Using appropriate definitions axioms or theorems give a rigorus proof of the following;
1) $\displaystyle \frac{1}{b}.\frac{1}{d} = \frac{1}{bd}$
2)$\displaystyle \frac{a}{b} + \frac{c}{d} = \frac{ad+cb}{bd}$
Hi xalk.
If $\displaystyle a,b,c,d$ are elements of a ring $\displaystyle R,$ the field of fractions of $\displaystyle R$ is the field $\displaystyle \mathrm{Frac}(R)=R\times(R\setminus\{0_R\}),$ where the element $\displaystyle (x,y)\in\mathrm{Frac}(R)$ is written $\displaystyle \frac xy,$ with addition and multiplication defined as in (2) and (1) respectively. This is how the field of rationals is constructed from the ring of integers. If you go along this line, then the results you want to prove are merely definitions of addition and multiplication in the field of fractions, and so there is nothing to prove!
that R has to be a commutative domain and x/y is actually the equivalence class of (x,y) under the equivalence relation ~ defined over Frac(R) by (x,y) ~ (z,t) iff xt = yz. then the equality of
two equivalence classes x/y and z/t becomes: x/y = z/t if and only if xt = yz. i guess they basically want xalk to prove that addition and multiplication defined on Frac(R) is well-defined.
I already explained this. The meaning of $\displaystyle \tfrac{1}{b}$ it simply $\displaystyle b^{-1}$ where the inverse is the multiplicative inverse from the group of non-zero elements over a field. You should know the result from group theory that says that $\displaystyle (ab)^{-1} = b^{-1}a^{-1}$, and so $\displaystyle \tfrac{1}{ab} = \tfrac{1}{b}\cdot \tfrac{1}{a}$.