Is the function $\theta(a,b) = a-2ab+b$ a bijection from $\{0,1\}\times\mathbb{N}$ to $\mathbb{Z}$? Consider the function $\theta:\{0,1\}\times\mathbb{N}\rightarrow\mathbb{Z}$ defined as $\theta(a,b) = a-2ab+b$. Is this function bijective?
For injective, I tried doing the contrapositive by supposing $\theta(a,b)=\theta(c,d)$, then $a-2ab+b=c-2cd+d$, but I have no idea where to go from there. I tried solving for a and b separately and plugging it back in, but that just turned into a huge algebraic mess.
I haven't figured what I'm going to do for surjective yet.
 A: Formulas are nice, but let's find out what's really going on: $\theta(0,b)=b\,$ and $\theta(1,b)=1-b$. Much clearer. 
Could $x=1-y\,$ where $x$ and $y$ are natural numbers? Depends on what one means by natural number. It can certainly happen if we allow $0$ to be a natural number.  I am inclined to allow that, out of loyalty to logic, where it is a standard convention.  But I am in a minority. Certainly it cannot happen if $x$ and $y$ are positive.  
Finally, for surjectivity (love that word) is every integer of the form $x$ or $1-y$ where $x$ and $y$ are positive integers? Sure.
A: Hint $\ $ Glue the bijections $\rm\ b = f_{\,0}(b):\Bbb N\to \Bbb N\ $ and $\rm\ 1\!-\!b = f_{\!\ 1}(b): \Bbb N\to\, {-}\Bbb N\cup \{0\}\ $ as below:
If $\rm\:S_0,\,S_1\:$ are disjoint subsets of $\,\Bbb Z\,$ and $\rm\:f_{\,i}(b) :\Bbb N\to S_i\,$  are bijections for $\rm\,i\in\{0,1\}\,$ then $\rm\:f(a,b):\{0,1\}\times\Bbb N \to S_0\!\cup S_1\,$ is a bijection for  $\rm\ f(a,b)\ =\ a\,f_{\!\ 1}(b) + (1\!-\!a)\,f_{\,0}(b)$
