Geometrical interpretation of the identity $\operatorname{lcm}(a,b) \operatorname{gcf}(a,b) = ab$

Question

Does anyone know a good geometrical representation of the fact that $\DeclareMathOperator{lcm}{lcm}\DeclareMathOperator{gcf}{gcf}\lcm(a,b) \gcf(a,b) = ab$? Because $\lcm$ and $\gcf$ are abstract concepts, it's often difficult to provide insight to the geometry-inclined student and one can easily get lost in mathematical paths that are not familiar. As I see, defining $\lcm(a,b) = \min (\langle a \rangle \cap \langle b \rangle)$ is not enough for the formation of the idea around $\lcm$, which should include the concept of the join for $a,b$ in the lattice $(\mathbb{N},|)$, as oppose to $\gcd(a,b)$, which is their meet.

Let me give one (poor) example: If a rectangle has sides $a$ and $b$, then $\gcf(a,b)$ is the size of the largest square that can evenly cover the rectangle and, counting the number of squares and multiplying by the length of size of the square, one gets the result. Nevertheless, this is not a geometrical operation, this multiplication does not represents clearly any area or segment, and so it's a poor illustration of the kind of appropriate visualization I seek.

Answer 1

I don't see anything wrong with what you were trying to say. In fact, I think that is a good way of saying it. Let's draw some pictures with some concrete numbers. Let's take $a=6$ and $b=9$. Then we can draw an $a\times b$ grid as shown below.

Now as you said we can show the greatest common divisor $d$ is the side length of the largest square that can evenly cover the rectangle. This is shown in the following picture.

Below, the area of the black region, which contains only the diagonal of each $d \times d$ subsquare and is therefore $1/d$ of the total area, i.e., $ab/d$, illustrates the least common multiple.

That the black area is a multiple of $a$ and $b$ can be seen from the fact that they comprise an integer number of rows and columns as seen in the following gifs.

These pictures are only an illustration and do not prove that the quantities shown are in fact the greatest common divisor and least common multiple.

Answer 2

The OP describes a geometrical illustration using the fact that the GCD is the side of the greatest square that tiles a $A \times B$ rectangle. I don't understand why the following is not a "geometrical operation." But perhaps it is slightly different than what the OP had in mind, since he claimed "this multiplication does not represent...any area...," but it does below.

You can even find the GCD square with Euclid's algorithm using a compass and straightedge.

Answer 3

I'll hazard a probably uninteresting answer. Here's an example:

Let's run through how this goes in general. Suppose $a,b$ are positive integers such that they share prime-power divisors $p_1^{r_1},\dots, p_m^{r_m}$ (we assume each prime power is as large as is possible). Furthermore, suppose $$ a = p_1^{r_1+s_1}\cdots p_m^{r_m+s_m}a' \qquad \& \qquad b = p_1^{r_1+t_1}\cdots p_m^{r_m+t_m}b' $$ where $a',b'$ are not divided by any of the primes $p_1,\dots , p_m$ and $s_1,\dots, s_m,t_1, \dots, t_m \geq 0$. Notice that $$ gcd(a,b) = p_1^{r_1}\cdots p_m^{r_m} $$ Also, $$ a = p_1^{r_1}\cdots p_m^{r_m}p_1^{s_1}\cdots p_m^{s_m}a' \qquad \& \qquad b = p_1^{r_1}\cdots p_m^{r_m}p_1^{t_1}\cdots p_m^{t_m}b' $$ The least common multiple of $a,b$ needs to carry all the prime powers in the factorizations of $a,b$. However, the lcm should not carry more prime power factors than are needed. In particular, we see $p_1^{r_1}\cdots p_m^{r_m}$ only needs to appear once. In my current notation we find: \begin{align} lcm(a,b) &= p_1^{r_1}\cdots p_m^{r_m}p_1^{s_1}\cdots p_m^{s_m}a'p_1^{t_1}\cdots p_m^{t_m}b' \\ &= \left(p_1^{r_1}\cdots p_m^{r_m}p_1^{s_1}\cdots p_m^{s_m}a' \right) \left(p_1^{t_1}\cdots p_m^{t_m}b' \right) \\ &= a \left( p_1^{t_1}\cdots p_m^{t_m}b' \right) \\ &= \left(p_1^{s_1}\cdots p_m^{s_m}a'\right)\left(p_1^{r_1}\cdots p_m^{r_m}p_1^{t_1}\cdots p_m^{t_m}b'\right) \\ &= \left(p_1^{s_1}\cdots p_m^{s_m}a'\right)b. \end{align} it is clear from the above calculation that $p_1^{r_1}\cdots p_m^{r_m}p_1^{s_1}\cdots p_m^{s_m}a'p_1^{t_1}\cdots p_m^{t_m}b'$ is a multiple of both $a$ and $b$. Minimality requires further analysis which I omit here. In any event, it is clear that: \begin{align} ab &= \left( p_1^{r_1}\cdots p_m^{r_m}p_1^{s_1}\cdots p_m^{s_m}a' \right)\left( p_1^{r_1}\cdots p_m^{r_m}p_1^{t_1}\cdots p_m^{t_m}b' \right) \\ &= \left( p_1^{r_1}\cdots p_m^{r_m}\right)\left(p_1^{r_1}\cdots p_m^{r_m}p_1^{s_1}\cdots p_m^{s_m}a'p_1^{t_1}\cdots p_m^{t_m}b' \right) \\ &= gcd(a,b)lcm(a,b). \end{align}

Geometrically: if we take an $a \times b$ rectangle and form another rectangle of the same area whose height is $gcd(a,b)$ then the length of the squished rectangle is exactly the $lcm(a,b)$.

Answer 4

During lockdown we hoarded toilet papers, so I piled up some like this and explained $gcd\cdot lcm=ab$ to my kid.