Solution of $\frac{a}{b}=\frac{a'}{b'}$ if $a,b,a',b' \in \mathbb{N}$ Let $\dfrac{a}{b}=\dfrac{a'}{b'}$ , $a,b,a',b' \in \mathbb{N}$ s.t. $a$ and $b$ have no common factors.
How can we show that the only solution to this equality is $a'=na$ and $b'=nb$, $n$ is a natural number. I know that $a'=na$ and $b'=nb$ is a valid solution of equation-1, because $(an)b=(bn)a$. But how can we prove that this is the only solution? 
What I have tried is this:  

Version 1.
  We know the two fractions $\dfrac{a}{b},\dfrac{a'}{b'}$ are equal iff $ab'=a'b\tag{1}.$
  Let $ab'=a'b=c$. Now, $a$ and $b$ have no common factors means that they are coprime, so we have, $a=p_1p_2\cdots p_m$ and $b=q_1q_2\cdots q_n$. $p's$ and $q's$ are prime numbers and no $p$ is equal to any $q$. Equation 1 becomes $$(p_1p_2\cdots p_m)b'=(b_1b_2\cdots b_n)a'$$
  By the Fundamental theorem of Arithmetic, both the sides should have same prime numbers, so $b'$ should contain $(b_1b_2\cdots b_n)$ and $a'$ should contain $(p_1p_2\cdots p_m)$, i.e. $b'=r(b_1b_2\cdots b_n)=rb$ and $a'=(p_1p_2\cdots p_n)s=sa$, where $r,s \in \mathbb{N}$. Now equation 1 becomes, $a(rb)=(sa)b$. This leaves the only possibility, $r=s$.
Version 2.
   The idea is basically the same as of version 1, but now I use Euclid's lemma in place of FTA. We know the two fractions $\dfrac{a}{b},\dfrac{a'}{b'}$ are equal iff $ab'=a'b\tag{1}.$
  Let $ab'=a'b=c$. $a$ and $b'$ are the factors of $c$, so both of them divide $c$, i.e. $a|c$ and $b'|c$. Now $a|c \implies a|{(a'b)}$, by Euclid's lemma, $a$ divides at least one of $a'$ and $b$, but we know that $a$ doesn't divide $b$, so $a$ has to divide $a'$.
Similarly we can show that $b$ divides $b'$. So we have $a'=\alpha a$ and $b' = \beta b$, where $\alpha, \beta \in \mathbb{N}$. Substituting these in equation 1 gives $\alpha = \beta$.

Question


*

*Is my method correct?   

*Is there any other method to prove that $a'=na$ and $b'=nb$ is the only solution of $\dfrac{a}{b}=\dfrac{a'}{b'}$; If yes then please explain.  

*Also if $a$ and $b$ have common factor then for what condition there does not exist any $n$ s.t. $a'=na$ and $b'=nb$ where $n$ is a natural number. E.g. $\dfrac 39 = \dfrac 26$

P.S:  I've just improved my question. Barry's comment points a crucial error in my previous proof. I needed to justify $a|a'$ and $b|b'$. In previous proof I proceeded with proof by contradiction. The proof by contradiction was ok, but unnecessary, so I've replaced it with a direct proof. At the time of asking the question I lacked the knowledge of some introductory number theory concepts. Now I've learned them and soon I will ask in comments for further clarifications. Thank you.

$a'=ma+r, b'=nb +s\implies \dfrac{a'}{b'}=\dfrac{ma+r}{nb+s}$

$$\begin{align}
\dfrac{a}{b}= \dfrac{ma+r}{nb+s} & \implies a(nb+s)=b(ma+r)\\
 & \implies a\mid (ma+r)\land b\mid (nb+s) \\ 
 & \implies r=s=0 \\
 & \implies a'=ma, b'=nb \\ 
 & \implies\dfrac{a}{b}= \dfrac{ma}{nb}\\
 & \implies m=n\end{align}$$

