Prove $ (\mathbb{R} − \mathbb{Z}) × \mathbb{N}=( \mathbb{R} × \mathbb{N} ) − ( \mathbb{Z} × \mathbb{N} ) $ The statement $ (\mathbb{R} − \mathbb{Z}) × \mathbb{N}=( \mathbb{R} × \mathbb{N} ) − ( \mathbb{Z} × \mathbb{N} ) $ is true, or false?
I have the following assumptions:


*

*$ A - B = \{ x : x \in A, x \notin B \} $ 

*$ A \times B = \{ (a,b): a \in A, b \in B \} $

*$ \mathbb{N} \subset \mathbb{Z} \subset \mathbb{R} $

My attempt:
Let 
$$
 C = (\mathbb{R} - \mathbb{Z}) = \{ c: c \in \mathbb{R}, c\notin \mathbb{Z} \} 
$$ 
then the left part of initial equation is equal to
$$
C \times \mathbb{N} = \{ (c,n): c \in \mathbb{R}, c \notin \mathbb{Z}, n \in \mathbb{N} \}
$$
Lets expand brackets in the right part of equation.
$$
D = \mathbb{R} \times \mathbb{N} = \{ (d, n): d \in \mathbb{R}, n \in \mathbb{N} \}
$$
$$
E = \mathbb{Z} \times \mathbb{N} = \{ (z, n): z \in \mathbb{Z}, n \in \mathbb{N} \}
$$
then
$$
D - E = \{ (x,y): x \in \mathbb{R}, x\notin Z, y \in \mathbb{N} \} 
$$
If the last statement true I think the proof is complete. But I am not sure.
Is this valid?
 A: Let me stress that the way you show set equality is almost always by the following method:
Let $A$ and $B$ be sets. Let $a \in A$, and $b \in B$. If you can show that $a \in B$ and $b \in A$, then $A = B$, since $a$ and $b$ were chosen arbitrarily. With that in mind, here is how to solve your question with this method:
Let $x = (a,b) \in (\mathbb{R}-\mathbb{Z})\times \mathbb{N}$. Since $(\mathbb{R}-\mathbb{N}) \subset \mathbb{R}$, we have that $a \in \mathbb{R}$ and $b \in \mathbb{N}$, so $x \in \mathbb{R} \times \mathbb{N}$.Since $a \notin \mathbb{Z}$, we have that $x = (a,b) \notin (\mathbb{Z} \times\mathbb{N})$, and we conclude that $x \in (\mathbb{R} \times \mathbb{N}) - (\mathbb{Z} \times \mathbb{N})$. 
For the other way around, let $x = (a,b) \in   (\mathbb{R} \times \mathbb{N}) - (\mathbb{Z} \times \mathbb{N})$, and do similar procedures as above.
A: To prove $(A\setminus B)\times C=(A\times C)\setminus(B\times C)$, note both sides have only ordered pairs as elements, so we just need the following: $$(x,\,y)\in(A\setminus B)\times C\iff x\in A\setminus B\land y\in C\iff x\in A\land x\notin B\land y\in C\\\iff (x\in A\land y\in C)\land\neg(x\in B\land y\in C)\\\iff((x,\,y)\in A\times C)\land((x,\,y)\notin B\times C)\iff(x,\,y)\in(A\times C)\setminus(B\times C).$$
