
Showing That it is onto?
Define a binary operation http://www.mathhelpforum.com/mathhe...7a0461831.gif on the rationals so that the map http://www.mathhelpforum.com/mathhe...9cf5ab481.gif defined by http://www.mathhelpforum.com/mathhe...c4a927061.gif is an isomorphism between sets with binary operations. Prove that http://www.mathhelpforum.com/mathhe...2cc525ec1.gif 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 onetoone 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 onetoone
3) I dont even know were to start (Doh)
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?

3) Let s'\in S'=\mathbb{Q}. The function \phi is onto if we can find x so that \phi(x)=7x+1=s'. But that easy x should be the solution to 7x+1=s', i.e., x=\frac{s1}{7} and it is in Q. Hence \phi is onto
For the whole thing your idea is OK. But to be in order you need to
1) define what u*v means (well you guess this from your calculation in 4)
2) shows \phi(u*v)=\phi(u)+\phi(v) (i.e, \phi is a homomorphism). This calculation guides you to answer 1, but in writing you need to answer 1 first and somehow you you define u*v out of thin air.
3) show that it is injective
4) show it is onto.

Thanks, i think i understand now.