# How to prove a claim about LCM?

Prove that LCM$$(\frac{n}{a}, \frac{n}{b}) = n$$ if $$(a,b)=1$$.

Let us assume the factorization of $$n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$$ and $$a=q_1^{\beta_1} q_2^{\beta_2} \cdots q_k^{\beta_k}$$ and $$b = r_1^{\gamma_1} r_2^{\gamma_2} \cdots p_k^{\gamma_k}$$.

Please note that both $$a$$ and $$b$$ divide $$n$$.

Case 1: If both $$a$$ and $$b$$ divide $$n$$ then I am getting that $$\frac{n}{a}$$ will divide $$n$$ and $$\frac{n}{b}$$ will also divide $$n$$ but why it will be the minimum multiple?

• I only know the lcm of integers. what is the lcm of two fractions? Feb 7 '18 at 13:50
• @miracle173 this doesn’t arise, since $a,b|n$ Feb 7 '18 at 13:55
• @StellaBiderman I see now that this was introduced in the second version of post. Feb 7 '18 at 14:06
• What do you mean? $2$ and $3$ divide $6$. Feb 7 '18 at 14:06
• @StellaBiderman sorry, I misread the post Feb 7 '18 at 14:08

I find it easier to work with gcd. The following 2 lemmas are useful and not hard to prove.

Lemma 1. If $(a,b) = 1$, then $(ak,bk) = k$ for any integer $k$.

Lemma 2. $\gcd(a,b) \text{ lcm}(a,b) = ab$.

Given these two results, we can now prove as follows: let $k = \frac{n}{ab}$ in Lemma 1. Note that $ab$ divides $n$ since $a$ and $b$ are coprime. Then $\gcd(\frac{n}{a}, \frac{n}{b} ) = ( \frac{bn}{ab}, \frac{an}{ab}) = \frac{n}{ab}$. Now use lemma 2: $$\frac{n^2}{ab} = \frac{n}{a} \frac{n}{b} = \gcd(\frac{n}{a}, \frac{n}{b}) \text{lcm}(\frac{n}{a}, \frac{n}{b}),$$ and since $\gcd = \frac{n}{ab}$ we get $\text{lcm} = n$.

Suppose $n/a$ divides $x$ and $n/b$ divides $m$, say $m=xn/a=yn/b$, for some integers $x,y$.

Then we have $xb=ya$. Since $a$ is co-prime to $b$, this implies $a$ divides $x$; similarly $b$ divides $y$. Consequently $m=xn/a=n\cdot x/a$ is a multiple of $n$.

Thus $n$ is the least common multiple of $n/a$ and $n/b$.

Hope this helps.