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Prove: $$ \text{lcm}(x,y)=\frac{|x\cdot y|}{\text{gcd}(x,y)}$$

I used many ways to do it, all failed. One of them was to represent $|x\cdot y|$ as a sum of primes then $\text{gcd}(x,y)$ as a sum of primes and do the operation but I ended up with a false result since some prime factors were left.

So what is the simplest proof for the gcd to lcm relation?

Martin Sleziak
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    http://math.stackexchange.com/questions/44835/for-integers-a-and-b-ab-textlcma-b-cdot-texthcfa-b – njguliyev Oct 20 '13 at 16:08
  • This is in all the books and there are duplicates on the related links to the right. – Git Gud Oct 20 '13 at 16:08
  • @GitGud I have a poor textbook that's the main problem, I will try to buy another in addition to the one obligated by my stupid school. –  Oct 20 '13 at 16:31

1 Answers1

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Let $\gcd(x,y)=d$ and, $x=dx_1$ and $y=dy_1$.

We show that $\text{lcm}(x,y)=dx_1y_1$.

Let $x$ and $y$ divide $m$.

Then $m$ is divisible by $d$.

Write $m=dm_1$.

So, $x_1,y_1|m_1$.

Since $\gcd(x_1,y_1)=1$, there exists integers $a$ and $b$ such that $ax_1+by_1=1$.

This gives $m_1ax_1+m_1ay_1=m_1$.

But $x_1y_1$ divides the LHS and hence the RHS too.

So $x_1y_1$ divides $m_1$.

Thus $dx_1y_1|dm_1$.

Hence $dx_1y_1|m$.

So $m\geq dx_1y_1$.

We have show than any multiple of both $x$ and $y$ is greater than or equal to $dx_1y_1$.

Noting that $x$ and $y$ divide $dx_1y_1$, we conclude that $dx_1y_1$ is the lcm of $x$ and $y$.

From here the rest is trivial.

caffeinemachine
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