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I tried to look at the cases and find a function, but I could not find a bijective function. I know that we should check the cases when x is a positive number and when x is negative. Can you help me to find one?

  • An idea: map the positive integers to twice their value. Any idea how to map the non-positive integers...? – DonAntonio Nov 20 '17 at 22:35
  • map them to -x, so they will become positive. But still this function is not bijective –  Nov 20 '17 at 22:37
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    Don't map them to just $-x$, map them to the odd integers. – Rellek Nov 20 '17 at 22:38
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    Have you heard the story of the Hilbert's Grand Hotel? https://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel What does the desk man do if the hotel is full and a bus load with an infinite number of tourists arrive? – Doug M Nov 20 '17 at 22:39
  • @BellePepper Nop. You've already covered the even naturals, so what is there left? The odd ones...so how can you map the negative integers (and zero) onto them? – DonAntonio Nov 20 '17 at 22:42
  • @DonAntonio can I map neagtive integers to -(x+1)/2? –  Nov 20 '17 at 22:43
  • @BellePepper That's close...but not quite: what about $;x=-4;$ , for example? That'd go to $;-\frac{-4+1}2=\frac32\notin\Bbb N;$ ... yet you're really close... – DonAntonio Nov 20 '17 at 22:46

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Consider the following: $$f(n) = \begin{cases} 2n, \quad n\geq0 \\ -2n-1, \quad n <0 \\ \end{cases}$$

Rellek
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$$ \begin{array}{rrrrrrrrrrrrrrrrrrrrrrrrrrrrrr} \mathbb Z: & 0 & & & 1 & -1 & & & 2 & -2 & & & 3 & -3 & & & 4 & -4 & & & 5 & -5 & & & \cdots \\ & \updownarrow & & & \updownarrow & \updownarrow & & & \updownarrow & \updownarrow & & & \updownarrow & \updownarrow & & & \updownarrow & \updownarrow & & & \updownarrow & \updownarrow & & & \cdots \\ \mathbb N: & 0 & & & 1 & 2 & & & 3 & 4 & & & 5 & 6 & & & 7 & 8 & & & 9 & 10 & & & \cdots \end{array} $$

Michael Hardy
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