Why doesn't this test work on this isomorphism? The map $\phi: \mathbb{Q} \to \mathbb{Q}$ defined by $\phi (x) = 3x-1$ for $x \in \mathbb{Q}$ is one to one and onto $\mathbb{Q}$, Give the definition of a binary operation * on $\mathbb{Z}$ such that $\phi$ is an isomorphism mapping 
a) <$\mathbb{Q}$,+> onto <$\mathbb{Q}$, * >
b) <$\mathbb{Q}$,*> onto <$\mathbb{Q}$, + >
The solution said 

Question
For (a), I tried doing it like this
$\phi(x+y) = 3(x+y) -1=3x + 3y - 1$
$\phi(x) *\phi(y)=(3x-1)*(3y-1)$
Setting them equal to each other, I get $(3x-1)*(3y-1)= 3x + 3y - 1 \iff (3x)*(3y) = 3x + 3y + 1 \iff x * y = x + y + 1/3$. It doesn't match up...
The method worked for (b). Why doesn't it work on (a)?
 A: For (a), $\phi(x + y) = 3(x+y) - 1 = 3x + 3y - 1$, and $\phi(x)\ast\phi(y) = (3x-1)\ast(3y-1)$ so $(3x-1)\ast(3y-1) = 3x + 3y - 1$, which is what you have written. 
In order to determine $(3x)\ast(3y)$ from $(3x-1)\ast(3y-1) = 3x + 3y -1$, we need to write $3x$ in the form $3x'-1$ and $3y$ in the form $3y'-1$. We can do this as follows:
$$(3x)\ast(3y) = \left(3\left(x+\frac{1}{3}\right)-1\right)\ast\left(3\left(y+\frac{1}{3}\right)-1\right) = 3x + 3y + 1,$$
which you have also written. The last step is to determine $x\ast y$ from $(3x)\ast(3y) = 3x+3y+1$, so we need to write $x$ in the form $3x'$ and $y$ in the form $3y'$. We can achieve this using $x = 3\left(\frac{1}{3}x\right)$ and $y = 3\left(\frac{1}{3}y\right)$. Then we have:
$$x\ast y = \left(3\left(\frac{1}{3}x\right)\right)\ast \left(3\left(\frac{1}{3}y\right)\right) = 3\left(\frac{1}{3}x\right) + 3\left(\frac{1}{3}y\right) + 1 = x + y + 1.$$

This really is the same method as what the solution uses, but in my opinion, this is more direct. We want to define a new binary operation $\ast$ on $\mathbb{Q}$, so we need to determine what $x\ast y$ is. In order to do this, we need to write $x$ in the form $3x'-1 = \phi(x')$ and similarly for $y$. Then we can use the fact that we want $\phi$ to be an isomorphism.
