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Okay so I know that I asked this already but I want to ask how much progress I have made. so the question is

Construct a bijection between $[1,2]$ and $[3,5)$

So I have: \begin{equation} f(x) = \left \{ \begin{array}{ll} 3 + 2^{1-n} & \textrm{ if } x = 1+2^{1-n} \textrm{ for } n \in \mathbb{N} \\ 2x+1 & \textrm{ if } x \neq 1+2^{1-n} \end{array} \right. \end{equation}

Sorry about the formatting, but is this correct?

Edit: the linked post is mine. I'm just asking since I didn't get a really satisfactory answer if the solution I have is going towards the right direction.

Edit2: changing the equation to what improvements I'm getting, still any feedback is helpful. Thanks!

rickster38
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  • So I think maybe it should be \begin{equation} f(x) = \left { \begin{array}{ll} 3 + 2^{1-n} & \textrm{ if } x = 3+2^{2-n} \textrm{ for } n \in \mathbb{N} \ 2n+1 & \textrm{ if } x \neq 3+2^{2-n} \end{array} \right. \end{equation} because so I want the first n=1 to equal to 5 to replace it to the next term and the next term. am i on the right track? – rickster38 Nov 24 '15 at 00:28
  • Umm, what's the $n$ in case 2 referring to? – Alan Nov 24 '15 at 00:31
  • @Alan n is all natural number – rickster38 Nov 24 '15 at 00:34
  • @RodrigoZepeda Zepeda, that questoin was submited by me actually, i"m just asking since I come a little farther and I want to know if I am on the right tracks. – rickster38 Nov 24 '15 at 00:35
  • actually whoops I think the second one equation should be 2x+1, my bad – rickster38 Nov 24 '15 at 00:44
  • \begin{equation} f(x) = \left { \begin{array}{ll} 3 + 2^{-n} & \textrm{ if } x = 3+2^{1-n} \textrm{ for } n \in \mathbb{N} \ 2x+1 & \textrm{ if } x \neq 3+2^{1-n} \end{array} \right. \end{equation} – rickster38 Nov 24 '15 at 00:44
  • I am not sure I am understanding your notation. If $x \in [1,2]$ then all $x$ fall in the second case of $f(x)$. In particular $x = 2$ maps to $x = 5$ which is not included in the set. – Rodrigo Zepeda Nov 24 '15 at 00:52
  • @joshua23 In your previous question you had an answer by user254665. I fail to see what is wrong with that answer. Could you please elaborate more? – Rodrigo Zepeda Nov 24 '15 at 00:54
  • I think you are right, i"m think that now instead it should be $x = 1+2^{1-n}$ – rickster38 Nov 24 '15 at 00:59
  • @RodrigoZepeda changed it does it make sense now? – rickster38 Nov 24 '15 at 01:11
  • @joshua23 Check the case $x = 2$ it falls in the first case with $n = 1$ and $f(x) = 3$. On the other hand, $x = 1$ falls in the second case and $f(x) = 3$. This is not a bijection either. – Rodrigo Zepeda Nov 24 '15 at 01:20
  • @RodrigoZepeda if n=1 then f(x) =4, because $3+2^{1-1} = 3+2^0=4$ – rickster38 Nov 24 '15 at 01:22
  • @joshua23 Right! Sorry about that... – Rodrigo Zepeda Nov 24 '15 at 01:24
  • @RodrigoZepeda It's okay, I think this equation work now right? – rickster38 Nov 24 '15 at 01:27
  • @joshua23 I think so... – Rodrigo Zepeda Nov 24 '15 at 01:36

1 Answers1

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The function $f$ is onto:

Case 1

Let $y \in [3,5) $ be of the form: $3 + 2^{1-n}$ for some $n \in \mathbb{N}$. Then $x = 1 + 2^{1-n}$ is such that $f(x) = y$. In particular, if $n$ is such that $3 + 2^{1-n} \in [3,5)$ then $n \geq 0$. This in turn implies that $1 \leq 1 + 2^{1-n} \leq 2$.

Case 2

Let $y \in [3,5) $ not be of the form: $3 + 2^{1-n}$ for some $n \in \mathbb{N}$. Then $x = (y-1)/2 \in [1,2]$ is such that $f(x) = y$. In particular if $y \in [3,5) $ then $(y-1)/2 \in [1,2]$.

The function $f$ is inyective:

Let $a,b \in [1,2]$ be such that $f(a) = y = f(b)$

Case 1

If $y \neq 3 + 2^{1-n}$ then $f(a) = 2a + 1 = 2b + 1 = f(b)$. It follows that $a = b$.

Case 2 If $y = 3 + 2^{1-n}$ then $f(a) = 3 + 2^{1-n_a}$ and $f(b) = 3 + 2^{1-n_b}$ where $a = 1 + 2^{1-n_a}$ and $b = 1 + 2^{1-n_b}$ respectively. Then: \begin{equation} \begin{aligned} 3 + 2^{1-n_a} = 3 + 2^{1-n_a} & \Leftrightarrow 2^{1-n_a} = 2^{1-n_a} \\ & \Leftrightarrow log_2\left(2^{1-n_a}\right) = log_2\left(2^{1-n_b}\right) \\ & \Leftrightarrow n_a = n_b \\ & \Leftrightarrow a = b. \end{aligned} \end{equation}

Rodrigo Zepeda
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