Thursday, May 26, 2011

Fractions From Digits

This week marked my baptism by fire into the twitter world. It was not long until I was neck deep in tweets, favorites, re-tweets, and followers. The eternal nerd awoke inside me when I was confronted with my first NCTM "Problem of the Day". A simple, yet dangerously deep, question was posed. Wanting to cement my reputation as a responsible twit, I sat down and began to tinker with the theory.

The question was as follows:

How many different fractions can you write using only the digits 1,2,3 & 4?
Be sure to include fractions greater than 1.

Immediately, I changed the word "fractions" into "rational numbers". That way there would be no debate whether a number with a denominator of 1 is a fraction. As I began to experiment with the obvious nature of the problem, certain problem solving strategies emerged. I began to list observations and questions. Every digit, when placed over itself, will create a fraction equal to 1/1. I then became concerned with the reducibility of the rationals involved. What operations (+,-,x,/) are allowed on the digits? Can we repeat digits? It is not hard to see that the problem was becoming very large, very fast.

I decided that ordinary operations were not possible, because there was no way to know if a multiplication of two numbers in the set X={1,2,3,4} would yield a number whose digits were also in X. I then began to explore any operations that I could think of that may be acceptable on this very unique set. After addition, subtraction, multiplication, division, and exponentiation failed, I decided to take a step outside of the box. I decided to show that there are an infinite amount of rational numbers that can be created with the digits by "smushing" digits together.

I called this new operation "composition". If you compose 1/2 and 3/4 you simply get 13/24. This way I was ensured that every subsequent "composed" fraction would have only digits in the original set X. From there I created a list of 11 rationals that had a single digit in the numerator and denominator. (1/1, 2/1, 3/1, 4/1, 1/2, 1/3, 1/4, 2/3, 3/2, 3/4, 4/3). I used the fact that these 11 were unique to begin the proof of this new infinite set of numbers.

From this point on the "@" symbol will represent composition. I now had the fact that:

a/b @ c/d = (10a + c)/(10b + d)

The first thing to be noticed that both a/b and c/d are in lowest terms and unique as rationals. So the newly composed fraction is also in lowest terms. The new fraction would be reducible if a common factor was available from all 4 constituents (a,b,c,and d). If such were the case, c and d would have a common factor, and wouldn't have been in lowest terms to begin with. This contradicts the construction of the new fraction.

The second thing to notice is the uniqueness of the newly composed fraction. We know no a/b = c/d = e/f. Let's assume that for some reason, we find two composed fractions where:

a/b @ c/d = a/b @ e/f


(10a + c)/(10b + d) = (10a + e)/(10b + d)

so breaking apart the fractions we get:

10a/10b + c/d = 10a/10b + e/f

From here it is easier to see that this can only be true if c/d = e/f. We know this is not the case. So from the original 11 fractions, we can now create a set of two-digit fractions that are all unique and irreducible. This list continues to grow when we consider "composing" these two-digit numbers together to get 4-digit ones that follow the same rules. Only now the 2-digit numbers p, q, r, and s follow the form:

p/q @ r/s = (100p + r)/(100q + s)

In general, let 'x' be the number of digits of {p,q,r,s}, then:

p/q @ r/s = [(10^x)p + q]/[(10^x)r + s]

This pattern continues to create an infinite set of rational numbers from the digits 1,2,3, and 4. Of course, this list is not exhaustive--like most lists of infinite length. In fact, using this method, every fraction must have 2^n digits, where 'n' is a natural number.

Today's problem has been yet another example of how an innocent problem, can lead in various directions. How many possible 4 digit combinations are possible in this set of numbers? 8-digit? Such are problems for another day.


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