Suppose all you had to work with was the set of numbers $\{n \in \mathbb{N}\}$ and $\{1/n\}$. Could you construct all other positive rational numbers from this set under addition using the finite number of elements in the basis of $\{1,2,3,...\}$ and $\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4}...\}$?
Asked
Active
Viewed 96 times
1
-
1Yes, it is possible. If the number is m/n in lowest terms, multiply 1/n by m. – DavidW Oct 25 '20 at 02:07
-
2Maybe you should specify what you mean by "kind of basis." For example, the rational number $p/q$ is $1/q$ times the integer $p$; is this valid in your interpretation of "basis"? – angryavian Oct 25 '20 at 02:07
-
What construction rules are you allowing, JRRs? – Gerry Myerson Oct 25 '20 at 02:35
-
Through addition only, not multiplication. – JRRs Oct 25 '20 at 03:48
-
1Of course. $r =\frac ab = \underbrace{\frac 1b + \frac 1b + ..... + \frac 1b}_{a\text{ times}}$. – fleablood Oct 25 '20 at 05:25
-
"Through addition only, not multiplication" Multiplication (by natural numbers) is addition a finite number of times. – fleablood Oct 25 '20 at 05:26
1 Answers
1
If you allow repeated use of a number then you can construct a positive rational number $\frac{m}{n}$ as the sum
$$\underbrace{ \frac{1}{n} + \dots + \frac{1}{n} }_{m \text{ times}}.$$
So the interesting question is what happens if you are only allowed to use the available numbers at most once each. In that case it is still possible, and in fact you don't need the positive integers, you can use just the unit fractions $\{ \frac{1}{n} \}$. This is the classical result about the existence of Egyptian fraction representations. A simple way to prove that these exist is to start from the above sum and repeatedly apply the identity
$$\frac{1}{k} = \frac{1}{k+1} + \frac{1}{k(k+1)}$$
until all the unit fractions that appear are distinct, although this gives many terms with very large denominators and I think a greedy method (subtracting the largest possible unit fraction at each step) usually does better.
(Sketch of a proof that this works: initially there are $m$ identical unit fractions in the sum. After applying the above identity to $m-1$ of them, there are $2$ sets of $m-1$ identical fractions in the sum. After applying the identity to $m-2$ of each of these two sets there are $4 = 2^2$ sets of $m-2$ identical fractions... at some point in this process there may be "accidental" collisions but you can just keep going until those disappear too.)
Qiaochu Yuan
- 419,620