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Given a function $f: \mathbf{N}_0 \to \mathbf{N}_0$, defined $$ f(x) = \begin{cases} x+3 & \text{if } x \in \mathbf{N}_{\text{even}} \\ x-1 & \text{if } x \in \mathbf{N}_{\text{odd}} \end{cases} $$

How can I find the image $f( $$\mathbf{N}_{\text{even}}$ )?

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  • $\begingroup$ I am guessing $\mathbb{N}_0$ refers to the natural numbers including $0.$ Also, have you tried anything? Do you have any ideas of where to start? $\endgroup$ – Aurel Jan 13 '18 at 17:17
  • $\begingroup$ @Aurel Yes, it refers to that. I think the image is all the even natural numbers but I could be wrong. $\endgroup$ – Marco Carta Jan 13 '18 at 17:17
  • $\begingroup$ A hint: think about what the function does to even numbers. For example, what is $f(4), f(6), f(8), etc.$ $\endgroup$ – Aurel Jan 13 '18 at 17:19
  • $\begingroup$ @Aurel the result is an odd number. $\endgroup$ – Marco Carta Jan 13 '18 at 17:21
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Note: Every even number $n$ can be written as

$n=2k, k=0,1,2 ......$.

Hence:

$f(2k) = 2k+3, k=0,1,2,.....$.

Finally:

$f(\mathbb{N_{even}}) = $

{$2k+3| k \in \mathbb{N}$} $= ${$3,5,7,........$}.

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Well, $\ f(\mathbb{N}_{even})\ $ is simply just $\ f|_{\mathbb{N}_{even}}:\mathbb{N}_{even} \rightarrow \mathbb{N}_{odd}\ $ defined by $\ f(x) = x+3\ $. So it follows that, $\ f(\mathbb{N}_{even})=\{ 0+3, 2+3, 4+3, \dots \} \ $ so then the image is simply $\ \{3,5,7,\dots \}$.

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