 = \frac{b^2 -1}{(b-1)b^2}, z'_3 = \frac{b^3 -1}{(b-1)b^3}, ...$$
And $$ \lim_{n\to\infty} z_n = \lim_{n\to\infty} z'_n = \frac{1}{b-1} \lim_{n\to\infty} \left( 1 - \frac{1}{b^n} \right) = \frac{1}{b-1}$$

So 0.111... = 0.111... = 1/9.

So Quantum Quack, I don't see that you have a reasonable position because there is no n such that $$a_n$$ or $$a'_n$$ is the last digit of the representation of $$\frac{p}{q}$$ . You can consistently create a mathematics where you stop at $$\tilde{\omega}$$ but such a mathematics is predicated on only quasi-infintesimals and that
$$\frac{p}{q} = \sum_{k=0}^{\infty} a_k b^{-k} = \sum_{k=0}^{\tilde{\omega}} a_k b^{-k} + \sum_{k= 1 \; + \; \tilde{\omega}}^{\infty} a_k b^{-k} = \sum_{k=0}^{\tilde{\omega}} a_k b^{-k} + \frac {\sum_{k= 1}^{\ell } a_{k + \; \tilde{\omega} } b^{\ell-k} }{\sum_{k= 1}^{\ell } (b-1) b^{\ell-k}} \epsilon = \frac{p'}{q'} + \frac{p''}{q''} \epsilon $$
thus $$\epsilon = b^{-\tilde{\omega} }$$
which solves precisely nothing while crippling your mathematics to the point you can't say $$\sqrt{2}$$ or $$\pi$$ is a number. Without infinity and continuity, there is no such concept as the real numbers.

//edit: And with infinity and continuity, 0.999... = 1.
Also, solved for $$\epsilon$$ in a way only hinted at in earlier post
Also, added text to the section marked Limits and fixed primes in Example 2.
Also, fixed invariants.
 
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Hi Pete. :)

Your example was 7 x 10. Shifting the decimal point to the right means we then have to write a zero we didn't bother to before, to get 70.

The aspect I am addressing/coming from is the actual irreducible fundamentalities entities/actions/properties involved at the most basic level before all the 'add on' higher level abstractions, conventions and trivial treatments/proofs come into it. That (from my) perspective makes the "moving the decimal point" a trivial consequence of the fundamental action rather than the other way round.

For further clarification of my approach/observation please see my previous post (#200) in reply to Trippy. Thanks.


If you don't mean it ["Bye."], don't say it.

Understood. Indeed, I had actually resolved to log out then and not come back into the thread for some time because I was getting short of time. I was just about to log out after a final check for typos and/or any posts I had missed from others. That is when I saw your post to QQ touching upon the very same point of 'trivial leading/trailing zeros' etc which was the subject of my previous post to you; so naturally I changed my mind about logging out, deciding to make that last post to further clarify my position in that context so that I would not leave that aspect 'hanging'.

I will take your advice to heart for future reference, Pete, and just leave and log out whenever I must, without making any indication as to when/whether I would return to a thread/discussion, if at all. No more "Byes." from me from now on, then, mate! Sorry if that habit bugged you. Not meant to. Cheers, Pete, Trippy, QQ, everyone! :)
 
Notation Guide
$$ \left[ \psi \right] $$ is 1 if $$\psi$$ is true and 0 otherwise.
$$ \psi \wedge \phi $$ is true only if both $$\psi$$ and $$\phi$$ are true.
$$ x | y $$ is true only if $$\frac{y}{x}$$ is an integer.
$$ \lfloor x \rfloor$$ is the largest integer equal or smaller than x.

Shared Assumptions

$$p \in \mathbb{N}_0, q \in \mathbb{N}, b \in \mathbb{N}, b > 1$$

Division algorithm 1 for positive rationals
Let $$a_0 = \lfloor p / q \rfloor, \quad s_0 = p - q a_0 , \quad z_0 = a_0$$ then it follows that $$a_0 \in \mathbb{N}_0, \; s_0 \in \mathbb{N}_0 , \; s_0 < q, \; z_0 = \sum_{k=0}^{0} a_k b^{-k}, \; b^0 p = s_0 + b^0 q z_0 $$
Let $$a_{n+1} = \lfloor b s_n / q \rfloor, \quad s_{n+1} = b s_n - q a_{n+1} , \quad z_{n+1} = z_n + b^{-n-1} a_{n+1}$$ then it follows that $$a_{n+1} \in \mathbb{N}_0, \; s_{n+1} \in \mathbb{N}_0 , \; s_{n+1} < q, \; a_{n+1} < b, \; z_{n+1} = \sum_{k=0}^{n+1} a_k b^{-k}, \; b^{n+1} p = s_{n+1} + b^{n+1} q z_{n+1} $$

Division algorithm 2 for positive rationals
Let $$s'_0 = ( p \; \textrm{mod} \; q ) + q \left[ q|p \; \wedge \; p \neq 0 \right] , \quad a'_0 = \frac{ p - s'_0 }{q}, z'_0 = a'_0$$ then it follows that $$a'_0 \in N_0, \; s'_0 \in N_0 \; s'_0 \leq q, \; z'_0 = \sum_{k=0}^{0} b^{-k} a'_k , \; b^0 p = s'_0 + b^0 q z'_0 $$
Let $$s'_{n+1} = \left( (b s'_n) \; \textrm{mod} \; q \right) + q \left[ q|(b s'_n) \; \wedge \; s'_n \neq 0 \right] , \quad a'_{n+1} = \frac{ b s'_n - s'_{n+1} }{q}$$ then it follows that $$a'_0 \in \mathbb{N}_0, \; s'_0 \in \mathbb{N}_0 \; s'_{n+1} \leq q, \; a'_{n+1} < b, \; z'_{n+1} = \sum_{k=0}^{n+1} b^{-k} a'_k , \; b^{n+1} p = s'_{n+1} + b^{n+1} q z'_{n+1}$$

Induction
$$\forall n \in \mathbb{N}_0 \quad a_n, a'_n, s_n, s'_n \in \mathbb{N}_0 \; \wedge \; s_n < q \; \wedge \; s'_n \leq q \; \wedge \; z_n = \sum_{k=0}^{n} a_k b^{-k} \; \wedge \; z'_n = \sum_{k=0}^{n} a'_k b^{-k} \; \wedge \; b^n p = s_n + b^n q z_n \; \wedge \; b^n p = s'_n + b^n q z'_n
\forall n \in \mathbb{N} \quad a_n < b \; \wedge \; a'_n < b
p < b q \Rightarrow \forall n \in \mathbb{N}_0 \quad a_n < b \; \wedge \; a'_n < b$$

Thus
$$\forall n \in \mathbb{N}_0 \quad \frac{p}{q} \; - \; z_n \; = \; \frac{s_n}{q b^n} \\ \forall n \in \mathbb{N}_0 \quad \frac{p}{q} \; - \; z'_n \; = \; \frac{s'_n}{q b^n} \\ \forall n \in \mathbb{N}_0 \quad 0 \; \leq \; \frac{p}{q} \; - \; z_n \; < \frac{1}{b^n} \quad \wedge \quad 0 \; \leq \; \frac{p}{q} \; - \; z'_n \; \leq \frac{1}{b^n}$$

Limits
Since we define $$\sum_{k=0}^{\infty}$$ as $$ \lim_{n\to\infty}\sum_{k=0}^{n}$$ we have $$\sum_{k=0}^{\infty} a_k b^{-k} = \lim_{n\to\infty} z_n $$ and likewise for $$z'_n$$.

$$ \lim_{n\to\infty} z_n = \lim_{n\to\infty} z'_n = \frac{p}{q}$$
Thus both division algorithms deliver a valid sequence of digits that sum to the desired rational number.

Repetition
By the pigeonhole principle it follows that
$$\exists \ell \in \left{ 1, ..., q \right} \quad n > q \quad \Rightarrow \quad s_{n+\ell} = s_{n}
\exists \ell \in \left{ 1, ..., q \right} \quad n > q \quad \Rightarrow \quad s'_{n+\ell} = s'_{n}$$
Thus all rational numbers lead to repetition of in s or s' which means that a and a' (the digits of the expansion) also repeat with period $$\ell$$.
$$\exists \ell \in \left{ 1, ..., q \right} \quad n > q \quad \Rightarrow \quad a_{n+\ell} = a_{n}
\exists \ell \in \left{ 1, ..., q \right} \quad n > q \quad \Rightarrow \quad a'_{n+\ell} = a'_{n}$$
Thus all rational numbers have only repeating expansions when represented as digits although the digits that repeat may be all zeros.

Example 1
Let $$p = 1, q = b$$
Then $$a_0 = 0, a_1 = 1, a_2 = 0, a_3 = 0, ..., s_0 = 1, s_1 = 0, s_2 = 0, s_3 = 0, ... , z_0 = 0, z_1 = 1/b, z_2 = 1/b, z_3 = 1/b, ...$$
and $$a'_0 = 0, a'_1 = 0, a'_2 = b-1, a'_3 = b-1, ..., s'_0 = 1, s'_1 = b, s'_2 = b, s'_3 = b, ... , z'_0 = 0, z'_1 = b^{-1} - b^{-1}, z'_2 = b^{-1} - b^{-2}, z'_3 = b^{-1} - b^{-3}, ...$$
And $$ \lim_{n\to\infty} z_n = \lim_{n\to\infty} z'_n = \frac{1}{b}$$

So 0.1 = 0.0999... = 1/10.

Example 2
Let $$p = 1, q = b-1$$
Then $$a_0 = 0, a_1 = 1, a_2 = 1, a_3 = 1, ..., s_0 = 1, s_1 = 1, s_2 = 1, s_3 = 1, ... , z_0 = 0, z_1 = \frac{b^1 -1}{(b-1)b^1}, z_2 = \frac{b^2 -1}{(b-1)b^2}, z_3 = \frac{b^3 -1}{(b-1)b^3}, ...$$
and $$a'_0 = 0, a'_1 = 1, a'_2 = 1, a'_3 = 1, ..., s'_0 = 1, s'_1 = 1, s'_2 = 1, s'_3 = 1, ... , z'_0 = 0, z'_1 = \frac{b^1 -1}{(b-1)b^1}, z'_2 = \frac{b^2 -1}{(b-1)b^2}, z'_3 = \frac{b^3 -1}{(b-1)b^3}, ...$$
And $$ \lim_{n\to\infty} z_n = \lim_{n\to\infty} z'_n = \frac{1}{b-1} \lim_{n\to\infty} \left( 1 - \frac{1}{b^n} \right) = \frac{1}{b-1}$$

So 0.111... = 0.111... = 1/9.

So Quantum Quack, I don't see that you have a reasonable position because there is no n such that $$a_n$$ or $$a'_n$$ is the last digit of the representation of $$\frac{p}{q}$$ . You can consistently create a mathematics where you stop at $$\tilde{\omega}$$ but such a mathematics is predicated on only quasi-infintesimals and that
$$\frac{p}{q} = \sum_{k=0}^{\infty} a_k b^{-k} = \sum_{k=0}^{\tilde{\omega}} a_k b^{-k} + \sum_{k= 1 \; + \; \tilde{\omega}}^{\infty} a_k b^{-k} = \sum_{k=0}^{\tilde{\omega}} a_k b^{-k} + \frac {\sum_{k= 1}^{\ell } a_{k + \; \tilde{\omega} } b^{\ell-k} }{\sum_{k= 1}^{\ell } (b-1) b^{\ell-k}} \epsilon = \frac{p'}{q'} + \frac{p''}{q''} \epsilon $$
thus $$\epsilon = b^{-\tilde{\omega} }$$
which solves precisely nothing while crippling your mathematics to the point you can't say $$\sqrt{2}$$ or $$\pi$$ is a number. Without infinity and continuity, there is no such concept as the real numbers.

//edit: And with infinity and continuity, 0.999... = 1.
Also, solved for $$\epsilon$$ in a way only hinted at in earlier post
Also, added text to the section marked Limits and fixed primes in Example 2.
Also, fixed invariants.
Given that I have absolutely no idea of what you have formulated above I shall just accept that you know what you are doing. This is not so much "agreement" but more a case of knowing my own limits. For one can not agree with something if one does not understand what he is agreeing with...so that said, I can accept that you know what you are doing but doubt I can ever agree with it due to my own ignorance of sophisticated [IMO] mathematics.

ok..

Can I ask you Rppener?
note: You may have already covered it in your presentations.

Can I direct you to post #191
ok... I did... however did you read why I reject the result... of


9.999...- 0.999... = 9 = 9 x 0.999...

with out the addition of an infinitesimal

Gosh this is hard to explain... sorry...
The nine in the middle is not the same nine depending on which side of the equation you are focussed on....

try this as proof of my point.

Theoretically the proof should also work as follows
Compare:

9.999... - 0.999... = 9 [exact with out the need for any "rounding"]

9 - 0.999... = 8 ["rounding" needed]


Contention:
The value of the distance between
9.999... and 9 is less than the distance between 9 and 8 [therefore "rounding" is being applied inconsistently]

If they read as follows
10 - 0.999... = 9
and
9 - 0.999... = 8
I would have no real problem and accept that rounding is being applied consistently.
However this would not be able to be used as proof that 0.999.. = 1

Do you understand what I am suggesting with the above..?

Where I make the value distance comparison between

9.999... and 9
9 and 8
The proof uses the multiplication of 0.999.... x 10 and then the subtraction of 0.999.... as a significant part of proving that 0.999... = 1

I would argue that the mere fact that it requires the proof to use the 0.999... or the 9.999... to prove the value of 1 self defeats the proof.
because by using 9.999... and subtracting a "clean" 0.999.... all this is stating is that 0.999... = 0.999... and not one.
To then use this as justification for 1=0.999... is apparently not logically valid.
To help here is the image file from wiki:

680fee112b7c09afa53b3f35eea46f9c.png


I would further contend that if the proof was valid and not a "trick" of binary manipulation.
the proof could be extended to include further subtracted evidence.
10 x 0.999... = 9.999...

9.999... - 0.999... = 9
9 - 0.999... =/= 8 if one wishes to apply a consistent value to 0.999...

Or simply ask:
if 9.999... - 0.999... = 9
then how does
9 + 0.999... = 10 ?

edit: I do understand that if I was really smart and really clever I could create a mathematical number system that would make 0.8888 = 1. So I appreciate we are working with in the limits and rules of the given number system in use. The absence of non-zero infinitesimals in this particular number system [Real] IMO forces the inability to use infinitesimals to support any contradicting proof.
Yet we are still left with the argument above.
 
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QQ: your attached image is full of mistakes.

9 - 0.999... = 9 - 1 = 8, contrary to what's in your post. You then repeat this basic error, why?
 
QQ: your attached image is full of mistakes.

9 - 0.999... = 9 - 1 = 8, contrary to what's in your post. You then repeat this basic error, why?
to show reason why
0.999... can be subtracted to achieve 9
yet can not be added to achieve 10 with out proving inconsistency. refer my post to rpenner #203 for details

"because 9.999...-0.999... = 9
9 + 0.999... can not equal 10"
 
It is not a "trick" but rather a completely valid way of manipulating an infinite series.


First -- do you know how to divide?
Look instead at $$\frac{41}{333} = 0.123123123...$$. How do we know this. By working through the division algorithm with p = 41, q = 333, b=10
then
$$\quad \quad \quad k \quad \quad \quad $$$$s'_k$$$$a'_k$$$$z'_k$$Check
04100$$\frac{41}{333} = 0 + 10^{-0} \frac{41}{333}$$
17710.1$$\frac{41}{333} = 0.1 + 10^{-1} \frac{77}{333}$$
210420.12$$\frac{41}{333} = 0.12 + 10^{-2} \frac{104}{333}$$
34130.123$$\frac{41}{333} = 0.123 + 10^{-3} \frac{41}{333}$$
47710.1231$$\frac{41}{333} = 0.1231 + 10^{-4} \frac{77}{333}$$
510420.12312$$\frac{41}{333} = 0.12312 + 10^{-5} \frac{104}{333}$$
64130.123123$$\frac{41}{333} = 0.123123 + 10^{-6} \frac{41}{333}$$
............because $$\ell = 3$$, we have for any integer n:
3 n + 1771$$z'_{3n} + 1 \times 10^{-3n-1}$$$$\frac{41}{333} = z'_{3n} + \times 10^{-3n-1} + 10^{-3 n - 1} \frac{77}{333}$$
3n + 21042$$z'_{3n} + 12 \times 10^{-3n-2}$$$$\frac{41}{333} = z'_{3n} + 12 \times 10^{-3n-2} + 10^{-3n -2} \frac{104}{333}$$
3n+3413$$z'_{3n} + 123 \times 10^{-3n-3}$$$$\frac{41}{333} = z'_{3n} + 123 \times 10^{-3n-3} + 10^{-3n -3} \frac{41}{333}$$

Thus $$\frac{41}{333} = 0.123123123...$$ which repeats forever.

You can even check the result: Wolfram Alpha Symbolic Computation


Next -- can you repeat the former success?
Multiply by 10 $$\frac{410}{333} = 1 + \frac{77}{333} = 1 + 0.231231231... = 1.231231231...$$
Multiply by 10 again $$\frac{4100}{333} = 12 + \frac{104}{333} = 12 + 0.312312312... = 12.312312312...$$
Multiply by 10 yet again $$\frac{41000}{333} = 123 + \frac{41}{333} = 123 + 0.123123123... = 123.123123123...$$
None of this depends on shifting the repeating decimal, it is simply ordinary division.

Thirdly -- can you do arithmetic with rational numbers?
Because $$\frac{41}{333} = 0.123123123...$$ it is perfectly legal to add 123 to both sides thus
$$123 + \frac{41}{333} = 123 + 0.123123123...$$
Because $$123 \times 333 = 40959$$ it is perfectly legal to write $$\frac{40959}{123} + \frac{41}{333} = 123 + \frac{41}{333}$$.
Because $$40959 + 41 = 41000 = 1000 \times 41$$ it is perfectly legal to write $$1000 \times \frac{41}{333} = 123 + \frac{41}{333} $$.

Last -- can you cancel identical terms via subtraction?
So it follows that $$999 \times \frac{41}{333} = 1000 \times \frac{41}{333} - \frac{41}{333} = 123 + 0.123123123... - 0.123123123... = 123$$.
But this means $$999 \times 0.123123123... = 123$$ in any self-consistent mathematics where the ... means repeat forever.

For the same precise reason $$9 \times 0.999... = 9$$.
 
@Rpenner,
How does that deal with the argument?

"because 9.999...-0.999... = 9
9 + 0.999... can not equal 10"

In this case we haven't even got to the rest of the proof...

for surely if 9.999...- 0.999... = 9 then logically
9 + 0.999.... must equal 9.999...

just to be absolutely clear..
it is similar to
for surely if 10 - 5 = 5 then logically
5 + 5 = 10

Understanding I am looking for and yet to be found:
The point I think is that the proof is only valid if numbers are increasing in value and not decreasing in value. [ or vice verse, but symmetry is absent.]
 
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A reverse proof would fail. I am almost sure of it....
therfore whilst
0.999... may be proved to equal 1
1 may not be proved to equal 0.999...

If you can accept that 9.999... - 0.999... = 9 and that 9 + 0.999... = 10

then I am just as entitled to accept that

0.999.. = 1 =/= 0.999... :p
 
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QQ said:
for surely if 9.999...- 0.999... = 9 then logically
9 + 0.999.... must equal 9.999...
That's correct, and 9.999... = 10.

This says there is no infinitesimal gap between 9.999... and 10. They are equal in size.
 
That's correct, and 9.999... = 10.

This says there is no infinitesimal gap between 9.999... and 10. They are equal in size.
prove it!
I tell you what.. a great idea!!! Start a thread on it called "1 is 0.99999999999999..." and when it gets to 15 pages let me know...:p:D
 
Or simply ask:
if 9.999... - 0.999... = 9
then how does
9 + 0.999... = 10 ?
If:
9.999... - 0.999... = 9
And:
9 + 0.999... = 10

Then:
1) Rearranging to find 0.99(9)
a) 0.99(9) = 9.99(9) - 9
And:
b) 0.99(9) = 10-9

Substuting a) into b) it follows that:
10 - 9 = 9.99(9) - 9
If we add 9 to both sides of the equation we find that:
10 = 9.99(9)

2) Rearranging to find 9
a) 9 = 9.99(9) - 0.99(9)
And
b) 9 = 10 - 0.99(9)

Substituting a) into b) we find that:
10 - 0.99(9) = 9.99(9) - 0.99(9)
If we ad 0.99(9) to both sides we find (again) that:
10 = 9.99(9) recurring.

Thus we are forced to accept that those two equations are consistent.

Further more, i f we assume that (i) 10=9.99(9), as we have just demonstrated twice using the two equations you believed to be inconsistent then we get to:

3) Rearranging to find 10 and 9.99(9)
a) 9.99(9) = 9 + 0.99(9)
b) 10 = 9 + 0.99(9)
Assuming (i) to be true and substituting a) into b) leads us to:
9 + 0.99(9) = 9 + 0.99(9)

Which hopefully you can recognize as being a tautology thus demonstrating (hopefully) the answer to your question: "If 9.999... - 0.999... = 9 then how does 9 + 0.999... = 10 ?" as well as alluding to the other answe "Because 0.99(9) = 1.


I do understand that if I was really smart and really clever I could create a mathematical number system that would make 0.8888 = 1.
No, because numbers exist between 0.88(8) and 1.
 
prove it!
I tell you what.. a great idea!!! Start a thread on it called "1 is 0.99999999999999..." and when it gets to 15 pages let me know...:p:D

I see what is causing the controversy, I think I can explain it.
In math , the notion of 0.(9) has the precise meaning:
$$0.(9)=\lim_{n \to \infty} (\frac{9}{10^1}+....+\frac{9}{10^n})$$
In the above context, $$0.(9)$$ is obviously equal to 1.
This explain why $$9.(9)=10$$ and why $$9.(9)-0.(9)=9$$
You, on the other hand, are using $$0.(9)$$ to mean $$\frac{9}{10^1}+....+\frac{9}{10^n}$$. In this context, obviously $$0.(9) < 1$$.
For as long as you use the notion in an incorrect way, this thread will continue forever.
 
I really don't have time to read through your exceedingly long posts that largely just seem to state the same things eight different ways and somehow manage to come close to (or even crossing the threshold of word salad.
Hi Trippy. :)

Yes, mate; already understood long since. As indicated in my post (#165, quoted below) to Pete mentioning that very aspect in connection with use for trivial leading/trailing zeros in certain contexts. I have highlighted the bit indicating that aspect was already understood...
And yet, evidently you have not.

The mere FORMATTING conventions which would bring in more zeros as purely formatting symbols is non-fundamental in that case, and is therefore essentially a trivial manipulation like "moving a decimal point" which does not come into it fundamentally in that case, but only as a non-fundamental 'overlay' FORMATTING device in the decimal system. This trivial 'overlay' is not fundamental or needed at all in some other systems. Yes?
As usual, you've got it arse about face.

First off, the zeroes are not trivial - that's not what significant means in this context.
Secondly the 'formatting device' is the agreement not to write the non-significant zeroes because it's cumbersome, not the other way around. If you want to deal in fundamentals then you have to realize that it 1.0(0) not 1. The zeroes exist whether you write them or not. Do I need to explain to you what numbers actually mean?

The decimal point is fundamental even though the convention is not to write it unless the zeroes after it. If we follow your logic, then rather than writing "A circle of 565m has an area of 1.00 km[sup]2[/sup]" I should just write "A circle of 565m has an area of 1km[sup]2[/sup]". There's a problem with this - a circle with a radius of 560m also has an area of 1km[sup]2[/sup].

So "moving the decimal point" instead of just recognizing the fundamentality of the numbers/entities/operations involved is mere convention and choice and not what actually happens: ie, the 10 brings the zero which is added to the target 7 to make it 70 as the result of that multiplication. No decimal point involved fundamentally.

That was my point. That is where I was coming from: fundamental versus convention choices and trivial notation/proofs based on same. I can't say it any more directly than that, Trippy. :)
As I said before, you've got it arse about face.

The fundamental fact is that the zeroes before and after the 1 exist whether we choose to write them or not.
The convention is to not write zeroes unless they're significant.

I can't say it any more directly than that.
 
If:
9.999... - 0.999... = 9
And:
9 + 0.999... = 10

Then:
1) Rearranging to find 0.99(9)
a) 0.99(9) = 9.99(9) - 9
And:
b) 0.99(9) = 10-9

Substuting a) into b) it follows that:
10 - 9 = 9.99(9) - 9
If we add 9 to both sides of the equation we find that:
10 = 9.99(9)

2) Rearranging to find 9
a) 9 = 9.99(9) - 0.99(9)
And
b) 9 = 10 - 0.99(9)

Substituting a) into b) we find that:
10 - 0.99(9) = 9.99(9) - 0.99(9)
If we ad 0.99(9) to both sides we find (again) that:
10 = 9.99(9) recurring.

Thus we are forced to accept that those two equations are consistent.

Further more, i f we assume that (i) 10=9.99(9), as we have just demonstrated twice using the two equations you believed to be inconsistent then we get to:

3) Rearranging to find 10 and 9.99(9)
a) 9.99(9) = 9 + 0.99(9)
b) 10 = 9 + 0.99(9)
Assuming (i) to be true and substituting a) into b) leads us to:
9 + 0.99(9) = 9 + 0.99(9)

Which hopefully you can recognize as being a tautology thus demonstrating (hopefully) the answer to your question: "If 9.999... - 0.999... = 9 then how does 9 + 0.999... = 10 ?" as well as alluding to the other answe "Because 0.99(9) = 1.



No, because numbers exist between 0.88(8) and 1.
Trippy , I appreciate your post...
however it is also a tautology or self evident that the following is true also

9.999.... - 0.999 = 9 & 9 + 0.999... = 10

same as exampled with 10-6 = 4 & 4 + 6 = 10
We have asymmetry, therefore reversibility to prove the equations.

The proof for 0.999... = 1 can not be reversed and is asymmetrical as far as I can tell..
I believe poster Tach may have something to say about this.
 
I see what is causing the controversy, I think I can explain it.
In math , the notion of 0.(9) has the precise meaning:
$$0.(9)=\lim_{n \to \infty} (\frac{9}{10^1}+....+\frac{9}{10^n})$$
In the above context, $$0.(9)$$ is obviously equal to 1.
This explain why $$9.(9)=10$$ and why $$9.(9)-0.(9)=9$$
You, on the other hand, are using $$0.(9)$$ to mean $$\frac{9}{10^1}+....+\frac{9}{10^n}$$. In this context, obviously $$0.(9) < 1$$.
For as long as you use the notion in an incorrect way, this thread will continue forever.


yes, although I can not read the notations [not familiar with the notation] you have made I can see that the limit function may be applied only in one direction...

Could you please explain why the result would end up being asymmetrical if you think this is the case?
Or do you believe the proof can be reversed where 1.000... can be proved to = 0.999...

Basically I am suggesting
that if 0.999... can be proved to = 1
then 1.000.... should be able to be be proved to = 0.999... using the same process from start to finish

and I believe at this stage this is not possible.

btw I do not believe this lack of symmetry, if proven, necessarily invalidates the proof. But it certainly makes it an interesting one to me, for many reasons.
 
Quantum Quack said:
Basically I am suggesting
that if 0.999... can be proved to = 1
then 1.000.... can be proved to = 0.999...
But that's what the "=" means. If a = b, then b = a. Equality is symmetric, by definition.
This is something you usually get taught fairly early in your education. although it might not be explained as a symmetric relation between elements in a set.
 
But that's what the "=" means. If a = b, then b = a. Equality is symmetric, by definition.
This is something you usually get taught fairly early in your education. although it might not be explained as a symmetric relation between elements in a set.
true, but you are ignoring the context of my post...

btw the = sign is often used ambiguously as well.
= can mean equal
= can mean same
= can mean IS

Often the equal sign can shift context midway in a formulation with out any strong indication of doing so thus invalidating the result.

"Equal" does not mean "same" nor does it mean "IS"
And in proofs like this one in discussion, I would imagine, the meaning of the signs are very important. "Contextual Axioms" as poster Undefined mentioned earlier.
This unfortunately often doesn't get much of a mention in high school at all as far as I can tell..
ie.
0.999... equals 1
or
0.999... same as 1
or
0.999... is 1

the difference between a quality and quantity come to the fore as well... blah blah blah... Philosophy and Math are strange bedfellows.. :)
 
Last edited:
QQ said:
btw the = sign is often used ambiguously as well.
Not in mathematics, or at least any of the mathematics I know about. The point of mathematical statements is in fact the exact opposite: non-ambiguity, or precision, even exactness.
 
Given that I have absolutely no idea of what you have formulated above I shall just accept that you know what you are doing.
If you have no idea what my math notation is saying, when what I am writing is accessible to any freshman math student, why would you have an opinion about any mathematical topic? Isn't the responsible thing to do after admitting ignorance is to give up all pretense at having an informed opinion?
Can I direct you to post #191
Where I make the value distance comparison between
9.999... and 9
9 and 8

Here is post #191:
ok... I did... however did you read why I reject the result... of
9.999...- 0.999... = 9 = 9 x 0.999...
with out the addition of an infinitesimal
Gosh this is hard to explain... sorry...
It is hard to explain because you don't have a consistent mathematical formulation of number so you insist 1 ≠ 0.999... because they look different and you insist 1 > 0.999... because 1 looks larger but this is predicated only experience with finite decimals and rational numbers, not any rigorous formulation of geometry or the real numbers. But the concept of an infinite number of repetitions can only be dealt with using the concept of continuity, thus the concept of Cauchy sequences and limits.

The nine in the middle is not the same nine depending on which side of the equation you are focussed on....
This contradicts the one of the earliest definitions of "equals." From the beginning of Euclid's Elements of Geometry, Book I:
Euclid said:
Κοιναι ἔννοιαι
αʹ. Τὰ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα.

Common notion 1. Things which equal the same thing also equal one another.
So if you agree A = C and either A = B or B = C then A = B = C.
http://farside.ph.utexas.edu/euclid/elements.pdf
http://aleph0.clarku.edu/~djoyce/java/elements/bookI/cn.html

try this as proof of my point.
Proof has a very different meaning than you are capable of using. You can't have a proof until you have axioms. Specifically, you can't prove 1 > 0.999... by assuming 1 > 0.999... That's not a proof, that's evasion of meaningful argument.

Theoretically the proof should also work as follows
Compare:

9.999... - 0.999... = 9 [exact with out the need for any "rounding"]

9 - 0.999... = 8 ["rounding" needed]
Not so. For 9.999... = 10, 0.999... = 1 and these are exact, without any rounding. Here you have assumed 1 > 0.999... in contradiction to the definition of the real numbers and in contradiction to the theory of limits and continuity. Having assumed that you conclude that 9 - 0.999... > 9 - 1 = 8 and try to flim-flam us that accepting the latter proves the hypothesis. That is an invalid form of argument called begging the question.

What is needed is not the assumption that 1 > 0.999... but a form of arithmetic that allow calculation of the magnitude of 1 - 0.999....

That is precisely what limits are good for. $$\lim_{n\to\infty} 1 - (1 - 10^{-n}) = \lim_{n\to\infty} 10^{-n} = 0$$ which says in English:
"As the number of digits used increases without bound, the difference between 1 and the repeating decimal expansion 0.999... shrinks to be smaller in magnitude than any finite number, thus the magnitude of the difference in the limit of an infinite number of steps must be zero."

http://en.wikipedia.org/wiki/Limit_of_a_sequence
http://en.wikipedia.org/wiki/Cauchy_sequence

Contention:
The value of the distance between
9.999... and 9 is less than the distance between 9 and 8 [therefore "rounding" is being applied inconsistently]
Again, you are still begging the question, for you are assuming 1 > 0.999... instead of calculating it.

If they read as follows
10 - 0.999... = 9
This is true as written in the real numbers.
and
9 - 0.999... = 8
This is true as written in the real numbers.
I would have no real problem and accept that rounding is being applied consistently.
Doing calculations with infinite decimals and therefore using limits of convergent sequences is in no way rounding.
However this would not be able to be used as proof that 0.999.. = 1
So, not only are you saying that you assume 1 > 0.999..., but you are saying if we prove 9 - 0.999... = 9 - 1, then you will not accept that as a demonstration that 0.999... = 1 even though Euclid writes:
Euclid said:
Common notion 3. And if equal things are subtracted from equal things then the remainders are equal.

Do you understand what I am suggesting with the above..?
That you are ignorant, opinionated and use a double-standard where you are allowed to make logical fallacies to defend your preconceptions but we are not allowed to use mathematics to prove theorems that follow from the definition of the number systems that all understanding of our universe is modeled in? If you don't like what I think you are suggesting, try making a formal argument without contradicting yourself or begging the question. And quit the conversational style. You don't have body language on a forum, so if you want me to know what your idea is then you need to take responsibility for expressing it in clear language.

This concludes my look at your post #191

The proof uses the multiplication of 0.999.... x 10 and then the subtraction of 0.999.... as a significant part of proving that 0.999... = 1
Thereby eliminating the need to consider the decimal positions which hold zeros, which are all of them to the right of the decimal point.

I would argue that the mere fact that it requires the proof to use the 0.999... or the 9.999... to prove the value of 1 self defeats the proof.
because by using 9.999... and subtracting a "clean" 0.999.... all this is stating is that 0.999... = 0.999... and not one.
To then use this as justification for 1=0.999... is apparently not logically valid.
The strength of the claim is obvious.
You haven't identified the mistake. Your naked claim that the argument is "apparently not logically valid" is not supported in any way.
So you have agreed that 9 times 0.999... is the same number as 10 times 0.999... minus 0.999... . Call this number x.
And you agree than 9 times 1 is 9. And you agree that x looks like 9. In fact x looks exactly like 9. That's because according to both this intuitive procedure of saying 9.999... = 9 + 0.999... and therefore that
9.999... - 0.999... = 9 + 0.999... - 0.999... = 9 + 0 = 9​
AND the formal procedure of taking limits
$$9.999... - 0.999... = \lim_{n\to\infty} \left( \left(10 - 10^{-n} \right) - \left(1 - 10^{-n}\right) \right) = \lim_{n\to\infty} ( 10 - 1 - 10^{-n} + 10^{-n} ) = 9$$​
it is 9.
Against the clearest possible cancelation and formal math you offer us ... nothing. No argument, just empty assertions.

To help here is the image file from wiki:
Poor form to link to "temporary" Wikipedia images. It is trivial to recreate it in TeX.
$$\Large \begin{eqnarray} x & = & 0.999... & & \textrm{Our premise} \\ 10 x & = & 9.999... & & \textrm{Valid because ... goes on without end} \\ 10x -x & = & 9.999... - 0.999... & & \textrm{Common Notion 3} \\ 9x & = & 9 & & \textrm{Arithmetic} \\ x & = & 1 & & \textrm{Q.E.D.} \end{eqnarray}$$

I would further contend that if the proof was valid and not a "trick" of binary manipulation.
The word you are looking for is algebraic, not binary.
the proof could be extended to include further subtracted evidence.
Mathematical proofs don't use evidence. There is no need to "extend" a proof, once it hits its target it is valid for all time.
10 x 0.999... = 9.999...

9.999... - 0.999... = 9
9 - 0.999... =/= 8 if one wishes to apply a consistent value to 0.999...
The Wikipedia proof proves x = 1 starting from x = 0.999... and without you pointing out an invalid step, therefore 9 - 0.999 = 8 if one wishes to be mathematically consistent.

Or simply ask:
if 9.999... - 0.999... = 9
then how does
9 + 0.999... = 10 ?
It's trivial if you accept 1 = 0.999... and deeply confusing if you deny it. My question is, why would you deny it? It's like you are are asserting because limes and green varieties of apples look similar to unripe citrus and apples, that they are unripe, when actually they are simply green.
1 looks larger that 0.999... because 0.999... looks like 0.999 but there is an infinite difference between 3 nines and nines repeating forever.

edit: I do understand that if I was really smart and really clever I could create a mathematical number system that would make 0.8888 = 1.
As written that statement is false. If you assume 0.8888 is a finite expansion in some non-base-ten number system, then you are asking to solve $$b^4 = 8 (b^2 + 1) (b + 1)$$ which has no integer solution, but does have an approximate solution near 8.99878
$$( ( ( 8 / 8.99878 + 8 ) / 8.99878 + 8) / 8.99878 + 8 ) / 8.99878 \approx 1$$
That's because in base 9, $${0.111...}_9 = \frac{1}{8}$$, $${0.888...}_9 = \frac{8}{8} = 1$$ as per division algorithm 2. Perhaps you meant to write 0.888....
So I appreciate we are working with in the limits and rules of the given number system in use. The absence of non-zero infinitesimals in this particular number system [Real] IMO forces the inability to use infinitesimals to support any contradicting proof.
Incorrect. You may use infintesimals if you want, but then you are responsible for the monster you create. You have to describe all the formal algebraic rules for manipulating your new type of "number" and you have to take responsibility if your system lacks internal consistency by falling on your sword. See the quote by Hamming at the end of this post.

Yet we are still left with the argument above.
What argument? Your naked assertions and begging the question don't amount to an argument.

I find making an image file is a great way to consolidate and clarify the issue.
Just another propaganda poster, not an argument.
Let me mark it up so it is a useful instructional tool.
$$\begin{eqnarray} x & = & 0.999... & & & \textrm{Our premise} \\ 10 x & = & 9.999... \\ 10x -x & = & 9.999... - 0.999... & = & 9 & \textrm{Common notion 3 and Arithmetic} \\ 9 - 0.999... & \neq & 8 & & & \textrm{Naked and invalid assertion -- Non sequitur}
\\ & & \textrm{So...} \\
9.999... - 0.999... & = & 9 & & & \textrm{Proved above}
9 - 0.999... & \neq & 8 & & & \textrm{Non sequitur}
\\ & & \textrm{Why?} & & & \textrm{because you cheated by begging the question} \\
9.999... - 0.999... & = & 9 & & & \textrm{Proved above}
\\ & & \textrm{yet} \\
9 + 0.999... & \neq & 10 & & & \textrm{Naked and invalid assertion -- Non sequitur}
\\ & & \textrm{because} \\
9.999... - 0.999... & = & 9 & & & \textrm{Proved above}
0.999... & = & x & = & 1 & \textrm{Q.E.D}
9 - 0.999... & = & 8 & & & \textrm{Corollary}
9 + 0.999... & = & 10 & & & \textrm{Corollary}
\\ \end{eqnarray}$$

@Rpenner,
How does that deal with the argument?

"because 9.999...-0.999... = 9
9 + 0.999... can not equal 10"
Noone ever needs to deal with that argument because it is a non sequitur. There is not the slightest chain of logic between the first and second part of that sentence. What you wrote is semantically equal to "Because D - B = A then A + B ≠ C" but at no point did you eliminate the possibility that C = D.

In this case we haven't even got to the rest of the proof...

for surely if 9.999...- 0.999... = 9 then logically
9 + 0.999.... must equal 9.999...
Correct. If D - B = A, then A + B = D.

just to be absolutely clear..
it is similar to
for surely if 10 - 5 = 5 then logically
5 + 5 = 10
The correspondence is exact. A = 5, B = 5, D = 10. But it doesn't rule out A + B = C or C = D.

Understanding I am looking for and yet to be found:
The point I think is that the proof is only valid if numbers are increasing in value and not decreasing in value. [ or vice verse, but symmetry is absent.]
This is muddled writing which I take as evidence of muddled thinking. There is a whole mathematical discipline to the study of symmetry, but you have yet to master the fundamentals of division algorithms, limits of sequences, what separates the real numbers from the rational numbers, how to write a proof, how to make a logical argument, etc.

So why you consider your opinion that 1 > 0.999... is a valid opinion even when the definition of real numbers says it is not and there are formal proofs of this on tertiary resources like Wikipedia and primary sources like the computer-validated proof at http://us.metamath.org/mpegif/0.999....html make it inescapable that the normal rules of math require 1 = 0.999...

If you want to invent new math to work with your expectations, you are welcome to try. But you don't get to win this debate which is based on the normal rules of math.
Repeating your naked assertions without argument of any kind is not a contribution to a dialogue or discussion; it's authoritarian posturing.
Richard Hamming "Mathematics on a Distant Planet." American Math Monthly 105 640-650 (1998)
"In mathematics we do not appeal to authority, but rather you are responsible for what you believe."​
Thus the burden is on you to either use the notation used by the rest of the world or take responsibility and actually explain what your notation means in detail
 
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