Let X=Z×(Z{0}), and ~ be the relation defined on X by (a,b)~ (c,d) whenever ad=bc. Let R=X/~={[x]:x∈X}. Define the operation * on R, by setting {(a1,b1)}*{(a2,b2)}={(a1b2+a2b1,b1b2)}. show that this operation is well-defind.
I'm trying to learn the algebra structure by myself, and I got an assignment online and I don't how to solve it, There is no similar exercise on the book.
What you have to do is show is that if:
$[(a_1,b_1)]\sim [(c_1,d_1)]$ and $[(a_2,b_2)] \sim [(c_2,d_2)]$
then:
$[(a_1b_2+a_2b_1,b_1b_2)] \sim [c_1d_2+c_2d_1,d_1d_2)]$.
(What this means is that the operation doesn't depend on the specific representative pairs $(a_1,b_1), (a_2,b_2)$ used, but just on their equivalence classes).
That is, if $a_1d_1 = b_1c_1$ and $a_2d_2 = b_2c_2$ that:
$(a_1b_2+a_2b_1)d_1d_2 = (c_1d_2 + c_2d_1)b_1b_2$
Expanding the left side:
$(a_1b_2+a_2b_1)d_1d_2 = a_1b_2d_1d_2 + a_2b_1d_1d_2$
$= (a_1d_1)b_2d_2 + (a_2d_2)b_1d_1 = (b_1c_1)b_2d_2 + (b_2c_2)b_1d_1$
$= c_1d_2b_1b_2 + c_2d_1b_1b_2 = (c_1d_2 + c_2d_1)b_1b_2$,
and Bob's your uncle.