$1-1$ functions
- Tolaso J Kos
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$1-1$ functions
Prove that every function \( f:\mathbb{Q} \rightarrow \mathbb{Q} \) can be written as the sum of two \( 1-1 \) functions \( g, h: \mathbb{Q} \rightarrow \mathbb{Q} \).
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Re: $1-1$ functions
Let \(q_1,q_2,\ldots\) be an enumeration of the rationals. Suppose we have already defined \(g,h\) on \(q_1,\ldots,q_k\) such that their sum is equal to \(f\) on those points and they are 1 to 1 so far.
When defining \(g\) and \(h\) on \(q_{k+1}\) it is enough to pick any \(q \in \mathbb{Q}\) such that \(q \notin \{g(q_1),\ldots,g(q_k)\}\) and \(f(q) - q \notin \{h(q_1),\ldots,h(q_k)\}\). We then take \(g(q_{k+1}) = q\) and \(h(q_{k+1}) = f(q)-q\). This is possible as there are only \(2k\) bad choices for \(q\).
When defining \(g\) and \(h\) on \(q_{k+1}\) it is enough to pick any \(q \in \mathbb{Q}\) such that \(q \notin \{g(q_1),\ldots,g(q_k)\}\) and \(f(q) - q \notin \{h(q_1),\ldots,h(q_k)\}\). We then take \(g(q_{k+1}) = q\) and \(h(q_{k+1}) = f(q)-q\). This is possible as there are only \(2k\) bad choices for \(q\).
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