Easiest and most complex proof of $\gcd (a,b) \times \operatorname{lcm} (a,b) =ab.$

Question

I'm looking for an understandable proof of this theorem, and also a complex one involving beautiful math techniques such as analytic number theory, or something else. I hope you can help me on that. Thank you very much

Answer 1

Let $\gcd(a,b)=d$. Then for some $a_0,b_0$ such that $a_0$ and $b_0$ are relatively prime, we have $a=da_0$ and $b=d b_0$. If we can show that the lcm of $a$ and $b$ is $da_0b_0$, we will be finished.

Certainly $da_0b_0$ is a common multiple of $a$ and $b$. We must show that it is the least common multiple.

Let $m$ be a common multiple of $a$ and $b$. We will show that $da_0b_0$ divides $m$.

Since $m$ is a multiple of $a$, we have $m=ka=ka_0d$ for some $k$. But $b$ divides $m$, so $db_0$ divides $ka_0d$, and therefore $b_0$ divides $ka_0$. Since $a_0$ and $b_0$ are relatively prime, it follows that $b_0$ divides $k$, and we are finished.

Answer 2

First notice that $$ \dfrac{ab}{\gcd(a,b)} = a\dfrac{b}{\gcd(a,b)} = b\dfrac{a}{\gcd(a,b)} $$ is a common multiple of $a$ and $b$. By the minimality of the $\operatorname{lcm}$, $$ \frac{ab}{\gcd(a,b)}\ge\operatorname{lcm}(a,b)\Longrightarrow ab\ge\operatorname{lcm}(a,b)\gcd(a,b)\tag{1} $$ By division, we can write $$ ab = q\operatorname{lcm}(a,b) + r\quad\text{where}\quad0 \le r \lt \operatorname{lcm}(a,b) $$ Because $ab$ and $\operatorname{lcm}(a,b)$ are common multiples of $a$ and $b$, so is $r$. By the minimality of the $\operatorname{lcm}$, $r = 0$. Therefore, $\operatorname{lcm}(a,b)$ divides $ab$. Notice that $$ \frac{ab}{\operatorname{lcm}(a,b)} = \frac{a}{\operatorname{lcm}(a,b)/b} = \frac{b}{\operatorname{lcm}(a,b)/a} $$ is a common divisor of $a$ and $b$. By the maximality of the $\gcd$, $$ \frac{ab}{\operatorname{lcm}(a,b)} \le \gcd(a,b)\Longrightarrow ab\le\operatorname{lcm}(a,b)\gcd(a,b)\tag{2} $$ Combining $(1)$ and $(2)$, we get $$ ab = \operatorname{lcm}(a,b)\gcd(a,b) $$

Answer 3

The following is more general than for the integers, and therefore simpler (but longer than a proof using unique factorisation without proving it; here we start from scrap).

Let $R$ be an integral domain, where $d=\gcd(a,b)$ is defined to mean that $d\mid a,b$ and $d'\mid a,b\implies d'\mid d$ for all $d'\in R$, while $\def\lcm{\operatorname{lcm}}m=\lcm(a,b)$ is defined to mean that $a,b\mid m$ and $a,b\mid m'\implies m\mid m'$ for all $m'\in R$ (in both cases it is not implied that $\gcd(a,b)$ or $\lcm(a,b)$ always exist, and if they do they are only unique up to multiplication by invertible elements; as a consequence in this setting the equality $\gcd(a,b)\times\lcm(a,b)=ab$ can only be asserted up to such multiplication, or for properly chosen values on the left hand side).

Lemma. Let $r\in R\setminus\{0\}$, and put $D_r=\{\, d\in R: d\mid r\,\}$, the set of divisors of $r$. Then $f_r:d\mapsto r/d$ defines an involution of $D_r$ which is an anti-isomorphism for the divisibility relation: for $a,b\in D_r$ one has $a\mid b\iff f(b)\mid f(a)$.

Proof. Since by definition $d f(d)=r$ for all $d\in D_r$ one has $f(d)\in D_r$ and $f(f(d))=d$. Suppose $a,b\in D_r$ satisfy $a\mid b$, so there exists $c\in R$ with $ac=b$, then $r=bf(b)=acf(b)$ so $f(a)=cf(b)$ and $f(b)\mid f(a)$. Conversly if $f(b)\mid f(a)$ applying this result gives $f(f(a))\mid f(f(b))$ which simplifies to $a\mid b$. QED

Proposition. If $ab\neq0$ and $m=\lcm(a,b)$ exists, then $ab/m=\gcd(a,b)$.

Proof. One has $a,b\mid ab$ so $m\mid ab$ by definition of the $\lcm$; therefore $a,b,m\in D_{ab}$. One has $f_{ab}(a)=b$ and $f_{ab}(b)=a$, and since $a,b\mid m$ one has $f_{ab}(m)\mid b,a$ by the lemma. Also if $d'\in R$ satisfies $d'\mid a,b$ then $d'\in D_{ab}$ so $b,a\mid f_{ab}(d')$ by the lemma, whence $m\mid f_{ab}(d')$ by definition of the $\lcm$, and once again by the lemma $d'\mid f_{ab}(m)$. Thus $$ab/m=f_{ab}(m)=\gcd(a,b). \qquad\text{QED}$$

Concluding $\gcd(a,b)\times \lcm(a,b)=ab$ needs the precaution that it only holds if $\lcm(a,b)$ exists, and then the left hand side is defined up to invertible factors only, so the equality should be interpreted in this sense. For the case $ab=0$ not covered by the proposition one has $0=\lcm(a,b)$ and $\{a,b\}=\{0,\gcd(a,b)\}$, so the equality holds without any difficulty.

Note that the existence of $\gcd(a,b)$ does not imply the existence of $\lcm(a,b)$ in general.

Answer 4

Let's prime factorize a and b.Let $a=p_1^{x_1}p_2^{x_2}\cdots\cdot q$ and $b=p_1^{y_1}p_2^{y_2}\cdots r$ where $p_i$'s are distinct primes and GCD$(r,q)=1$ Then

GCD$(a,b)=p_1^{\min(x_1,y_1)}p_2^{\min (x_2,y_2)} \cdots$

LCM$(a,b)= qrp_1^{\max(x_1,y_1)}p_2^{\max(x_2,y_2)}\cdots$

Then since $\min(x, y) + \max(x, y) = x+y$, we have LCM$(a,b)$GCD$(a,b)=ab$

Answer 5

I think this is a simple one:

By definition, a least common multiple of a pair of integers $a$ and $b$ is an integer $m$ such that $a|m$, $b|m$, and $m$ divides every common multiple of $a$ and $b$.

Just look that if $c$ is a common multiple of $a$ and $b$, we have that $c=ax=by$ for some integers $x$ and $y$.

Then $\frac{a}{(a,b)}x=\frac{b}{(a,b)}y$, and because $\left(\frac{a}{(a,b)},\frac{b}{(a,b)}\right)=1$, we have that $\frac{a}{(a,b)}$ divides $y$.

So $y=\frac{a}{(a,b)}n$ for some integer $n$ and $c=\frac{ba}{(a,b)}n$. This shows that every time you have a common multiple of $a$ and $b$ it can be divisible by $\frac{ba}{(a,b)}$, then $[a,b]=\frac{ba}{(a,b)}$.

Answer 6

The following simple duality-based proof works in any integral domain.

Theorem $\rm\quad gcd(a,b)\, =\, ab/lcm(a,b)\ \ $ if $\ \ \rm lcm(a,b) \;$ exists, and $\rm\ ab\ne 0$

Proof $\rm\quad d\mid a,b\!\color{#c00}\iff\! a,b\mid ab/d \!\iff\! lcm(a,b)\mid ab/d \color{#c00}\iff d\mid ab/lcm(a,b)$

Remark $\ $ The red equivalences are $\rm\:x\mid y\color{#c00}\iff y'\mid x'\:$ for $\rm\ x'\! = ab/x\ $ being reflection on the divisors of $\rm\:ab,\:$ highlighting the $\rm\ gcd = lcm' \ $ duality, namely

$$\rm gcd(a,b)\, =\, \frac{ab}{lcm(b,a)}\, =\, lcm(a',b)'\qquad\quad $$

See here for a proof emphasizing this reflection (involution) and the innate duality.

Answer 7

I don't know that you are familiar to the Group theory, but if you consider groups $\mathbb Z_a$ and $\mathbb Z_b$ then the following homomorphism can do what you are looking for. I mean: $$\phi: \mathbb Z\to\mathbb Z_a\times\mathbb Z_b,~~~~n\mapsto(n|_{\text{mod}~a},n|_{\text{mod}~b})$$

Answer 8

Brute Force Approach
We can write $a$ and $b$ in the following manner: $$a=d x_0,\;b=d y_0$$ $$m=a p_0=b q_0$$ Here $d=gcd(a,b)$ and $m=lcm(a,b)$. And both $\{x_0,y_0\}$ and $\{p_0,q_0\}$ are sets of co-prime numbers.

By multiplication, we can write: $$ab=d^2x_0y_0$$ $$ab\cdot p_0 q_0=m^2$$ Multiplying both equations: $$(ab)^2\cdot p_0q_0=(md)^2 \cdot x_0y_0$$ Now we only need to prove that $p_0q_0=x_0y_0$, in fact we can prove that $x_0=q_0$ and $y_0=p_0$.

Using the first two equations: $$\frac{a}{b}=\frac{x_0}{y_0}=\frac{q_0}{p_0}$$ Because $\{x_0,y_0\}$ and $\{p_0,q_0\}$ are sets of co-prime numbers, both $\frac{x_0}{y_0}$ and $\frac{q_0}{p_0}$ are the simplest or irreducible fractions of $\frac{a}{b}$. Hence both the numerators and denominators are equal i.e. $x_0=q_0$ and $y_0=p_0$.
$$\require{cancel} (ab)^2\cdot \cancel{p_0q_0}=(md)^2 \cdot \cancel{x_0y_0}$$ $$(ab)^2=(md)^2 \Rightarrow ab=md$$

---

Thanks for reading!

Answer 9

Easy proof:

$$[p,q] = p*q, \space if \space (p, q) = 1$$ [p, q] = lcm(p, q) and (p, q)=gcd(p, q). This can be proved using only simple property of relatively prime numbers.

Another fact: $$ [mp, mq] = m[p, q] \space if \space m\in N $$ Since common multiple is all about dividable relation, multiply everything by m won't change that.
let $(p, q) = d, so \space (p/d,q/d)=1$ $$ (p,q)[p,q] = d*d*[p/d,q/d]=d*d*\frac{p}{d}*\frac{q}{d}=pq $$

Answer 10

Let $\ell= \text{lcm}(a,b), g=\text{gcd}(a,b)$ for some $a,b$ positive integers.

Division algorithm: exists $q,r$ integers with $0\leq r < \ell$ such $ab = q\ell + r$. Observing that both $a$ and $b$ divide both $ab$ and $q\ell$ we conclude they both divide $r$. As $r$ is a common multiple, we must have $\ell \leq r$ or $r\leq 0$ so $r=0$.

Therefore $s=\frac{ab}{\ell}$ is an integer. Observe that $\frac{a}{s}=\frac{\ell}{b}$ and $\frac{b}{s}=\frac{\ell}{a}$ are both integers, so $s$ is a common divisor. Thus $g\geq s = \frac{ab}{\ell}$.

On the other hand, $\frac{ab}{g}=a\frac{b}{g}=b\frac{a}{g}$ where $\frac{b}{g}$ and $\frac{a}{g}$ are both integers, so $\frac{ab}{g}$ is a common multiple of $a$ and $b$. As $\frac{ab}{g}>0$ we conclude $\frac{ab}{g}\geq \ell$, therefoire $\frac{ab}{\ell}\geq g$ and we are done.

Answer 11

I don't know if it is the easiest proof but here an other solution using the Gauss lemma.

1-
$GCD(a;b)=d \Rightarrow a=dx, b=dy$ with $x,y,d \in \mathbb{N}$ and $GCD(x;y)=1$
More over we can writte$\frac{ab}{d}=ay=bx$ so it cames that the integer $ay=bx$ is a commom multiple of $a$ and $b$ by definition.

2-
Now let note $m=LCM(a;b) \Rightarrow \exists k,l \in \mathbb{N} $ s.t. $ m = ak = bl \Rightarrow \frac{m}{d}=\frac{xdk}{d}=\frac{ydl}{d}=xk=yl \in \mathbb{N}$

3-
Reminder: Gauss lemma: If $a,b,c$ are three integers such that $a$ and $b$ are coprime and $a$ divides $bc$ then $a$ divises $c$.
We take $x,y$ two coprime integers and $l$. As we wrotte above $xk=yl \Rightarrow \frac{yl}{x}=k \in \mathbb{N}$ so $x$ divides $yl=\frac{m}{d}$. Hence by the Gauss lemma we get that $x$ divides $l$. Equivalently we get by the same justification that $y$ divides $k$.
So we can writte: $m=bl=dyl=dyxl'$ with $l=xl'$ and on the other side $m=ak=dxk=dxyk'$ with $k=xk'$

4-
From "-3" we got that $dyxl'= m = dxyk'\Rightarrow l'=k'$. But $m$ is by def the smallest commom multiplier so $l'=1=k'$ and we get that $LCM(a;b)=dxy$.
Rem: Indeed $dxy$ is all ready by def a commom multiplier of $a$ and $b$ because $dxy$ is all ready a multiple of $a$ and $b$ as it is divisible by them!

5-
So: $GCD(a;b)LCM(a;b)=ddxy=d^2xy$ while $ab=dxdy=d^2xy$
Q.E.D.

Answer 12

Easy proof by set theory: (I'm a beginner and thought of this, please correct if me if I'm wrong)

Let $a=xy, b=xz$.

So $a = \{x,y\}$ and $b = \{x, y\}$,

where $x$ is the product of the common prime factors of $a, b$ which is $\gcd(a,b)$ and the product $xyz$ produces $\text{lcm}(a,b)$

By Set Theory,

$P(a \cup b) \cdot P(a \cap b) = P(a) \cdot P(b)$.

$xyz \cdot x = xy \cdot xz$

lcm(a,b) * gcd(a,b) = ab