Code to recognize repeating numbers.

lets say we have a fraction were the denominator is always bigger than the numerator. For example 2/3 or 7/12. How would you create code that identifies the repeating part of the decimal for example. 7/12 is 0.583333333 and 1/2 as 0.5000000

I would like code that returns 7/12 as a= 5 8 b = 3

Further 1/2 a = 5 b = 0

so a would be the non repeating part and b would be the repeating part. Thanks for the help on this one. I'd really appreciate it.

[Jan]

How many digits do you want to consider? What about 1.0000000000000001? What is the wanted output for 1.2323232323? Does "a = 5 8" mean "a = [5, 8]"?

[Max]

I wouldn't like digits above 1 to be output so the value should always be below 1(so I would never divide say 6/5 or 3/2 only things like 5/6 and 2/3)as the denominator should always be bigger than the numerator. Yes sorry it should be a = [5,8]

[Image Analyst]

Any fraction that is limited to the number of decimal places it can have, as are all numbers in a computer, can be represented as a rational number b/a. How many decimal places do you want to inspect before you decide it's a rational number? Or, equivalently, what is the largest denominator you're willing to accept? Would an answer of 18463878207 / 918298769838690 be okay with you?

This sounds so much like homework. If it is, read this

[Walter Roberson]

How odd, just yesterday I was sent the following in email, which is obviously homework:

--- being quote ---

Hey I would really like your help for this one. I was given this question could you solve it for me please, I would be so grateful.

If i have a function

function [a,b]=periodicfraction(p,q)

How would I write code to express divisions in array form. Where (p/q) where 0<q and a is an array of non repeated decimals and b are the repeated decimals. For example, if p=5 and q=6 p/q would be 0.83333333. From this result a (the non repeated part of the decimal) would be 8 and b the repeating part of the decimal would be 3. so my answer should appear as a=(8) b=(3). [a,b]=periodicfraction(1,2) % 1/2 a = 5 b = 0

as 1/2 is 0.50000 b would equal 0.

A final example would be 119/990= 0.120202020202020

[a,b]=periodicfraction(119,990) % 119/990
a = 1
b = 2 0

Answer 1

[Image Analyst]

Try this:

clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 20;

% If the fraction is within this, say it's a match
tolerance = 0.1;
denominators = 2:1000;
theFraction = .123456;  % Whatever number you're examining
% Find potential numerators, rounded to the closest integer:
numerators = round(theFraction * denominators);
% We need to get rid of zeros.
zeroIndexes = numerators == 0;
numerators(zeroIndexes) = [];
denominators(zeroIndexes) = [];
% Now get the ratios of integers.
ratios = numerators ./ denominators;
differences = abs(ratios - theFraction);
subplot(2,1,1);
plot(ratios);
title('Ratios', 'FontSize', fontSize);
grid on;
subplot(2,1,2);
plot(differences);
title('Differences', 'FontSize', fontSize);
grid on;
% Set up figure properties:
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
% Get rid of tool bar and pulldown menus that are along top of figure.
set(gcf, 'Toolbar', 'none', 'Menu', 'none');
% Give a name to the title bar.
set(gcf, 'Name', 'Demo by ImageAnalyst', 'NumberTitle', 'Off') 

% Find the min difference:
[minDifference, indexOfMin] = min(differences)
% Print the fraction that is closest:
message = sprintf('%.8f is approximately equal to %d / %d, which equals %.9f.\nThe difference is %.9f.',...
theFraction, round(numerators(indexOfMin)), denominators(indexOfMin), ...
numerators(indexOfMin)/denominators(indexOfMin), minDifference)
uiwait(helpdlg(message));