Equivalence classes and rational numbers

Question

We defined $\mathbb{Q}$ as the set of equivalence classes for the relation $\sim$. Tentatively define operations $+,\cdot:\mathbb{Q}\times\mathbb{Q}\rightarrow\mathbb{Q}$ by

$[(a,b)]+[(c,d)] = [(ad+bc,bd)],$ $[(a,b)]\cdot[(c,d)]=[(ac,bd)]$

These definitions have a potential problem. They read: "Given equicalence classes $x$" and $y$, choose elements $(a,b)$ and $(c,d)$ of them, perform such and such operations, and let $x+y$ or $x\cdot y$ be the equivalence class of the result." But what if one chose different representatives $(a',b')$ and $(c',d')$? Might those choices produce a different class?

Prove that the definitions are indeed sound by showing that, for all pairs in $A$,

$(a,b)\sim(a',b')$ and $(c,d)\sim(c',d') \implies \begin{cases} (ad+bc,bd)\sim(a'd'+b'c',b'd')\\ (ac,bd)\sim(a'c',b'd') \end{cases}$

I am given that the equivalence relation is $(m,n)\sim (p,q)$ if $mq=np$ in the set $A=\{(m,n) \in \mathbb{Z}\times\mathbb{Z}:n\ne0 \}$.

I am not sure how using representatives $(a',b')$ and $(c',d')$ would product

Answer 1

Remember, $(a, b) \sim (a', b')$ means that $ab' = a'b$.

You want to show that

$(a,b)\sim(a',b') \text{ and } (c,d)\sim(c',d') \implies \begin{cases} (ad+bc,bd)\sim(a'd'+b'c',b'd')\\ (ac,bd)\sim(a'c',b'd') \end{cases} $

I'll do the second one.

We want to show that, if $(a,b)\sim(a',b') \text{ and } (c,d)\sim(c',d') $ then $(ac,bd)\sim(a'c',b'd') $.

From the definition of "$\sim$", this means that $acb'd' = a'c'bd $. Since $(a,b)\sim(a',b') $, $ab' = a'b$. Similarly, $cd' = c'd$. Multiplying these, $ab'cd' = a'bc'd$, which is what you want, rearranged.

The first one, for addition, is more complicated, so I'll leave it for you.

Answer 2

You need to do just what the problem says. Let $(a,b)$ and $(a',b')$ be two elements in the same class, which means that $ab'=a'b$, and similarly for $(c,d),(c',d')$ Your addition says that $(a,b)+(c,d)=(ad+bc,bd)$ and $(a',b')+(c',d')=(a'd'+b'c',b'd')$ Your addition is well defined on the classes if the two results represent the same rational, which means they are in the same equivalence class. As an example, let $(a,b)=(1,2), (a',b')=(3,6), (c,d)=(1,7), (c',d')=(2,14)$. Then $(a,b)+(c,d)=(1\cdot 7+2\cdot 1,2\cdot 7)=(9,14)$ and $(a',b')+(c',d')=(3\cdot 14+6\cdot 2,6\cdot 14)=(54,84)$ You can check that they are in the same equivalence class. You have to prove the general case, then do the same for multiplication.