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I have that A is a set of $2k^2$ so it equals $\{2,8,18,32,50...\}$ How do you Construct two distinct bijections $f, g : \mathbb{Z}^{+} \to A$.

I was able to get $f(x)=2x^2$ what would $g(x)$ be?

Thank You

ml0105
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Yamen Alghrer
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  • If there are no restrictions on the map, you could swap two element. So sent 1 to 8 and 2 to 2 and leave the rest equal to $f$. – Marc Apr 16 '14 at 20:23
  • How about $g(1)=8$, $g(2)=2$, and $g(x)=2x^2$ for $x\geq 3$. – DKal Apr 16 '14 at 20:25
  • That would work but I do not think that I would be able to do that. I need a whole new function of some sort that is also a bijection. Thank you though – Yamen Alghrer Apr 16 '14 at 20:42
  • Any non-identity bijection $p:\mathbb Z^+\to\mathbb Z^+$ lets you choose $g(x)=f(p(x))$. It's not clear what you mean by "whole new function," but every bijection from $\mathbb Z^+$ and $A$ will be of this form. – Thomas Andrews Apr 16 '14 at 20:42
  • @ThomasAndrews I don't understand what you mean exactly sorry

    By whole function I mean something like g(x)=something that ends up being a bijection with a different map than f(x)

     – Yamen Alghrer Apr 16 '14 at 20:43
  • Which part don't you understand? – Thomas Andrews Apr 16 '14 at 20:44
  • @ThomasAndrews By your inference, what would be a valid g(x)? – Yamen Alghrer Apr 16 '14 at 20:46

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As written DKal's comment is an answer to the question. It is a function, as it assigns a unique element of the range to each element of the domain. It is a different function, as it disagrees with $f(x)$ for $x \in \{1,2\}$ Many posters (and it seems you) believe that a function has to have a "simple formula". This is simply not true. In particular, there is resistance to definition by cases. We could show DKal's example as $$g(x)=\begin {cases} 8&x=1\\1&x=2\\2x^2& \text{otherwise} \end {cases}$$ To follow up on Thomas Andrews' suggestion define $$p(x)=\begin {cases} x+1 & x \text { odd}\\ x-1 & x\text { even} \end {cases}$$ Then $x \leftrightarrow f(p(x))$ is a different bijection than $x \leftrightarrow f(x)$

Ross Millikan
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