My Math Forum An Anomaly of Decimal Representation

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June 28th, 2018, 04:21 PM   #31
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Quote:
 Originally Posted by zylo The theorem on term by term multiplication only applies if you think of "infinite decimals" as limits rather than sums. A sum does not have a limit, it depends on the number of terms. A sum is literal. Limit is a defined $\displaystyle \epsilon$ $\displaystyle \delta$ operation on a sum.
I used to think you spent all of this time coming here and never learned a thing. At this point I am convinced you are actually regressing in knowledge, becoming more ignorant the more you "learn" math. This post certainly doesn't discredit that theory.

As usual \popcorn. Carry on.

June 28th, 2018, 07:40 PM   #32
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 Originally Posted by Maschke a decimal expansion as a function from the naturals to the digits. I regard that as essential to understanding why, for example, there is no such real number as .000....00001. People always have an intuition that there is such a thing.
Doesn't that function restrict you to finite (but arbitrarily long) decimals?

 June 28th, 2018, 07:58 PM #33 Global Moderator   Joined: Dec 2006 Posts: 20,933 Thanks: 2207 No. I assume you misinterpreted what Maschke posted.
 June 29th, 2018, 07:26 AM #34 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra That's quite possible.
June 29th, 2018, 07:40 AM   #35
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 Originally Posted by zylo The theorem on term by term multiplication only applies if you think of "infinite decimals" as limits rather than sums.
This could make some sort of sense if you then divorce decimal representation entirely from the concept of real numbers. In this sense you can deal with them in the way that Maschke outlined. But they no longer represent the real numbers - you are into a non-standard analysis.

As soon as you bring the real numbers into consideration, you have to include the concept of limits because you need the limit of convergent series of rationals. This is because that is what the real numbers are in the most fundamental sense.

In the case that we aren't considering the real numbers, we have the problem that an infinite sum no longer has a value. So we need some other way to assign a value to that infinite sum. You haven't done this, and it's causing all kinds of problems because without it your sums don't mean anything.

Last edited by skipjack; July 3rd, 2018 at 09:36 AM.

June 29th, 2018, 09:10 AM   #36
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Quote:
 Originally Posted by v8archie Doesn't that function restrict you to finite (but arbitrarily long) decimals?
No, why? You input a natural and get out a digit. One digit for each natural.

 July 3rd, 2018, 05:19 AM #37 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Divide a line into halves. Label origin and midpoint 0,1 respectively. Divide each half in halves again and label the resulting points 00,01,10,11. Proceeding in this manner an n place binary sequence is a point on a line divided in segments n times. Divide a line into tenths. Label origin and tenth points 0,1,2,3,4,5,6,7,8,9 respectively. Divide each tenth in tenths again and label the resulting points 00,01,....09,10,11,12,13,...........99. Proceding in this manner an n place binary sequence is a point on a line divided in tenths n times. And now the meaning of digits becomes clear. They represent the points of the first division. Divide a line in half. Digits: 0,1 Divide a line in thirds: Digits: 0,1,2 Divide a line in tenths: Digits: 0,1,2,....9 Divide a line in sixteenths: Digits: 0,1,2,......,9,A,B,...,F For an alphabetical number system: A,B: Divide the line into two equal parts and label the points A,B A,B,C: Divide the line into three equal parts and label the points A,B,C. After dividing each segment into three equal parts again, the points on the line are AA,AB,AC,BA,BB,BC,CA,CB,CC A,B,C,......Z.: Divide the line into twenty seven equal parts. AA,BB,........,ZZ: Divide the line into 54 equal parts. After dividing again, the points are (AA)(AA),(AA)(BB),.......,(ZZ)(ZZ). Observations: 1) There are no anomalies. Every sequence (AZEQ.....M) in a number system represents a unique point on the line no matter how many places you use. 2) Thinking in decimal notation again: A decimal represents a rational number no matter how many places in the decimal. So for any n, a sequential list of decimals contains only rational numbers. Then by induction, ANY list of decimals (binary digits) contans only rational numbers. So real numbers don't have a decimal representation? The way out is that a rational number represented by a non-repeating decimal as n-> infinity is called real.
July 3rd, 2018, 05:32 AM   #38
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Quote:
 Originally Posted by zylo Divide a line into halves. Label origin and midpoint 0,1 respectively. Divide each half in halves again and label the resulting points 00,01,10,11. Proceeding in this manner an n place binary sequence is a point on a line divided in segments n times. Divide a line into tenths. Label origin and tenth points 0,1,2,3,4,5,6,7,8,9 respectively. Divide each tenth in tenths again and label the resulting points 00,01,....09,10,11,12,13,...........99. Proceding in this manner an n place binary sequence is a point on a line divided in tenths n times. And now the meaning of digits becomes clear. They represent the points of the first division. Divide a line in half. Digits: 0,1 Divide a line in thirds: Digits: 0,1,2 Divide a line in tenths: Digits: 0,1,2,....9 Divide a line in sixteenths: Digits: 0,1,2,......,9,A,B,...,F For an alphabetical number system: A,B: Divide the line into two equal parts and label the points A,B A,B,C: Divide the line into three equal parts and label the points A,B,C. After dividing each segment into three equal parts again, the points on the line are AA,AB,AC,BA,BB,BC,CA,CB,CC A,B,C,......Z.: Divide the line into twenty seven equal parts. AA,BB,........,ZZ: Divide the line into 54 equal parts. After dividing again, the points are (AA)(AA),(AA)(BB),.......,(ZZ)(ZZ). Observations: 1) There are no anomalies. Every sequence (AZEQ.....M) in a number system represents a unique point on the line no matter how many places you use. 2) Thinking in decimal notation again: A decimal represents a rational number no matter how many places in the decimal. So for any n, a sequential list of decimals contains only rational numbers. Then by induction, ANY list of decimals (binary digits) contans only rational numbers. So real numbers don't have a decimal representation? The way out is that a rational number represented by a non-repeating decimal as n-> infinity is called real.

Sure, but your process of subdividing the interval only really represents numbers with finite decimal expansion, which are rational. You won't even get 1/3.

And you can't generalize this by induction really. Induction doesn't work like that. Induction makes you able to prove that for all natural numbers $n$, the decimal expansion with length $n$ is finite. But nothing is ever said this way about infinite decimal expansion. Such a thing can never be done by induction.
Here is a quite different example to show the induction won't work: Take a set $A_n = \{1,...,n\}$. This set is finite. Thus by induction, we can obtain that for all $n$, we have that $A_n$ is finite. Fine. But we can't get by induction that the "limit" (which would intuitively be $\{1,2,3,....\}$) is infinite.

Last edited by skipjack; July 3rd, 2018 at 09:40 AM.

July 3rd, 2018, 11:29 AM   #39
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Quote:
 Originally Posted by zylo So for any n, a sequential list of decimals contains only rational numbers. Then by induction, ANY list of decimals (binary digits) contans only rational numbers.
Misuse/misunderstanding of induction.

Induction says that if something's true for 1 and true for n+1 whenever it's true for n, it's true for all natural numbers. It says nothing about transfinite numbers.

As an example, each natural number 1, 2, 3, 4, ... is finite. But you can't therefore conclude that the transfinite ordinals or cardinals are finite. In fact they're not.

July 6th, 2018, 04:22 AM   #40
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Quote:
 Originally Posted by Micrm@ss 1) Sure, but your process of subdividing the interval only really represents numbers with finite decimal expansion, which are rational. You won't even get 1/3. 2) But nothing is ever said this way about infinite decimal expansion. Such a thing can never be done by induction. 3) Here is a quite different example to show the induction won't work: Take a set $A_n = \{1,...,n\}$. This set is finite. Thus by induction, we can obtain that for all $n$, we have that $A_n$ is finite. Fine. But we can't get by induction that the "limit" (which would intuitively be $\{1,2,3,....\}$) is infinite.
1) You are right. The process leaves open the possibility of non-repeating, or repeating, decimals (or binaries) which get arbitrarily close to, say, pi or $\displaystyle \sqrt{2}$ or 1/3, though you will never hit them.

So if .333...... is rational and 1/3 is rational, what is the real number between them? The anomaly is still unresolved.*

$\displaystyle \sqrt \neq p/q$ but you can get as close as you like.

You can't hit every point on a line with a radix number system, but you can get arbitrarily close, and it's the best nature gives you.

A decimal (or binary) system of rational numbers gets you as close as you want to a point on the line. So now v8archie says, aha, so now you admit there are points on the line not counted by the rational (decimal or binary) number system. I reply, if you reduce the distance between points to zero, you in effect, hit all the points. And counting all decimal or binary representations of points on the line is not necessarily the same as counting all points on the line.

And anyhow, we are talking about points on a line, and there is no such thing as a continuous line in nature. There is continuous motion (as far ae we know), but you can only measure position with a real ruler. Unless you somehow imagine a light ray travelling along the line and marking points continuously.

2) Induction is the best nature gives you. Sure, you can imagine the moon is made of green cheese, or beyond the beyond.

3) Circular reasoning.

Maschke, Sorry, to me trans=finite is double=talk. There is no trans-finite, there is only for any n (rather than all n, which is still a little flaky, but serviceable to combat trans-finite).

*Edit
The answer is that in the limit of infinite decimal places there is no number between them though they remain distinct.

And in deference to Maschke, one (you) might conceive that there are more points on a line than you can count with a number system.

Last edited by zylo; July 6th, 2018 at 04:48 AM.

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