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Let $a,b,c,d$ be positive integers. Suppose that $$\frac cd=\frac ab.$$ I want to prove that if $a$ and $b$ are relative primes, then $c/a=d/b$ is an integer.

That is, the only way a fraction can be represented in a way other than its simplest form is to multiply both the numerator and the denominator by the same integer.

I’m a little ashamed to seek help, because the statement seems so trivial, yet I’m stuck with a rigorous proof. Any input would be appreciated.

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    $\begingroup$ I think this is backwards, and you should have $c/a$: Take $a = 1, b = 2, c = 2, d = 4$ for a counterexample. $\endgroup$ – user296602 Feb 4 '16 at 0:24
  • $\begingroup$ @T.Bongers You’re right, correction made. $\endgroup$ – triple_sec Feb 4 '16 at 0:26
  • $\begingroup$ If a and b are relative primes then c =ka and d = kb and your integer is k. $\endgroup$ – Piquito Feb 4 '16 at 0:30
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    $\begingroup$ @Piquito that's what the OP is trying to prove. That the only way to write a/b is as ak/bk. $\endgroup$ – fleablood Feb 4 '16 at 0:31
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Let's restate the question without fractions. We want to show that if $bc = ad$ and $\gcd(a, b) = 1$, then there is some $k \in \mathbb Z$ such that $c = ak$ (and thus $d = bk$).

Indeed, using Bezout's Identity, we know that there exist $u,v \in \mathbb Z$ such that: $$ au + bv = 1 $$ Multiplying through by $c$, we get: $$ c = acu + (bc)v = acu + (ad)v = a\underbrace{(cu + dv)}_{k \in \mathbb Z} $$ as desired.

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  • $\begingroup$ Thank you, but I’m not familiar with Bezout’s identity (even though it is apparently very elementary). I have presented a little more complicated, but (for me) more transparent, proof based on the fundamental theorem of arithmetic (which I’m more comfortable with). $\endgroup$ – triple_sec Feb 4 '16 at 0:50
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I think I got it. By the fundamental theorem of arithmetic, each $x\in\{a,b,c,d\}$ can be represented as $$x=\prod_{i=1}^k p_i^{n^x_i},$$ where $(p_i)_{i=1}^k$ are distinct primes, $(n_i^x)_{i=1}^k$ are non-negative integers, and $k\in\mathbb N$. Since $a$ and $b$ are relative primes, $n_i^a n_i^b=0$ for each $i\in\{1,\ldots,k\}$ (otherwise both $a$ and $b$ would be divisible by $p_i$).

Now, $c/a=d/b$ means that $$\prod_{i=1}^k p_i^{n^c_i-n^a_i}=\prod_{i=1}^k p_i^{n^d_i-n_i^b}.$$ By the uniqueness of prime factorizations (which holds also when the exponents are potentially non-positive), one must have $$n^c_i-n_i^a=n^d_i-n^b_i\quad\forall i\in\{1,\ldots,k\}.$$ There are two cases. If $n_i^a>0$, then $n_i^b=0$, so that $n_i^c-n_i^a=n_i^d\geq 0$. If $n_i^a=0$, then $n_i^c-n_i^a=n_i^c\geq 0$. The point is that, in each case, $\Delta_i\equiv n_i^c-n_i^a\geq 0$. Therefore, $$w\equiv\prod_{i=1}^kp_i^{\Delta_i}$$ is a positive integer, and, by construction, $c/a=w$.

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