I am trying to show that the addition of rational numbers is well defined.
Does anyone know if this is a legitimate strategy?
Also, I am quite unfamiliar with coding on here.
Q is defined as $Q = \{m/n: m, n \in Z, n \ne 0\}$
where $m/n$ is the class of ordered pairs $(m,n)$ such that $(m,n)\equiv (p,q)$ if $mq = pn$ with $n, q \ne 0$.
Statement: If $(a,b) \equiv (a',b')$ and $(c,d) \equiv (c',d')$, then $(ad+bc,bd) \equiv (a'd'+b'c',b'd')$.
Proof:
Assumptions:
$(a,b) \equiv (a',b')$ implies $ab'=ba'$, which then implies $ab'-ba'=0$.
$(c,d) \equiv (c',d')$ implies $cd'=dc'$, which then implies $dc'-cd'=0$.
Since $ab'-ba'=0$, this implies $dd'(ab'-ba')=0$, also $bb'(dc'-cd')=0$ with similar reasoning.
Since $dd'(ab'-ba')=0$ and $bb'(dc'-cd')=0$, this implies $dd'(ab'-ba')=bb'(dc'-cd')$.
Now some algebraic manipulations
$ab'dd'-ba'dd'=dc'bb'-cd'bb'$
$ab'dd'-a'bdd'=c'dbb'-cd'bb'$
$ab'dd'+cd'bb'=a'bdd'+c'dbb'$
$adb'd'+bcb'd'=bda'd'+bdb'c'$
$(ad+bc)b'd'=bd(a'd'+b'c')$
Which implies, $(ad+bc,bd) \equiv (a'd'+b'c',b'd')$
Please let me know if this is legitimate. I worked backwards a lot in order to manipulate in just the right way.
I have trouble seeing all the primes but in general the method looks good to me. Somewhere you should mention the restriction that the denominators b, b', d, d' are not zero.