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.
$$ \left[ \psi \right] $$ 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.
Last edited:
