Jonathan wrote:

> **Puzzle 141.** Define \\(f : \mathbb{Z} \to \mathbb{Z}\\) by \\(f(x) = 2x\\). Then both \\(g(x) = \lfloor \frac{x}{2} \rfloor\\) and \\(h(x) = \lfloor \frac{x + 1}{2}\rfloor\\) are left inverses of \\(f\\).

Yes, that's nice! These functions \\(g\\) and \\(h\\) take different values on odd integers, but they agree on even integers - they just halve any even integer - so they both provide a left inverse to the function \\(f\\), which double integers.

You could in fact have chosen \\(g\\) to map the odd integers to whatever you wanted!

> **Puzzle 141.** Define \\(f : \mathbb{Z} \to \mathbb{Z}\\) by \\(f(x) = 2x\\). Then both \\(g(x) = \lfloor \frac{x}{2} \rfloor\\) and \\(h(x) = \lfloor \frac{x + 1}{2}\rfloor\\) are left inverses of \\(f\\).

Yes, that's nice! These functions \\(g\\) and \\(h\\) take different values on odd integers, but they agree on even integers - they just halve any even integer - so they both provide a left inverse to the function \\(f\\), which double integers.

You could in fact have chosen \\(g\\) to map the odd integers to whatever you wanted!