Proof that lcm(a,b) = ab/gcd(a,b)

Question

I'm having trouble completing a proof that for positive integers a and b, that the least common multiple of a and b is ab/gcd(a,b).This is how I've approached it so far: For s = lcm(a,b) we have the following definition: i) a|s and b|s ii) for any integer k where a|k and b|k, s|k

Thus the proof comes down to proving that s satisfies these two properties. For i) let gcd(a,b) = d. Thus s = ab/d. Since gcd(a,b) = d, it follows that b/d is an integer so a|ab/d since ab/d = ta for t = b/d. The same conclusion can be made for b so i) is satisfied. Now comes the confusion. I'm not not sure how to prove that if a|k and b|k then ab/d|k. I tried using am = k and bn = k for integers m and n but it lead to no avail. Should I approach the problem som other way?

Answer 1

Let $x,y$ positive integers and $\{p_1,...,p_r\}$ the primes which divide $x$ or $y$. We can write:

$$x=p_1^{\alpha_1},...,p_r^{\alpha_r}$$

$$y=p_1^{\beta_1},...,p_r^{\beta_r}$$

where $\alpha_1,...,\alpha_r,\beta_1,...,\beta_r$ are integers non-negatives and take $z_i:=\max\{\alpha_i,\beta_i\} \geq 1$ and $w_i:=\min\{\alpha_i,\beta_i\}$. Then:

$$\newcommand{\lcm}{\operatorname{lcm}}xy=\gcd(x,y) \cdot \lcm(x,y)$$ \

Well, we know that $d:=\gcd(x,y)$ divides $x$ or $y$. Moreover, if $c\in\mathbb{Z}$ is positive and divides $x$ and $y$, then we can take $e_1,e_2\in\mathbb{Z}$ such that:

$$ce_1=x=p_1^{\alpha_1},...,p_r^{\alpha_r}$$

$$ce_2=y=p_1^{\beta_1},...,p_r^{\beta_r}$$

So the prime divisors of $c$ are in ${p_1,...,p_r}$ and the exponent is compressed from $\alpha_i$ and $\beta_i$. Thus, $c$ divides $d$. It proofs that $d=\gcd(x,y)$. The argument for the $M=\lcm(x,y)$ is similar. Like $\alpha_i+\beta_i=w_i+z_i$, we have:

$$Md=p_1^{z_1},...,p_r^{z_r}p_1^{w_1},...,p_r^{w_r}=p_1^{z_1+w_1},...,p_r^{z_r+w_r}=p_1^{\alpha_1+\beta_1},...,p_r^{\alpha_r+\alpha_r}=xy$$