Define a binary operation on the rationals so that the map defined by is an isomorphism between sets with binary operations. Prove that is an isomorphism.

THIS IS WHAT I HAVE TO DO

"We now give an outline to show two binary structures (S,*) and (S,*') are isomorphic.

To prove it is an isomorphism i have to:

1) Define the function $\displaystyle \phi$ that gives the isomorphism of S with S'. Now this means that we have to describe, in some fashion, what $\displaystyle \phi (s)$ is for every s $\displaystyle \in$ S.

2) Show that $\displaystyle \phi$ is a one-to-one function. That is, suppose that $\displaystyle \phi (x) = \phi (y)$ in S' and deduce from this that x = y in S.

3) Show that $\displaystyle \phi$ is onto S'. that is, suppose that s' $\displaystyle \in$ S' is given and show that there does exist s $\displaystyle \in$ S such that $\displaystyle \phi(s) = s'$

4) Show that $\displaystyle \phi(x*y) = \phi(x) *' \phi(y)$ for all x,y $\displaystyle \in$ S. This is just a question of computation. Compute both sides of the equation and see whether they are the same."

----------------------------------------------------------------

I have defined $\displaystyle u \bullet v$ = u + v + $\displaystyle \frac{1}{7}$

1) The function has already been defined $\displaystyle \phi(x) = 7x + 1$

2) Lets Suppose $\displaystyle \phi(x) = \phi(y)$

then 7x + 1 = 7y + 1

7x = 7y

x = y

Therefore it is one-to-one

3) I dont even know were to start

4) $\displaystyle \phi(u \bullet v) = \phi(u) + \phi(v)$

7(u $\displaystyle \bullet$ v) + 1 = 7u + 1 + 7v + 1

7(u $\displaystyle \bullet$ v)= 7(u + v + $\displaystyle \frac{1}{7}$)

u $\displaystyle \bullet$ v= u + v + $\displaystyle \frac{1}{7}$

So what i need help is part 3) and i would like some1 to tell me if i had done the other parts correctly?