This of course uses much more code, but I think the spirit of the problem exonerates this bare-bones approach; for all we know, the internal rationalize() functions of modern languages have lots of internal loops.

This works by calling a rationalization function r() with a starting denominator of 1. The function begins incrementing a numerator, and checking at every increment whether the resulting number, when rounded to the same number of digits as the original, has the same string representation as the original. Once the numerator has been incremented so much that the result is greater than the original, the function increments the denominator and calls itself.

This of course uses much more code, but I think the spirit of the problem exonerates this bare-bones approach; for all we know, the internal rationalize() functions of modern languages have lots of internal loops.

C, 232

This of course uses much more code, but I think the spirit of the problem exonerates this bare-bones approach; for all we know, the internal rationalize() functions of modern languages have lots of internal loops.

Note that this doesn't work for an input of "0." because that is not a standard way to write a float, so when it re-writes the float to string, the result will never be a "0.".

The specs want a function that returns values instead of just printing to screen, hence the argument-passing.

Code (ungolfed):

void r(char* x, int* a, int* b) {
    int i = -1;
    char z[32];
    double v =atof(x);
    while(1) {
        i++;
        double y = ((double)i)/((double)(*b));
        double w;
        sprintf(z, "%.*f", strlen(strchr(x,'.'))-1, y);
        if(strcmp(x, z)==0) {
            *a = i;
            return;
        }
        w = atof(z);
        if(w > v) {
            (*b)++;
            r(x, a, b);
            return;
        }
    }
}

Usage:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char* argv[]) {
    int num;
    int denom = 1; // start with a denominator of 1
    r(argv[1], &num, &denom);
    printf("%d/%d\n", num, denom);
    return 0;
}

Golfed code:

#define double D
void r(char*x,int*a,int*b){int i=-1;char z[32];D v=atof(x);while(1){i++;D y=((D)i)/((D)(*b));D w;sprintf(z,"%.*f",strlen(strchr(x,'.'))-1,y);if(!strcmp(x,z)){*a=i;return;}w=atof(z);if(w>v){(*b)++;r(x,a,b);return;}}}

C, 233

This of course uses much more code, but I think the spirit of the problem exonerates this bare-bones approach; for all we know, the internal rationalize() functions of modern languages have lots of internal loops.

Note that this doesn't work for an input of "0." because that is not a standard way to write a float, so when it re-writes the float to string, the result will never be a "0.".

The specs want a function that returns values instead of just printing to screen, hence the argument-passing.

Code (ungolfed):

void r(char* x, int* a, int* b) {
    int i = -1;
    char z[32];
    double v =atof(x);
    while(1) {
        i++;
        double y = ((double)i)/((double)(*b));
        double w;
        sprintf(z, "%.*f", strlen(strchr(x,'.'))-1, y);
        if(strcmp(x, z)==0) {
            *a = i;
            return;
        }
        w = atof(z);
        if(w > v) {
            (*b)++;
            r(x, a, b);
            return;
        }
    }
}

Usage:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char* argv[]) {
    int num;
    int denom = 1; // start with a denominator of 1
    r(argv[1], &num, &denom);
    printf("%d/%d\n", num, denom);
    return 0;
}

Golfed code:

typedef double D;
void r(char*x,int*a,int*b){int i=-1;char z[32];D v=atof(x);while(1){i++;D y=((D)i)/((D)(*b));D w;sprintf(z,"%.*f",strlen(strchr(x,'.'))-1,y);if(!strcmp(x,z)){*a=i;return;}w=atof(z);if(w>v){(*b)++;r(x,a,b);return;}}}