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 May 31st, 2016, 05:47 PM #1 Newbie   Joined: May 2016 From: Somewhere, southamerica Posts: 17 Thanks: 7 Working over standard notation systems That is my focus on these days. Not much progress but less partners with wich share them. For some reason it seems that Peano forgot that \begin{align}\left (x \right )_b= a_nb^n+a\left_\left ( n-1\right )\right + ... \end{align} so every number is a polimomial equation \begin{align} F:X \mapsto Y , f\left ( x \right )=a_{n}x^{n}+a_{n-1}x^{n-1}+... \end{align} such that \begin{align}\left (a \right )_{n}=\left \{ a/a\in \mathbb{Z},\forall \left |a \right |< x \right \},x\in \mathbb{N}\end{align}= being \begin{align} \left ( a \right )_{n}\end{align} a series is natural that \begin{align}n\in \mathbb{Z}\end{align}. All these delivers us to think that is possible to deduce a simple algorithm for base convertion having account of the rest but what is rest? In fact everything what we learned at school is in fact complex math on the abstract algebra field. This post is not a diary but ill try to expand and complete it on further editions, meanwhile feel free to comment and send your opinion. Thanks from manus
June 1st, 2016, 01:37 PM   #2
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Quote:
 Originally Posted by Politician That is my focus on these days. Not much progress but less partners with wich share them. For some reason it seems that Peano forgot that \begin{align}\left (x \right )_b= a_nb^n+a\left_\left ( n-1\right )\right + ... \end{align} so every number is a polimomial equation \begin{align} F:X \mapsto Y , f\left ( x \right )=a_{n}x^{n}+a_{n-1}x^{n-1}+... \end{align} such that \begin{align}\left (a \right )_{n}=\left \{ a/a\in \mathbb{Z},\forall \left |a \right |< x \right \},x\in \mathbb{N}\end{align}= being \begin{align} \left ( a \right )_{n}\end{align} a series is natural that \begin{align}n\in \mathbb{Z}\end{align}. All these delivers us to think that is possible to deduce a simple algorithm for base convertion having account of the rest but what is rest? In fact everything what we learned at school is in fact complex math on the abstract algebra field. This post is not a diary but ill try to expand and complete it on further editions, meanwhile feel free to comment and send your opinion.
So I already know that most of this has been already solved by other skilled brains, but im doing this effort at doing it formyself with the tools that I have.
Being that \begin{align}\left (x \right )_{b}=a_{n}b^{n}+a_{n-1}b^{n-1}+...\end{align} is an \begin{align}n^{th}\end{align} grade polinomial $f\left (b \right )=a_{n}b^{n}+a_{n-1}b^{n-1}+...$ or more properly $f\left (x \right )=a_{n}x^{n}+a_{n-1}x^{n-1}+...$ Then how do we know based on previous defintions that for every $b \mapsto x/F:X,\: f\left (b \right )=a_{n}b^{n}+a_{n-1}b^{n-1}+...$ther is an unique ${b}' \mapsto x/G:X, \:g\left ( {b}' \right )=a_{m}{b}'^{m}+a_{m-1}{b}'^{m-1}+...$ such that $f\left ( b \right )=g\left ( {b}' \right )$?
If anybody has a clue on this and it is reading please submit it because i dont.
We can alternative define $b \mapsto x/F:X, \:f\left (b \right )=a_{n}b^{n}+a_{n-1}b^{n-1}+...$, $\left (x \right )_{b}$ as $\left (x \right )_{b}=\sum_{i=-\infty}^{i=n}a_{i}b^{i}$.

 June 1st, 2016, 01:44 PM #3 Newbie   Joined: May 2016 From: Somewhere, southamerica Posts: 17 Thanks: 7 A good property of this is a Goldstain machine for saying that $\mathbb{Q}=\frac{a}{b}$ with $a,\: b\: \in \: \mathbb{Z}$. I have had already demontrated thes ($\left ( x \right )_{b}\in \mathbb{R}$). Thanks from manus
 June 1st, 2016, 01:44 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra Are you asking "how can we be sure that the representation of a given number $x$ in a given base $b$ is unique?" Seeing as you are looking at the reals, the answer is that it is not for some reals. Generally speaking, mathematics is not overly concerned by this because it deals with numbers rather than the representation of numbers. Thanks from manus Last edited by v8archie; June 1st, 2016 at 01:48 PM.
 June 19th, 2016, 04:35 PM #5 Newbie   Joined: May 2016 From: Somewhere, southamerica Posts: 17 Thanks: 7 Representation of numbers are also mathemathic algebraic systems and effectively standard positional is a semigroup where $a+b=ab$
June 19th, 2016, 05:09 PM   #6
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Representation of numbers are also mathemathic algebraic systems and effectively standard positional is a semigroup where $a+b=ab$, so though as you say not overly concerned it is at least undirectly studied over monoids and probability and surely I am not the only miscelanous curious, also base conversion is very well studied though not the same with the rest and as i said it seems that they are reached over complex maths.
Quote:
 Originally Posted by v8archie Are you asking "how can we be sure that the representation of a given number $x$ in a given base $b$ is unique?"
Effectively yes that was on of my inquiries as I started on this.
Quote:
 Originally Posted by v8archie Seeing as you are looking at the reals, the answer is that it is not for some reals.
Can you be more specific? What real number has more than one representation on a given base over a standard positional system?
For sure we have that $f\left ( n,a,b... \right )=ak^{n}+bk^{n-1}+...$ being $k$ a constant and ${n,a,b...}$ an ungiven number of variables so how can be sure that for a number $f( n,a,b...)$ there isnt a $g( n',a',b'...)$ with ${a',b'...}$ such that $f( n,a,b...)=g(n',a',b'...)$ where the number of variables is undefined and so... cant we call It a polydimensional equation?
P.D. : I learn the html slang by stealing it so thanks.

Last edited by Politician; June 19th, 2016 at 05:14 PM.

 June 19th, 2016, 05:45 PM #7 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 The non-terminating decimal 0.999... = 1 (working in base 10 of course).
 June 19th, 2016, 06:55 PM #8 Newbie   Joined: May 2016 From: Somewhere, southamerica Posts: 17 Thanks: 7 Lets demonstrate that for \begin{align}\left ( x \right )_{b}=a_{n}x^{n}+a_{n-1}x^{n-1}+...,\: \left ( x \right )_{b}=f\left ( x \right ), \: F:X \mapsto Y , \:f\left ( x \right )=a_{n}x^{n}+a_{n-1}x^{n-1}+... ,\:\left (a \right )_{n}=\left \{ a/a\in \mathbb{Z},\:\forall \left |a \right |< x \right \},\: x\in \mathbb{N},\: S_{n}=\left ( a \right )_{n}\end{align}\rightarrow \left ( x \right )_{b}\in \mathbb{Q} Because its easy:$H\)\left\{\begin{matrix} \\\forall x \in \mathbb{N},\: \left ( x \right )_{b}=a_{n}x^{n}+a_{n-1}x^{n-1}+... \\ \: \left ( x \right )_{b}=f\left ( x \right ), \: F:X \mapsto Y, \:f\left ( x \right )=a_{n}x^{n}+a_{n-1}x^{n-1}+... \\ S_{n}=\left ( a \right )_{n}, \:\left (a \right )_{n}=\left \{ a/a\in \mathbb{Z},\:\forall \left |a \right |< x \right \} \\ \end{matrix}\right.\: T)\left\{\begin{matrix} x\in \mathbb{Q}\left ( x\right )_{b}=f\left ( x \right )\rightarrow \left ( x\right )_{b}=g\left ( x \right )+h\left ( x \right )+...$ $\left ( x\right )_{b}=a_{n}x^{n}+a_{n-1}x^{n-1}+...\rightarrow \left ( x\right )_{b}=...+a_{n-n-1}x^{n-n-1}+... \: x^{n-n-1}=x^{-1},\: x^{-1}=\frac{1}{x},\: x\in \mathbb{N}\rightarrow x^{n-n-1}\notin \mathbb{Z} \: \forall x\neq 1\rightarrow x^{n-n-1}\in \mathbb{Q}\rightarrow \left ( x\right )_{b}\in \mathbb{Q}$ sillyness but doesnt that tell us that $\left (x \right )_{b}\in \mathbb{Q}\: \forall b\neq 1$? not so useless after all. unless intuition tell us rhe contrary but that would be later, this is all for today. Last edited by Politician; June 19th, 2016 at 06:57 PM.
June 19th, 2016, 07:00 PM   #9
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Quote:
 Originally Posted by Azzajazz The non-terminating decimal 0.999... = 1 (working in base 10 of course).
It seems its a classic, someday ill try to demonstrate that if that is true then every number is equal to its subsequent, in fact i dont have to demontrate that, is pretty intuitive.
However you are right; $9\times 10^{-1}+9\times 10^{-2}+...=1\times 10^{0}$
and possible $\left ( 0,FFF... \right )_{16}=\left ( 1 \right )_{16}$ and even demonstrable for every base just try it.

Last edited by Politician; June 19th, 2016 at 07:13 PM.

June 20th, 2016, 12:04 AM   #10
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Quote:
 Originally Posted by Politician It seems its a classic, someday ill try to demonstrate that if that is true then every number is equal to its subsequent, in fact i don't have to demonstrate that, is pretty intuitive.
Relying on intuition is dangerous in maths. The point here is that $0.99...$ is not the number before $1$; it is EXACTLY EQUAL to $1$.

In fact, the term "subsequent number" is meaningless for the reals, since they are uncountable.

Quote:
 Originally Posted by Politician $\left ( 0,FFF... \right )_{16}=\left ( 1 \right )_{16}$
That's true. In general, $(0.(b - 1)(b - 1)...)_b = 1_b$ where $b - 1$ represents the largest 1-digit number in base $b$.

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