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Suppose that we are given reduced rational numbers $\,\dfrac{a}{k},\ \dfrac{b}{\ell},\ \dfrac{c}{q},\,$ i.e. $\gcd(a,k)=\gcd(b,\ell)=\gcd(c,q)=1$ such that $$\frac{c}{q}=\frac{a}{k}-\frac{b}{\ell}.$$

Then we have $\ q=k'\ell'e = {\dfrac{gk'l'}{f}}$ and $\,c=\dfrac{a\ell'-bk'}{f}$, where $$g=\gcd(k,l),\,\ k'\,=\frac{k}g,\,\ \ell'=\frac{\ell}g,\,\ e=\frac{g}f,\ f=\gcd(a\ell'\!-bk',g).$$

I have some troubles to prove that $\,q=\ell'k'e.\,$ I was trying something like that: $$\frac{c}{q}=\frac{a\ell-bk}{k\ell}.$$ If we assume that $t=\gcd(a\ell-bk,k\ell)$, then $c=\dfrac{(a\ell-bk)}{t}$ and $q=\dfrac{kl}{t}$. Then I performed some manipulations but I did not reach the needed equality.

Can anyone show it please? Thank you!

Bill Dubuque
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RFZ
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2 Answers2

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For simpler algebra, let $g = \gcd(k,\ell)$, which means $\ell = g\ell'$ and $k = gk'$. This gives that

$$\begin{equation}\begin{aligned} \frac{a\ell - bk}{k\ell} & = \frac{g(a\ell' - bk')}{(gk')(g\ell')} \\ & = \frac{a\ell' - bk'}{gk'\ell'} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Next, using that $g = ef$, and $f = \gcd(a\ell' - bk',g) \; \to \; f \mid a\ell' - bk'$ meaning there's an integer $h$ such that $h = \frac{a\ell' - bk'}{f}$, we then get

$$\begin{equation}\begin{aligned} \frac{a\ell' - bk'}{gk'\ell'} & = \frac{a\ell' - bk'}{efk'\ell'} \\ & = \frac{\left(\frac{a\ell' - bk'}{f}\right)}{ek'\ell'} \\ & = \frac{h}{ek'\ell'} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Note from $f = \gcd(a\ell' - bk',g)$ with $a\ell' - bk' = hf$ and $g = ef$, we have that $\gcd(h,e) = 1$. Also, since $\gcd(a,k) = 1$, then $\gcd(a,k') = 1$, plus $\gcd(k',\ell') = 1$ (due to their definitions involving dividing by $\gcd(k,\ell)$), thus $\gcd(k',a\ell' - bk') = 1$, so $\gcd(k',h) = 1$. Similarly, $\gcd(\ell',h) = 1$. This means $\gcd(h, ek'\ell') = 1$, so \eqref{eq2A} is in the reduced rational form of $\frac{c}{q}$ with $c = h = \frac{a\ell' - bk'}{f}$ and $q = \ell'k'e$.

John Omielan
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  • How did you obtain that $\dfrac{a\ell' - bk'}{gk'\ell'}= \dfrac{cf}{efk'\ell'}$? – RFZ Jul 05 '22 at 04:51
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    @ZFR Note $c = \frac{a\ell' - bk'}{f} ; \to ; a\ell' - bk' = cf$. Also, since I set $g = \gcd(k,l)$, and you have stated $\gcd(k,l) = ef$, then $g = ef$. I'll add the detail about the numerator to the answer. – John Omielan Jul 05 '22 at 04:54
  • Sorry but I am really confused. We actually need to prove that $q=\ell'k'e$ and $c=(a\ell'-bk')/f$. But you are already using the fact that $c=(a\ell'-bk')/f$. Do you see my point? – RFZ Jul 05 '22 at 14:06
  • @ZFR Thank you for explaining the source of your confusion, and I'm sorry for not understanding earlier what your actual issue was and explaining it to you better. The point is you're trying to determine the reduced rational form of $\frac{a}{k}-\frac{b}{\ell}$, i.e., in the form of one integer divided by another integer, with both integers being coprime to each other, by removing all common factors in the numerator and denominator. We do this step by step to reach that point in the second line of my $(2)$, as I show in the next paragraph. ... – John Omielan Jul 05 '22 at 15:35
  • @ZFR (cont.) Rather than use $\frac{a\ell' - bk'}{f}$ everywhere later on, I wanted to assign it a letter for simpler algebra. Since $c$ was already defined to be this, it's what I used. Nonetheless, I've now changed my answer to use a different variable, $h$, instead. In the end, the numerator is just that single variable, i.e., $h$, which is then your $c$, with the denominator being the expression $\ell'k'e$ which is assigned to be your $q$ (note that, if I thought it would help simplify or explain things, I could've also assigned a letter variable to that expression earlier as well). ... – John Omielan Jul 05 '22 at 15:35
  • @ZFR (cont.) I hope this now explains better what is going on & that I have resolved your issue. However, if not or there's anything else confusing you, please let me know. – John Omielan Jul 05 '22 at 15:35
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    Sorry for bothering you! I wrote all this on the paper and everything makes sense now! Thanks a lot for help! +1 – RFZ Jul 05 '22 at 15:38
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$\ \,\dfrac{c}q \,=\, {\dfrac{a}k-\dfrac{b}\ell} \,=\!\!\!\!\!\!\!\!\overbrace{\dfrac{a\ell -bk}{k\ell}}^{\!\textstyle\small {\rm cancel}\,\ g \!=\! (k,l) \Rightarrow }\!\!\!\!\!\!\!\!= {\dfrac{\color{#0a0}{a\ell' -bk'}}{\color{#0a0}g\:\!\color{#c00}{k'\ell'}}},\ $ for $\,\ \ell' = \dfrac{\ell}g,\,\ k' = \dfrac{k}g,\,\ (k',\ell')=1$

$\ \color{#c00}{k'}\,$ & $\,\color{#c00}{\ell'}$ are already coprime to $\,\color{#0a0}{a\ell'\! -bk'}$ (proof below), so to reduce $\rm\color{#0a0}{RH}\color{#c00}{S}$ fraction it suffices by $\small \rm\color{#0af}{EL}=$ Euclid's Lemma to cancel $\:\!(\color{#0a0}{a\ell'\!-\!bk',g}) =:\! \color{darkorange}f,\,$ yielding $\,\bbox[6px,border:1px solid #c00]{\begin{align}c &= (a\ell'\!-bk')/ \color{darkorange}f\\ q &= gk'\ell'/ \color{darkorange}f,\,\text{ as claimed}\end{align}}$

Proof: $\,\ (\color{#c00}{k'},\:\!a\ell'\! -b\color{#c00}{k'})=\!\!\! \underbrace{(k',a\ell')\overset{\rm\color{#0af}{EL}}= 1}_{\small\textstyle (k',a)\!=\!1\!=\!(k',\ell')}\!\!\!.\,$ By $\,k',\ell'$ symmetry $\,(\color{#c00}{\ell'},a\color{#c00}{\ell'}\! -bk')\!=\!1\,$ too.

Bill Dubuque
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  • Can you add more details? I am not following what are you doing here. – RFZ Jul 05 '22 at 16:21
  • @ZFR After persuing your comments, I rewrote my answer to highlight the key idea. Note that above I use the standard notation $,(m,n):=\gcd(m,n),,$ and the deduced equalities for $c$ and $q$ use the basic fact that two reduced fractions have equal top & bottom - see unique fractionization. – Bill Dubuque Jul 05 '22 at 19:17