I was surprised to discover that I hadn’t written about this topic before: converting a decimal value into a fraction. Of course, the solution is really stupid — which I’ll show in a moment. But the goal is to reduce or simplify the stupid way and end up with a fraction instead of a decimal.
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\nTo convert a decimal into a fraction you convert the decimal portion into an integer ratio and then reduce the fraction.<\/p>\n
For example, the value 0.256 can be expressed as 256<\/sup>\/1000<\/sub>. This fraction is then reduced to 32<\/sup>\/125<\/sub>. The value 0.25 becomes 25<\/sup>\/100<\/sub> and then 1<\/sup>\/4<\/sub>.<\/p>\n So, the process seems rather simple: Multiply the decimal to create a larger, integer value and then reduce this fraction.<\/p>\n To write the code, I borrowed from an Exercise<\/a> presented four years ago on reducing fractions. Here’s what I came up with:<\/p>\n The program prompts for a decimal value as input, stored in float<\/em> variable To reduce the fraction, and employ Euclid’s algorithm, I calculate the The rest of the code is lifted from the earlier Exercise’s solution. A printf(<\/em>) statement outputs the results.<\/p>\n Here are a few sample runs:<\/p>\n Alas, the program fails to properly convert thirds:<\/p>\n I can read this failure in that either I’m not doing the conversion properly or there’s some other trick that I’m missing to catch certain exceptions. Or, I suppose, the problem could be with Euclid’s algorithm not being able to reduce a continuing fraction. I dunno.<\/p>\n For the most part, my approach works. It solves a puzzle I hadn’t yet addressed in this blog. I’d be interested to know of any other approaches for converting decimal values into a rational representation.<\/p>\n","protected":false},"excerpt":{"rendered":" I’ve covered how to get from ¼ to 0.25 before. Now it’s time to explore how to get from 0.25 to ¼. Continue reading 2026_03_28-Lesson.c<\/a><\/h3>\n
\r\n#include <stdio.h>\r\n#include <stdlib.h>\r\n\r\nint main()\r\n{\r\n float decimal;\r\n int numerator,denominator,diff,larger,smaller;\r\n\r\n \/* obtain and validate input *\/<\/span>\r\n printf(\"Enter decimal value: \");\r\n scanf(\"%f\",&decimal);\r\n if( decimal > 1.0 || decimal < 0.0 )\r\n {\r\n puts(\"Please input a value less than 1.0\");\r\n puts(\"and greater than zero\");\r\n return 1;\r\n }\r\n\r\n \/* configure the numerator and denominator\r\n use 100000 based on 'float' precision *\/<\/span>\r\n denominator = 100000;\r\n numerator = decimal*denominator;\r\n\r\n \/*\r\n Use Euclid's algorithm to find the least common\r\n denomniator and reduce the fraction\r\n *\/<\/span>\r\n \/* calculate differences between the larger and smaller values *\/<\/span>\r\n larger = numerator>denominator ? numerator : denominator;\r\n smaller = numerator<denominator ? numerator : denominator;\r\n diff = larger-smaller;\r\n \/* keep calculating until the common denominator is found *\/<\/span>\r\n while( diff!=larger )\r\n {\r\n larger = smaller>diff ? smaller : diff;\r\n smaller = smaller==larger ? diff : smaller;\r\n diff = larger-smaller;\r\n }\r\n\r\n printf(\"%f is the fraction %d\/%d\\n\",\r\n decimal,\r\n numerator\/diff,\r\n denominator\/diff\r\n );\r\n\r\n return 0;\r\n}<\/pre>\ndecimal<\/code>. A test is made to confirm that the value is in the range of 1.0 to zero — a positive decimal without an integer portion. (Though the program does allow 1.0 to be input.)<\/p>\nnumerator<\/code> and denominator<\/code> values:<\/p>\ndenominator = 100000;
\nnumerator = decimal*denominator;<\/code><\/p>\nEnter decimal value: 0.875\r\n0.875000 is the fraction 7\/8\r\n\r\nEnter decimal value: 0.4\r\n0.400000 is the fraction 2\/5\r\n\r\nEnter decimal value: 0.212121\r\n0.212121 is the fraction 5303\/25000<\/pre>\n
Enter decimal value: 0.6666666\r\n0.666667 is the fraction 33333\/50000<\/pre>\n