0

Before defining the lcm I would like to prove following lemma about its properties:

Lemma If $a$ and $b$ are nonzero integers, then there is a unique positive integer $m$ such that

  1. $a,b\mid m$,
  2. $a,b\mid c\implies m\mid c$.

Proof Uniqueness: Suppose $m'$ is another such number, then from (2) follow $m\mid m'$ and $m'\mid m$, thus by Lemma ??? $|m|=|m'|$ and since both are positive, follows that $m=m'$.

Existence: We define $S=\{n\in\mathbb{N} : a,b\mid x\}$, since $|ab|\in S$, then $S\neq\varnothing$. Let $m$ be the minimum of the set (todo: existence from set theory), thus $a,b\mid m$, proving (1).

Now suppose $a,b\mid c$, set $d:=\gcd(a,b)$, then $$...$$ Conclusion $m\mid c$.

Is it possible to fill the dots using Bezout and the usual identities for divisors and gcd? Without primes and Division Algorithm (I think this is one of the standard way) and $\textrm{lcm}(a,b)\gcd(a,b)=ab$. If not, is there an explanation why not?

I tried starting from the second proof of this answer here, but I don't get anywhere.

EDIT

Completion of the proof based on the answer of @haran :

Now suppose $a,b\mid c$, set $d:=\gcd(a,b)$, then $a,b\mid m,c$, thus $a,b\mid \gcd(m,c)\le m$ (because $\gcd(a,b)\le a$). Thus $\gcd(m,c)\in S$, but since $m$ is the minimal element, $\gcd(m,c)=m$ and we can conclude that $m\mid c$.

PeptideChain
  • 829
  • Re "Lemma ???", if $m,m'\in \Bbb Z^+$ we have $m|m'\implies \exists n\in \Bbb Z^+;(m'=nm\geq m)\implies m'\geq m.$ Similarly $m'|m\implies m\geq m'$...... So $(m|m'\land m'|m)\implies m'\geq m\geq m'\implies m=m'.$ – DanielWainfleet May 31 '18 at 21:58

1 Answers1

1

The uniqueness proof as you have done above is straightforward and beautiful. For the proof of existence, first define LCM as the smallest number divisible by both inputs. We know that the product of two numbers is divisible by them individually which shows $lcm(a,b)<ab+1$.

Now it suffices to show that if there exists a number $x$ such that both the numbers divide, then the LCM which is $y$ divides it too. Assume the contrary, then since $a$ and $b$ divide both $x,y$, we have $a$ and $b$ divide $\gcd(x,y) =z$. Note that now, $z < y$ but this contradicts our definition of LCM being the smallest number divisible by both $a,b$.

Hence our initial assumption is false. Thus, the definition of LCM being the least number divisible by $a,b$ is the same as the definition given. Since we have already proved that if LCM is the least number divisible by $a,b$, then there exists and LCM for every $a$ and $b$, we are done!

Haran
  • 9,717
  • 1
  • 13
  • 47
  • For the contradiction, do you mind $z<y$ (and not $z<x$)? The idea is perfect. I'll edit my qst, adding your idea without proof by contradiction. If you could give me a feedback would be appreciated. – PeptideChain May 31 '18 at 17:16
  • Thanks a lot man. And yeah, I meant to write $z<y$. Will edit. – Haran May 31 '18 at 17:36
  • What do you want feedback on? – Haran May 31 '18 at 17:37
  • to the last part of my edited question, but I think it is fine. thank you – PeptideChain May 31 '18 at 17:40
  • Yeah, what you have done is really good. Avoiding contradiction statements is better. Well done. – Haran May 31 '18 at 17:41
  • We do not need to mention $\gcd$....Let $m=lcm(|a|,|b|).$ If $c\in \Bbb Z^+$ and $a,b|c$ then $c$ is a common multiple of $|a|,|b|$ so by def'n of $lcm$ we have $c\geq m.$ Now let $K=\min {n\in \Bbb Z^+: nm>c}.$ Then $K\geq 2$ and $(K-1)m\leq c<Km.$ ... Case (I). If $(K-1)m=c$ then $m|c.$... Case (II). If $(K-1)m<c$ then $c-(K-1)m$ is is a common multiple of |a|,|b|, with $0<c-(K-1)m<Km-(K-1)m=m$ contrary to $m=lcm(|a|,|b|).$. So case (II) is impossible. So it must be that $(K-1)m=c$ and $m|c.$ – DanielWainfleet May 31 '18 at 22:21