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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.

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  • $\begingroup$ What is your target audience? Elementary school students? High school students? $\endgroup$ – Joel Reyes Noche Apr 28 at 3:17
  • $\begingroup$ I was thinking in a broad perspective actually. A geometrical interpretation could be used with children (11+), high schoolers and adult learners alike. $\endgroup$ – Jonas Gomes Apr 28 at 3:46
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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.

enter image description here

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.

enter image description here

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.

enter image description here

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.

enter image description here enter image description here

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.

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    $\begingroup$ Thank you, that last two gifs were precisely what I was looking for as they give a geometric intuition of why the segment is equal the lcm. I'll adapt your argument and use in my class and possibly in a video. How can I cite your answer? $\endgroup$ – Jonas Gomes Apr 29 at 16:51
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    $\begingroup$ I don't know how to cite answers, and it doesn't really matter to me personally. The tikz source for the pictures and gifs are in my post as a comment, just click edit to see it. You can play around if you want. $\endgroup$ – Brian Moths Apr 29 at 21:06
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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.

enter image description here

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

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  • $\begingroup$ Thank you for your answer. My reservation towards this is because I didn´t have a clear justification as to why this segment has length equals LCM $\endgroup$ – Jonas Gomes Apr 29 at 16:46
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    $\begingroup$ @JonasGomes Ah, I thought about that. But you asked for an "interpretation," not a proof. I think the GCD is clear (or not?). This also seemed clear: The rows of $A\times B$ are put together to form LCM; similarly the columns of $A\times B$ can be stacked to form a rectangle of height LCM — well that's half the proof, but that's all the other answer gives, too. $\endgroup$ – user1027 Apr 29 at 19:36
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I'll hazard a probably uninteresting answer. Here's an example: enter image description here

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 $lcm(a,b)$.

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

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