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After defining the bijection $$ F: \mathbb{N}\to \mathbb{Z}$$ using floor function, and by defining bijections:

$$\mathbb{N} \to \mathbb{N} \setminus \{ 0 \}\\ \mathbb{Q} \to \mathbb{Q} \setminus \{ 0 \} $$

define bijection:

$$\mathbb{Z} \to \mathbb{Q} \setminus \{ 0 \}$$.

After a few attempts I got first one $F: \Bbb N \to \Bbb Z$ , $F(x) = (-1)^x \lfloor(x/2)\rfloor$ but I'm stuck with other ones. I'm new into that course at university - if I made some mistakes - I'm sorry.

MJD
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user3231387
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    $\mathbb N\to\mathbb N\setminus{0}:x\mapsto x+1$ is a bijection. Also, doesn't $F$ send $0$ and $1$ both to $0$? – Kenta S Mar 14 '21 at 13:27
  • Forget solving the first three for the time being. Let's supposed you paid some one and they came up with bijections $F:N\to Z$ and $G:N \to N\setminus {0}$ and $H:Q\to Q\setminus {0}$. And you were left to fine a bijection $J:Z\to Q\setminus{0}$... could you? – fleablood Mar 15 '21 at 02:00

1 Answers1

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  1. $F : \mathbb N \to \mathbb Z$ defined as $F(n)=(-1)^n \lfloor {n+1 \over 2} \rfloor$.
  2. $G : \mathbb N \to \mathbb N \backslash \{0\}$ defined as $G(n)=n+1$.
  3. $H : \mathbb Q \to \mathbb Q \backslash \{0\}$ defined as $H(n)=n+1$ for $n \in \mathbb N$ and identity function otherwise.
  4. You can find very nice detailed answers here.
MathIsNice1729
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