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This question is more concerned with understanding the reasoning behind mathematical proving rather than explaining this specific proof. I understand why this proof works.

This is the proof of Euclid Lemma from Wikipedia:

This states that if $x$ and $y$ are relatively prime integers (i.e. they share no common divisors other than 1) there exist integers $r$ and $s$ such that

$rx+sy=1$. Let $a$ and $n$ be relatively prime, and assume that $n|ab$. By Bézout's identity, there are $r$ and $s$ making

$rn+sa=1$. Multiply both sides by $b$:

$rnb+sab=b$. The first term on the left is divisible by $n$, and the second term is divisible by $ab$, which by hypothesis is divisible by $n$. Therefore their sum, $b$, is also divisible by $n$.

I am mostly concerned with this statement:

Multiply both sides by $b$:

My question is why this multiplication by $b$ happens. Is it because the person proving this observed that multiplying the equation by $b$ he would get a sum of two multiples of $n$? Or is there a mathematical rule by which we know that the next step is multiplying by $b$?

I am aware that this question might seem trivial to people here, but I just want to make sure.

Michael Munta
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    "Because we can" - You are free to multiply by $a^3$ or by $17$ instead. But that would not help you carry on. - How could one get the idea to multiply by $b$? Well, we are given a fact about $ab$, namely that it is a multiple of $n$. So any step that makes $ab$ appear and allows us to make use of that property might be helpful. – Hagen von Eitzen Oct 26 '18 at 11:50
  • @HagenvonEitzen Thank you, that is exactly what bothered me. – Michael Munta Oct 26 '18 at 11:59

1 Answers1

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My question is why this multiplication by b happens. Is it because the person proving this observed that multiplying the equation by b he would get a sum of two multiples of n? Or is there a mathematical rule by which we know that the next step is multiplying by b?

Yes, the "rule" is: $ $ an $\color{#c00}{\rm invertible}$ element is cancellable (simply by scaling by its inverse), and integers coprime to $\,n\,$ are invertible$\bmod n\,$ (by Bezout), see here. $ $ Therefore we have:

$\qquad\qquad\ \ \ \ n\mid ax\,\Rightarrow\, n\mid x\ \ $ if $\,\gcd(a,n) = 1,\, $ interpreted mod $n$ becomes

$\quad \iff\ \ \ ax\equiv 0\,\Rightarrow\, x\equiv 0\ $ if $\,\gcd(a,n) = 1.\ $ The coefficient $\,a\,$ is invertible by Bezout:

$\qquad\ \ \qquad \color{#c00}{sa\equiv 1}\,\,\ $ by $\ sa = 1 - rn\ $ for some $\,s,r,\,$ by Bezout identity for $\,\gcd(a,n) = 1$

$\qquad\, \Rightarrow\,\ \color{#c00}{sa}x \equiv s0\ $ by scaling 2nd last equation by $\,\color{#c00}s,\,$ valid by the Congruence Product Rule

$\qquad\, \Rightarrow\,\ \quad x\equiv 0\ $ by $\ \color{#c00}{sa\equiv 1}$

i.e. scale by $\,\color{#c00}{s\equiv a^{-1}}\,$ to cancel $\,\color{#c00}a\,$ from $\,\color{#c00}ax\equiv \color{#c00}a0\,$ to get $\,x\equiv 0\,$ (that's how to view it explicitly as cancelling $\,\color{#c00}a\,$ on both sides).

Remark $\ $ Thus reformulating the divisibility relation as arithmetical operations mod $\,n\,$ clarifies the arithmetical essence of the matter. In the same way many divisibility properties are greatly clarified and simplified when reformulated arithmetically in congruence form.

Bill Dubuque
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  • This is maybe too much for me to grasp. I am a beginner at this with very bad mathematical background. Also my question was probably very stupid to alot of people here. How would I go about reaching a level at math where all of this would be trivial to me at first sight? Should one simply have a knack for this? Right now it would take me multiple days just to understand what you wrote there. :) – Michael Munta Oct 26 '18 at 13:48
  • How can gcd of anything be 0? It should be at least 1? – Michael Munta Oct 26 '18 at 14:40
  • @Michael That's a typo, Any textbook on elementary number theory will explain congruences. Once you learn them then you can understand that the solution is trivial, just as it is for real numbers, viz. $, ax = 0,\Rightarrow, x = 0\ $ if $,a\neq 0,$ (i.e. if $a$ is invertible), which follows by multiplying through by $,a^{-1}.\ $ That's exactly what we do above when we multiply by $,s\equiv a^{-1},$ in the ring of $,\Bbb Z_n = $ integers $!\bmod n\ $ vs real numbers. – Bill Dubuque Oct 26 '18 at 14:53
  • What do you mean by $a$ being invertible? What does that mean exactly? – Michael Munta Oct 26 '18 at 14:59
  • @Michael Just as for real numbers, in the ring of integers $\bmod n,$ if $,ax\equiv 1,$ has a solution $,x \equiv s,$ then we say $,a,$ is invertible with inverse $,s,,$ i.e. $\ s\equiv a^{-1},$. It's easy to show that this is well-defined, and inverses are unique (when they exist). – Bill Dubuque Oct 26 '18 at 15:07