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CLAIM : Prove that LCM$(n/a , n/b) = n$ 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 $b$ divides $n$.

case 1: if both $a$ and $b$ divides $n$ then I am getting that $n/a$ will divid the $n$ and $n/b$ will also divide $n$ but why it will be the minimum multiple ?

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

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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 $.

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