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Show that if $fd'=f'd $ and the pair $f, d $ are coprime, that is gcd($d,f $)=1, as is the pair $f',d' $, then $f=f' $ and $d=d' $.

This should be simple, but I couldn't verify it. Thanks in advance!

$\begin{array}{}(f,\ d)=1\\(f',d')=1\end{array},\ \dfrac{f}d=\dfrac{f'}{d'} \Rightarrow\begin{align}f=f'\\d=d'\end{align}\ $ [Unique Fractionization]

$\begin{array}{}(f,\ d)=\!1\\ \!(f',d')\!=\!1\end{array},\,\ \dfrac{f}d=\dfrac{f'}{d'}\ \Rightarrow\,\begin{align} f=f'\\d=d'\end{align}\ $ [Unique Fractionization]

Bill Dubuque
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Alexander
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3 Answers3

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Hint

Use the Euclid's lemma to show that $f$ divides $f'$ and $f'$ divides $f$ and conclude that $f=f'$ and then $d=d'$.

  • Hi, I need help here. I tried as follows: if $a|f $ then $fd' =aqd'=f'd$. And so $a |f'd $ but $a $ cannot divide $f' $ so that we must have $a |d $, and $aqd'=f'ak $ so that $qd'=f'k $. But I get no where here... – Alexander Jun 08 '14 at 10:00
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    We have $f|f'd$ and $\gcd(f,d)=1$ so by the Euclid's lemma we have $f|f'$. Do the same thing to find $f'|f$. –  Jun 08 '14 at 10:05
  • Understod how to do it now, thanks! – Alexander Jun 08 '14 at 10:16
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Another solution, using Corollary 31.4 from CLRS:

For all integers a and b and any nonnegative integer n, $ (an, bn) = n*(a, b); $

$ (f, d) = 1 => (ff', df') = f' $ (multiplying by f')

$ (f', d') = 1 => (ff', d'f) = f $ (multiplying by f)

Now it's easy:

$ (ff', df') = (ff', d'f) => f' = f $ (because df' = d'f)

$ df' = df = d'f => d' = d $

limido
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Said in fractions $\ \begin{array}{c} (f,\,\ d)=\!1\\ (f',d')\!=\!1\end{array},\, \ \ \dfrac{f}d\, =\, \dfrac{f'}{d'}\ \Rightarrow\,\ \begin{array}{c} f\, =\, f'\\ d\, =\, d'\end{array}\,\ $ $ \bbox[5px,border:1px solid #c00]{\begin{align}&\bf{Unique\ Fractionization}\\ &\!\text{ reduced fractions are }\textit{unique}\end{align}}$

Proof $ $ It suffices to show the denominator of a reduced fraction $\rm\color{#0a0}{divides}$ every other denominator, since then we deduce that $\,d\mid d'$ and $\,d'\mid d,\ $ so $\,d = d',\,$ both being positive (cf. my comment).

Now if $\ \color{#c00}{(a,b)= 1}\ \,$ & $\,\ \dfrac{A}B\, =\, \dfrac{a}b\, $ then $\, Ab = aB \ $ so $\ \color{#c00}{b\mid a}B\,\Rightarrow\,\color{#0a0}{b\mid B}\ $ by Euclid's Lemma.

Remark $\,\ $ Further: $\ \ \bbox[5px,border:1px solid #c00]{\begin{align} A &\,=\, n\:\!a\\ B &\,=\, n\:\! b\end{align}\ }\ \,$ by $\,\ \dfrac{A}a = \color{#0a0}{\dfrac{B}b} =: n\in\Bbb Z$

By $\,n\mid A,B,\:$ if $\ (A,B)=1\,$ then $\, n\mid 1\,$ so $\, n= \pm 1,\,$ i.e. reduced fractions are unique up to sign. Usually we normalize fractions by requiring positive denominators, cf. unit normalization.

Conversely, this denominator divisibility property implies Euclid's Lemma (reverse above proof). Thus both are equivalent. Further, they are both equivalent to uniqueness of prime factorizations. As such, I call the fractional version unique fractionization. It is discussed further in various answers. If you know about ideals, then you may also find of interest various posts on denominator ideals (the denominator divisibility property is a special case of the fact that the set of denominators of a fraction forms an ideal, and $\,\Bbb Z\,$ is a PID, i.e. all ideals are principal, being generated by their least nonzero element). So reduced fraction uniqueness boils down to the principality of an ideal.

Alternatively, we can replace use of Euclid's Lemma by a descent using the Division Algorithm, which yields a more direct (but less conceptual proof), e.g. see here.. This can be employed to give proofs of irrationality of square-roots using only the division algorithm, using a descent achieved by taking fractional parts of fractions, e.g. see this answer (based on a discussion with John Conway on irrationality proofs).

More generally see Euler's Four Number Theorem for the non-coprime case.

Bill Dubuque
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