What it means? How to solved it?

Question

For any positive rational number $u$, let us agree to call the numbers $u + 1$ and $u/(u+1)$ the children of $u$. Show that every rational number is the descendant of $1$ in a unique way.

Answer 1

If $u = \frac{p}{q}$ is in its simplest terms, i.e. $\gcd(p, q) = 1$, then $u + 1 = \frac{p+q}{q}$ and $\frac{u}{u+1} = \frac{p}{q+p}$, and $\gcd(p+q,q) = \gcd(p,q+p) = \gcd(p,q) = 1$ so the children of $u$ come from adding its denominator to its numerator and vice versa, and doing so keeps it in simplest terms.

Now suppose that $u = \frac{p}{q}$ in simplest terms is a child of $w = \frac{r}{s}$, also in simplest terms. Then that means either:

• $p = r + s$ and $q = s$, i.e. $w = \frac{p - q}{q}$; or
• $p = r$ and $q = s + r$, i.e. $w = \frac{p}{q - p}$.

That means that every positive rational other than 1 is always the child of exactly one other positive rational, because you can only pick one of the two choices for $w$ and the other will give you something negative. So for $u \in \mathbb{Q} \setminus \{1\}$, the function $P(u)$ that returns the "parent" of $u$ is well-defined.

Then proving that every positive rational number is a descendant of 1 via a unique path is equivalent to proving that applying $P$ to $u$ some number of times will always get you to 1, which you can relate directly to the Euclidean algorithm.

Answer 2

Note that all natural numbers are trivially descendants.

We will prove by induction.

Base case: $u = 1$ is a descendant.

Inductive step: Suppose $u = \frac{a}{b}$ is a decendant for $ \forall a \in \aleph$, $b \leq n \in \aleph$

Proof: Let $a+b = n+1$

Hence $\frac{a}{b}$ will have descendants $\frac{a}{a + b} = \frac{a}{n+1}$

It follows that $\frac{a}{n+1} + k = \frac{a + k(n+1)}{n+1} $ for $\forall k \in \aleph$ are also descendants. $a$ can represent any number between $1$ and $n \ $ hence $u = \frac{a}{b}$ is a decendant for $ \forall a \in \aleph$, $b \leq n +1$

Hence by induction all rational numbers are descendants of 1.