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Lemma: Suppose $m\geq{n}>0$. Then $hcf(m,n) = hcf(m-n, n).$

Proof: Suppose $d|m$ and $d|n$. So there are $a,b$ with $da=m$, $db=n$ and $a\geq{b}$. So $d(a-b)=m-n$. Thus $d|m-n$. So $hcf(m,n)\leq{hcf(m-n,n)}$. The other inequality is similar so $hcf(m,n)=hcf(m-n,n)$ $\blacksquare$.

I have no idea where the inequality $hcf(m,n)\leq{hcf(m-n,n)}$ comes from, in fact I was thinking the inequality would be the other way round since $m-n<n$. Can someone please enlighten me on this inequality?

Note: For you American folks, $hcf(m,n)$ is the same as $gcd(m,n)$.

Bill Dubuque
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Nostradamus
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  • The key idea is that ${n,m}$ and ${n,m-n}$ have the same set of common divisors $d$ so the same greatest common divisor - see here in the linked dupe. – Bill Dubuque Jul 22 '22 at 17:34
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    @BillDubuque Why is r congruent to 0 in the second line of the linked dupe? – Nostradamus Jul 22 '22 at 17:40
  • If $,\color{#c00}{b\equiv 0},$ then $,0\equiv q\color{#c00}b+r\equiv q\color{#c00}0+r\equiv r,,$ i.e. $r\equiv 0,,$ by Congruence Sum & Product Rules. – Bill Dubuque Jul 22 '22 at 17:50
  • Geometrically, if we rename to customary variables, then the equivalent systems of equations is $\ x\equiv 0, y\equiv 0\iff x\equiv 0,, y+mx\equiv 0.,$ This says the the origin $, (x,y) \equiv (0,0),$ can be specified either as the intersection of the $y$ axis $(x\equiv 0)$ with the $x$ axis $(y\equiv 0)$ or, equivalently, as the intersection of the $y$ axis $(x\equiv 0)$ with the line $, y\equiv -mx.,$ – Bill Dubuque Jul 22 '22 at 17:50

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