Solution of $\frac{a}{b}=\frac{a'}{b'}$ if $a,b,a',b' \in \mathbb{N}$

Question

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

1. Is my method correct?

2. 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.

3. 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. $\dfrac 39 = \dfrac 26$

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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.

Answer 1

The proof boils down to a little lemma:

If $\gcd(a,b)=1$ and $a\mid bc$, then $a\mid c$.

Here's how it applies:

$$\begin{align} {a\over b}={a'\over b'}&\implies ab'=ba'\\ &\implies a\mid ba'\\ &\implies a\mid a'\qquad\text{(by the lemma)}\\ &\implies a'=an\\ &\implies ab'=ban\\ &\implies b'=bn \end{align}$$

One way to prove the lemma is by invoking the theorem that if $\gcd(a,b)=1$, then $am+bn=1$ for some integers $m$ and $n$. Combining this with $bc=ak$ (based on the assumption $a\mid bc$), we have

$$\begin{align} am+bn=1&\implies amc+bnc=c\\ &\implies amc+nak=c\\ &\implies a(mc+nk)=c\\ &\implies a\mid c \end{align}$$

Added 12/9/14 (in response to OP's edits): Version 1 looks fine to me (except the $b_i$'s should be $q_i$'s, or vice versa), but Version 2, I'd say, makes too liberal a use of Euclid's lemma, which lets you say $a\mid a'b$ and $a\not\mid b$ implies $a\mid a'$ only if $a$ is prime. What you need to say in that proof is that $a\mid a'b$ and $(a,b)=1$ implies $a\mid a'$. The hypothesis $(a,b)=1$ is much stronger than $a\not\mid b$. Basically a correct version of Version 2 amounts to the proof in my original answer above.

Answer 2

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

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

Answer 3

As the saying goes, just when you think you have life's answers they change the questions ! (this was posted before the caveat about using rational real numbers)

For any non-zero $a, a’, b, b’$ the ratios can be expressed as non-zero real numbers $R_a$ and $R_b$, so that $b’ = R_b \cdot b$ and $a’ = R_a \cdot a$

$a'/b' = a/b = \dfrac{R_a\cdot a}{R_b\cdot b}.$

So, $\dfrac{a}{b} = \dfrac{R_a}{R_b}\cdot \dfrac{a}{b}$.

So, $\dfrac{R_a}{R_b} = 1$ and therefore $R_a = R_b$.

If you chose, you can now look only at integer values of $R_a$ and $R_b$, so that $Ra = Rb = k$

and $a’ = k\cdot a$ and $b’=k\cdot b$

Answer 4

We use proof by contradiction. We have $\frac{a}{b}=\frac{a'}{b'}$, where $a$ and $b$ are relatively prime.

Suppose that $\frac{a'}{a}$ is not an integer, i.e. in simplest forms, it is in the form $\frac{p}{q}$ where $q>1$. Then $(a)(\frac{p}{q})$ and $(b)(\frac{p}{q})$ are both integers (equal to $a'$ and $b'$ respectively).

Since $p$ is relatively prime to $q$, $a$ and $b$ must both be divisible by $q$, contradiction as $q>1$ but $a$ and $b$ are relatively prime.

Answer 5

This is equivalent to Euclid's Lemma (EL), e.g. see the standard proof in Barry's answer However, it can also be proved directly using the foundation of EL, i.e. the (Euclidean) Division algorithm, used in the proof below to take the (unique!) fractional part $\rm\,r/B\in [0,1)\,$ of a fraction $\rm\,A/B,\,$ i.e.

by the Division Algorithm $\rm\ A = q B + r,\,\ 0\le r < B\!\! \overset{\large\ \ \times\ B^{-1}}\implies\, \dfrac{A}B = q + \dfrac{r}B,\,\ 0\le \dfrac{r}B < 1$

Unique Fractionization $\ $ The least denominator $\rm\:B\:$ of a fraction divides every denominator.

Proof $\rm\displaystyle\ \ \frac{A}B = \frac{C}D\ \Rightarrow\ \frac{D}B = \frac{C}A \:.\ $ Taking fractional parts $\rm\displaystyle\ \frac{b}B = \frac{a}A\ $ where $\rm\,\ 0 \le b < B.\ $ But

$\rm\displaystyle\ \:B\nmid D\ \Rightarrow\ b\ne 0\ \Rightarrow\ \frac{A}B = \frac{a}b\ \ $ contra leastness of $\rm\:B.\:$ Hence $\rm\ B\mid D.\quad $ QED

Further $\rm\,B\mid D\,\Rightarrow\, n = \dfrac{D}B = \dfrac{C}A\,\Rightarrow\,\begin{array}{}\rm nA = C\\ \rm nB = D\phantom{{I_I}^I}\end{array}\!\!\!\!$ In particular, if $\rm\,C,D\,$ are coprime then $\rm\,n=1\,$ therefore $\rm\,C,D = A,B,\,$ i.e. a reduced fraction is in least terms, which yields the sought result.

Answer 6

I think your method is correct.

I have an other method: Let $\frac{a}{b}=\frac{a'}{b'}$. From that we can write $\frac{a}{a'}=\frac{b}{b'}$, than there exist a rational number $k$ s.t. $\frac{a}{a'}=\frac{b}{b'}=k$, so $a=a'k$ and $b=b'k$ or $a'=a\frac{1}{k}$ and $b'=b\frac{1}{k}$ or for $n=\frac{1}{k}$ we finally have $a'=na$ and $b'=nb$.

Updated: If (a,b)=1, and a'>a, b'>b, than we have the following: Suppose that $\frac{a}{b}=\frac{a'}{b'}$. From $\frac{a}{b}=\frac{a'}{b'}$ and (a,b)=1, we conclude that (a',b')=n where n is natural. Now, from (a',b')=n we can write a'=pn and b'=qn where (p,q)=1.So we have (p,q)=(a,b)=1 and $\frac{a}{b}=\frac{p}{q}$ ($\frac{a}{b}=\frac{a'}{b'}=\frac{pn}{qn}=\frac{p}{q}$)which can be true iff a=p and b=q. So finally we have that a'=na and b'=nb must be the only solution.