Let $\mathbb{Z}_{(2)} = \{ \frac{a}{b} \in \mathbb{Q} \vert a \in \mathbb{Z}, b \not\in 2\mathbb{Z}\}$ Let $\mathbb{Z}_{(2)} = \{ \frac{a}{b} \in \mathbb{Q} \vert a \in \mathbb{Z}, b \not\in 2\mathbb{Z}\}$, I want to answer the following questions. 
1) Is $\mathbb{Z}_{(2)}$ a subring of $\mathbb{Q}$ that contains $\mathbb{Z}$
2) Determine a necessary and sufficient condition on $a \in \mathbb{Z}$ such that it is invertible in $\mathbb{Z}_{(2)}$.
3) Is $\mathbb{Z}_{(2)}$ a subfield of $\mathbb{Q}?$
4) Consider $$ I_2 = \{ 2\frac{a}{b} \in \mathbb{Q} \vert a \in \mathbb{Z}, b \not\in 2\mathbb{Z} \}$$
Prove that $I_2$ is a proper ideal of $\mathbb{Z}_{(2)}$.
The following is what I tried. 
1) $\mathbb{Z} \subseteq \mathbb{Z}_{(2)}$ because I can just take b = 1 and whatever a, b = -1 and whatever a. 
To prove that it is a subring I prove the 3 following conditions:
a) $\mathbb{Z}_{(2)} \neq \emptyset$
b)$\forall x,y \in \mathbb{Z}_{(2)}$ $x +(-y) \in \mathbb{Z}_{(2)}$
c) $\forall x,y \in \mathbb{Z}_{(2)}$ $x*y \in \mathbb{Z}_{(2)}$
These are trivial to prove, and 1) is done. 
2)Now I need to find some conditions on $a$ to make it invertible. 
  I need to find a' such that a * a' = 1. Let $a' = \frac{c}{b}$, then I have that $a = \frac{b}{c}$ with $c \not\in 2\mathbb{Z}$ and $c \not\in 2\mathbb{Z}$. Is this ok? I'm not really sure. 
3) To prove that $\mathbb{Z}_{(2)}$ is a subfield of $\mathbb{Q}$, I would need to prove that: 
a) $\mathbb{Z}_{(2)} \neq \emptyset$
b) $\forall x,y \in \mathbb{Z}_{(2)}$, $x - y \in \mathbb{Z}_{(2)}$
c) $\forall x\neq 0, y \neq 0\in \mathbb{Z}_{(2)}$, $x*y^{-1} \in \mathbb{Z}_{(2)}$
I do not think that c) is true because not all elements of $\mathbb{Z}_{(2)}$ have an inverse, this idea is based on answer 2).
4) To prove that $I_2$ is an ideal, I need to prove that:
$$x,y \in I_2, z \in \mathbb{Z}_{(2)}\rightarrow x-y, zx, xz \in I_2$$
This is pretty easy to prove so I am done. 
Is everything I did correct?
 A: You have a lot of the right ideas but you definitely need the details.
(1) To show that $\mathbb{Z}_{(2)}$ is a subring, we need to show it's nonempty, closed under addition, and closed under multiplication. 
The identity element $1 = \frac{1}{1} \in \mathbb{Z}_{(2)}$ where $1 \notin 2\mathbb{Z}$. So this set is nonempty. Moreover, the subring generated by $1$ is $\mathbb{Z}$ since $\mathbb{Q}$ has characteristic 0.
Since $\mathbb{Z}_{(2)}$ has conditions on its fractions, we need to show that those conditions hold together when we add and multiply elements from this set.
So for two elements $\frac{a}{2k+1}$ and $\frac{a'}{2j+1}$ in $\mathbb{Z}_{(2)}$, 
$$\frac{a}{2k+1}+\frac{a'}{2j+1} = \frac{a(2j+1) + a'(2k+1)}{(2k+1)(2j+1)}$$
Since $\mathbb{Z}$ is a ring, $a(2j+1) + a'(2k+1) \in \mathbb{Z}$. Now (2k+1)(2j+1) = 2(2kj+k+j)+1 is an odd number, i.e., not an element of $2\mathbb{Z}$. Hence the sum is an element of $\mathbb{Z}_{(2)}$.
Multiplicatively, $$\frac{a}{2k+1} \cdot \frac{a'}{2j+1} = \frac{aa'}{2(2kj+k+j)+1} \in \mathbb{Z}_{(2)}$$
You can avoid the details of the arithmetic in the denominator if you can prove that the complement of any prime ideal is a multiplicatively closed set. I'm not sure how much you know about localization and prime ideals though. 
(2) For an integer $a$, $a^{-1} = \frac{1}{a} \in \mathbb{Q}$, but $\mathbb{Z}_{(2)}$ has the condition that denominators are odd. Hence, the only integers that are invertible in $\mathbb{Z}_{(2)}$ are the integers that are odd. Furthermore, if $a,b \notin 2\mathbb{Z}$, then $\frac{a}{b}, \frac{b}{a} \in \mathbb{Z}_{(2)}$ and $\frac{a}{b} \cdot \frac{b}{a} = 1$.
(3) As you said question 2 gives us some intuition that $\mathbb{Z}_{(2)}$ may not be a subfield of $\mathbb{Q}$. The element $\frac{2k}{b} \in \mathbb{Z}_{(2)}$ is not invertible since the numerator is an element of $2 \mathbb{Z}$. 
(4) To prove $I_2$ is an ideal of $\mathbb{Z}_{(2)}$, we need to show it is a subring and that it absorbs rings elements, i.e., for all $r \in \mathbb{Z}_{(2)}$, $rI_2 \subseteq I_2$ (or $\mathbb{Z}_{(2)} I_2 \subseteq I_2$). 
Computations similar to (1) show $I_2$ is a subring.
For $r \in \mathbb{Z}_{(2)}$ and $i \in I_2$, let $r = \frac{a}{b}$ and $i = 2 \frac{a'}{b'}$. Then
$$ri = \frac{2aa'}{bb'} = 2 \frac{aa'}{bb'} \in I_2$$ recognizing that $aa' \in \mathbb{Z}$ and $bb'$ is odd, i.e.,  $\notin 2 \mathbb{Z}$.
Hope that helps.
