Problem
I've got this function:
$f(x,y)=(6-3\cdot x)\cdot(y+2)$, with $(x,y)\in\mathbb{N}^2$
Now I have to find $g=\mu f$.
Proposed solution
My solution was to find the smallest $n\in\mathbb{N}$ to find $f(n,y)=0$ and show that $\forall 0\leq m\leq n : f(m,y)$ is defined:
$f(0,y)=(6-0)(y+2) = 6y+12$, defined $\forall y\in\mathbb{N}$
$f(1,y)=(6-3)(y+2) = 3y+6$, defined $\forall y\in\mathbb{N}$
$f(2,y)=(6-6)(y+2) = 0$, defined $\forall y\in\mathbb{N}$
So I've found $g=\mu f=2$.
Questions
- Is above solution correct?
- Is it always the first parameter that becomes $n$?
- If $f(x,y)=(6-3\cdot x)\cdot(y-2)$ and $f(0,y) = (6-0)(y-2) = 6y-12$, wouldn't be $f(0,y)=0$, if $y=2$ and therefore my $n=0$ for this $y$, but for other $y$, my $n$ would be different?
