The inverse function The function $g :\mathbb{Q} → \mathbb{Q}$ is deﬁned by $g(r) = 4r + 1$ for each $r \in \mathbb{Q}$.
(a) Determine $g(\mathbb{Z})$ and $g(E)$, where $E$ is the set of even integers.
(b) Determine $g^{-1}(N)$ and $g^{−1}(D)$, where $D$ is the set of odd integers.
This is the answer:

I don't get why in (b)  $g^{-1}(N) = n/4$, i think it should be $(n-1)/4$, also for  $g^{-1}(D)$
Could any one explain this to me ?
 A: The sets $\{n/4 : n\in\mathbb{Z}, n\geq 0\}$ and $\{(n-1)/4 : n\in\mathbb{N}\}$ are the same set. Every element added to the first one will be added to the second as well. 
You can see some of what's going on by looking at a finite subset. Compare 
$\{n/4 : n\in\{1,2,\ldots,5\}\}=\{1, .25, .5, .75, 1.25\}$ 
$\{(n-1)/4 : n\in\{1,2,\ldots,5\}\}=\{0, .25, .5, .75, 1\}$
Because of the $(n-1)$ the second version runs behind the first by one element. However, if you compare
$\{n/4 : n\in\{0,2,\ldots,4\}\}=\{0, .25, .5, .75, 1\}$ 
$\{(n-1)/4 : n\in\{1,2,\ldots,5\}\}=\{0, .25, .5, .75, 1\}$
You get roughly what is going on in your answer sheet. You have one version which starts counting at zero and then divides the integer by four. The other starts counting at one, subtracts one (which will send one to zero, two to one, etc.), and then divides. 
(One of the easier ways to see they are equivalent might be by noticing that $\{n\in\mathbb{Z}, n\geq0\}$ is the same as $\{n-1 : n\in\mathbb{N}$). 
In the case of $g^{-1}(D)$ the fact that odd integers are of the form $2k+1$ means that, since $g^{-1}(r)=(r-1)/4$, $g^{-1}(2k+1)=(2k+1-1)/4=2k/4=k/2$. This is the same as the set $\{(n-1)/4 : n\in D\}$, but not the same as $\{(n-1)/4 : n\in \mathbb{Z}\}$. (Why?)
A: you are right, either 


*

*$(n-1)/4$ and $n\geq0$

*or $n/4$ and $n\geq-1$ (as $g(-1/4) = 0 \in N$)


for g-1(D), both sets (yours, and the one you mentioned) are equal
