Is $\mbox{lcm}(a,b,c)=\mbox{lcm}(\mbox{lcm}(a,b),c)$?

Question

$\newcommand{\lcm}{\operatorname{lcm}}$Is $\lcm(a,b,c)=\lcm(\lcm(a,b),c)$?

I managed to show thus far, that $a,b,c\mid\lcm(\lcm(a,b),c)$, yet I'm unable to prove, that $\lcm(\lcm(a,b),c)$ is the lowest such number...

Answer 1

Let the highest power of prime $p$ in $a,b,c$ be $A,B,C$ respectively.

The highest power of prime $p$ in lcm$(a,b,c)=$max$(A,B,C)$

The highest power of prime $p$ in lcm$(a,b)=$max$(A,B)$

The highest power of prime $p$ in lcm$($lcm$(a,b),c)=$max$($max$(A,B),C)$

Can you see max$($max$(A,B),C)=$max$(A,B,C)$ ?

This holds true for any prime that divides at least one of $a,b,c$

Answer 2

The "universal property" of the $\newcommand{\lcm}{\operatorname{lcm}}\lcm$ is

if $\lcm(a,b) \mid x$, and $c \mid x$, then $\lcm(\lcm(a,b),c) \mid x$.

Make a good choice for $x$, for which you can prove the hypothesis and/or the conclusion is useful.

Answer 3

What you still want to show is that $lcm(lcm(a,b),c)|lcm(a,b,c)$. So what you want to show is that the least, or actually any, common multiple of $a$, $b$, and $c$ is a multiple of the least common multiple of $lcm(a,b)$ and $c$. This follows from the fact that a common multiple of two numbers is a multiple of their least common multiple.

Answer 4

With this type of problem, it's often useful to be aware of a particular order on $\Bbb N_{>0}$ -- at least if one is familiar with posets (partially ordered sets).

Namely, the poset $(\Bbb N_{>0}, \mid)$, where $a \mid b$, as usual, denotes that $a$ divides $b$. (One may readily see/prove that this indeed forms a poset.)

This order has the following properties:

• $1$ is the least element;
• $\gcd(a,b)$ is the greatest lower bound or infimum of $a$ and $b$: If $m \mid a$ and $m \mid b$, then $m \mid \gcd(a,b)$;
• ${\rm lcm}(a,b)$ is the least upper bound or supremum of $a$ and $b$: If $a \mid n$ and $b \mid n$, then ${\rm lcm}(a,b) \mid n$.

which together make it a so-called lower-bounded lattice.

By the general theorem that in any lattice, for a collection of sets $(A_i)_i$ such that $\bigcup_i A_i$ is finite, we have $\sup_i \sup A_i = \sup \bigcup_i A_i$, we are done immediately:

Taking $A_1 = \{a,b\}$ and $A_2 = \{c\}$ (note that $\sup A_2 = c$), we find that: $${\rm lcm}({\rm lcm}(a,b),c) = \sup\{\sup\{a,b\},\sup\{c\}\} = \sup \{a,b,c\} = {\rm lcm}(a,b,c)$$

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Of course, to be able to use this argument, one requires some knowledge about lattices and posets. If you don't understand it at this point, don't worry: when you progress in mathematics, you're bound to encounter these concepts, after which you hopefully can appreciate this argument.

Answer 5

Lemma:
Let $l:=\operatorname{lcm}(a,b,\dots,z)$.
Let $m$ be a common multiple of $a,b,\dots z$.
Then, $m$ is a multiple of $l$.

Proof:
We can write $m=ql+r$, $0\leq r<l$.
Then, $r=m-ql$.

Both $m$ and $l$ are a multiple of $a$.
So, $r$ is a multiple of $a$.
Both $m$ and $l$ are a multiple of $b$.
So, $r$ is a multiple of $b$.
$\dots$
Both $m$ and $l$ are a multiple of $z$.
So, $r$ is a multiple of $z$.

So, $r$ is a common multiple of $a,b\dots, z$.
If $0<r<l$, then $r$ is a common multiple of $a,b\dots, z$ which is less than the least common multiple of $a,b\dots, z$.
This is a contradiction.
So, $r$ must be $0$.
So, $m=ql+r=ql$.
So, $m$ is a multiple of $l$.

$\operatorname{lcm}(a,b,c)=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$.
Proof:
Let $l:=\operatorname{lcm}(a,b,c)$ and $l^{'}:=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$.
$l$ is a common multiple of $a$, $b$ and $c$.
So, $l$ is a common multiple of $a$ and $b$.
So, by the above lemma, $l$ is a multiple of $\operatorname{lcm}(a,b)$.
$l$ is a multiple of $c$.
So, $l$ is a common multiple of $\operatorname{lcm}(a,b)$ and $c$.
So, by the above lemma, $l$ is a multiple of $l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$.

$l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$ is a multiple of $\operatorname{lcm}(a,b)$.
$\operatorname{lcm}(a,b)$ is a multiple of $a$.
So, $l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$ is a multiple of $a$.
$l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$ is a multiple of $\operatorname{lcm}(a,b)$.
$\operatorname{lcm}(a,b)$ is a multiple of $b$.
So, $l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$ is a multiple of $b$.
$l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$ is a multiple of $c$.
So, $l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$ is a common multiple of $a$,$b$ and $c$.
So, by the above lemma, $l^{'}=\operatorname{lcm}(\operatorname{lcm}(a,b),c)$ is a multiple of $l=\operatorname{lcm}(a,b,c)$.

$l$ is a multiple of $l^{'}$ and $l^{'}$ is a multiple of $l$.
So, $l=l^{'}$.