ITT I develop the Rational Fraction Finding Procedure (RFFP) and explore one of its uses in the current bleeding edge of the field mathematics.
Suppose x is some positive real number. Start with the fraction 1/1. If the fraction is less than 1/1, increase the numerator by 1. If the fraction is greater than x, decrease the numerator by 1. Repeat this process of incrementing the numerator or denominator until the fraction equals exacty x. For clarity I illustrate how RFFP finds a rational fraction for 1.666...:
1/1 -> 2/1 -> 2/2 -> 3/2 -> 4/2 -> 4/3 -> 5/3 = 1.666....
Now I use a slight modification of RFFP to settle the debate of whether or not 0.999=1. Instead of targeting some real number x, we will proceed as follows: start with the fraction 1/1. If the fraction is greater than or equal to 1, increase the denominator. If the fraction is less 1, increase the numerator. This means we will converge on a number less than 1. Using the power of computers, we get the following sequence of fractions:
1/1 = 1.000000
1/2 = 0.500000
2/2 = 1.000000
2/3 = 0.666667
3/3 = 1.000000
3/4 = 0.750000
4/4 = 1.000000
...
49997/49998 = 0.999980
49998/49998 = 1.000000
49998/49999 = 0.999980
49999/49999 = 1.000000
49999/50000 = 0.999980
50000/50000 = 1.000000
...
As you can see, the number of 9s in the decimal form of the fraction increases as the iterations continue. In fact, we can mathematically prove that the number of leading 9s increases. First note that the sequence of fractions alternates between 1 and something less than one. Exclude all the 1s, so we are left with an increasing sequence where the i'th fraction is given by S_i = i/i+1. As i increases, since S_i is strictly increasing, eventually it will become greater than 9/10 but less than 1. This means it has at least 1 leading 9 in the demical form, since we can subtract 9/10 from it. Continue increasing, and eventually we can substract 9/10+9/100, meaning it has two leading nines. So the number of leading 9s increases.
Suppose x is some positive real number. Start with the fraction 1/1. If the fraction is less than 1/1, increase the numerator by 1. If the fraction is greater than x, decrease the numerator by 1. Repeat this process of incrementing the numerator or denominator until the fraction equals exacty x. For clarity I illustrate how RFFP finds a rational fraction for 1.666...:
1/1 -> 2/1 -> 2/2 -> 3/2 -> 4/2 -> 4/3 -> 5/3 = 1.666....
Now I use a slight modification of RFFP to settle the debate of whether or not 0.999=1. Instead of targeting some real number x, we will proceed as follows: start with the fraction 1/1. If the fraction is greater than or equal to 1, increase the denominator. If the fraction is less 1, increase the numerator. This means we will converge on a number less than 1. Using the power of computers, we get the following sequence of fractions:
1/1 = 1.000000
1/2 = 0.500000
2/2 = 1.000000
2/3 = 0.666667
3/3 = 1.000000
3/4 = 0.750000
4/4 = 1.000000
...
49997/49998 = 0.999980
49998/49998 = 1.000000
49998/49999 = 0.999980
49999/49999 = 1.000000
49999/50000 = 0.999980
50000/50000 = 1.000000
...
As you can see, the number of 9s in the decimal form of the fraction increases as the iterations continue. In fact, we can mathematically prove that the number of leading 9s increases. First note that the sequence of fractions alternates between 1 and something less than one. Exclude all the 1s, so we are left with an increasing sequence where the i'th fraction is given by S_i = i/i+1. As i increases, since S_i is strictly increasing, eventually it will become greater than 9/10 but less than 1. This means it has at least 1 leading 9 in the demical form, since we can subtract 9/10 from it. Continue increasing, and eventually we can substract 9/10+9/100, meaning it has two leading nines. So the number of leading 9s increases.
