1) f3: Z × Z → Z × Z. f3(x, y) = (x + 1, 2y)
2) f4: Z+ × Z+ → Z+. f4(x, y) = 2x + y − 1
I understand the conditions for a function to be onto and one-to-one but in this case I'm having trouble understanding how to read the function since it is given x,y would I just find an x value in the cartesian product Z (same with y) and plug it in to determine if the conditions hold?
I'd check if $f(x,y)= f(w,z) \implies (x,y) = (w,z)$ or if it possible for $(x,y)= (w,z)$. If the implication follows it's injective other wise it's not.
And to show if it is surjective I'd try $f(x,y)=(w,z)$ and to solve for $x,y$. If $x,y$ always have a solution it's surjective. If not, the its not.
Example:
$f_3(x,y) = (x+1, 2y)$ is that injective?
Suppose $(x+1, 2y) = (w+1, 2z)$. Then $x+1 = w+1$ and subtracting $1$ from both side $x = w$ and $2y = 2z$ so dividing both sides by $2$ give $y = z$. SO yes: if $(x+1,2y) =(w+1, 2z)$ then $(x,y) = (w,z)$ and it is injective.
Is it surjective.
Suppose $(x + 1, 2y) = (w,z)$. The $x+1 =w$ and $x = w-1$. And $2y = z$ and $y =\frac z 2$. Is that always possible. Well, no. If $z$ is odd then $y=\frac z2$ is not an integer. So it is impossible to have $f_3(x,y) = (w,z)$ if $z$ is odd. So it is not surjective.