Prove by the Gauss lemma that if $a, b \in \mathbb{N} \Rightarrow $ the product $ab$ is equal to the multiplication of $GCD(a;b)$ with $LCM(a;b)$.

Question

Question:

Prove with the help of the Gauss lemma that if $a$ and $b$ are two integers then the product $ab$ is equal to the multiplication of $GCD(a;b)$ by $LCM(a;b)$.

My attempt:

1-
$GCD(a;b) = d \Rightarrow a=dx, b=dy$ with $GCD(x;y)=1$ and $x,y,d \in \mathbb{N}$
So we can writte $ab=dxdy=d^2xy$.

2-
Reminder: Gauss lemma: If $a,b,c$ are three integers such that $a$ and $b$ are coprime and $a$ divides $bc$ then $a$ divises $c$.
Now we writte $LCM(a;b)=lxy$ with $GCD(x;y)=1$ and $x,y,l \in \mathbb{N}$. As $x$ and $y$ are coprime i would like to use the Gauss lemma, but I can't continue since $x$ doesn't divide necessarly $ly$ and this in order to prove that $d=l$ and so finish the prove In short I am stuck any help will be appreciated.

Thank you.

Answer 1

I have tryed to re think this question from the beginning and this is what I get. I hope that it is correct.

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 2

I will exclude the trivial cases $a=0$ or $b=0$. If $a$ and $b$ are coprime, then the least common multiple coincides with $a \cdot b$. As for the general case, you can conclude by noticing that $\frac{a}{\text{GCD}(a,b)}$ and $\frac{b}{\text{GCD}(a,b)}$ are coprime. Thus,

$$ \frac{a}{\text{GCD}(a,b)} \cdot \frac{b}{\text{GCD}(a,b)} = \text{LCM}\left(\frac{a}{\text{GCD}(a,b)}, \frac{b}{\text{GCD}(a,b)}\right) $$

By multiplying both sides by $\text{GCD}(a,b)^2$, you obtain

$$ a \cdot b = \text{GCD}(a,b)^2 \cdot \text{LCM}\left(\frac{a}{\text{GCD}(a,b)}, \frac{b}{\text{GCD}(a,b)}\right) = \text{GCD}(a,b) \cdot \text{LCM}(a,b). $$

In the last step, I used the property

$$ \text{GCD}(a,b) \cdot \text{LCM}\left(\frac{a}{\text{GCD}(a,b)}, \frac{b}{\text{GCD}(a,b)}\right) = \text{LCM}(a,b), $$

which intuitively seems true. Although I haven't rigorously proven it, I visualized the prime factorization of the two integers and it appeared to hold. When you divide both $a$ and $b$ by their GCD, you eliminate the common powers of primes with the least exponent. The least common multiple between $\frac{a}{\text{GCD}(a,b)}$ and $\frac{b}{\text{GCD}(a,b)}$ has the same factorization as $\text{LCM}(a,b)$ except for those common powers of primes with the least exponent that were subtracted earlier. By multiplying by $\text{GCD}(a,b)$, you reintroduce those missing powers, resulting in the recovery of $\text{LCM}(a,b)$. Does this explanation make sense? I assumed you were open to any form of assistance based on your statement.