Geometrical proof that product of two numbers both relatively prime to $n$ is also relatively prime to $n$ I wonder if there is an elementary and geometrical proof of the following very basic fact.
Let $(k, n) = 1$ and $(a, n) = 1$ for some $a, k, n \in \mathbb{N}$. Then, it is also $(ak, n) = 1.$ (This is obviously connected with Euler's theorem whose proof I relearned yesterday for some reason.)
I tried this with lines (gcd of two lines, $x, y$, would then be the "longest" line which fits "nicely" in both lines $x, y$), but I cannot find obvious geometrical way to show this.
I would prefer you prove this with lines, because maybe I am missing something obvious, but other geometrical (visual) methods are also just fine.
I observed that I just love to torture myself with this kind of questions. :/
Anyway, thanks!
 A: Consider lines with lenghts $k = 18 \cdot 1$, $n = 245 \cdot 1, a = 24 \cdot 1.$ We can prove, case by case,  that line with maximal lenght that divides $k, n$ is the unit lenght. Simillary for $a, n.$
In other words, we know $(k, n) = (a, n) = 1.$
Now, consider line $p$ with lenght $ak = 24 \cdot 18$. We can visualize this as 24 distinct segments of line $p$ with each has $18$ unit lenghts. So, 24 sublines (groups) of 18 objects - unit lengths. Now, suppose $(a, n) = d > 1.$
We have 3 cases.

*

*$1 < d < 18$. Now it must be $d | 18$ or $d | 24$ because otherwise $d$ would not divide $p$. Namely, if $d$ does not divide $18$ then $2d$ $3d, 4d, ...$, must reach endpoints of $18 \cdot 1$,
or $18 \cdot 2$, ..., or $18 \cdot 24$ (divisors of $24$) - if not $d$ obviously wouldn't divide $p$. But then such maximal $d$ would be in form $18 \cdot 1$ or $18 \cdot 2$, ..., or $18 \cdot 24$ which is greater than 18, which gives contradiction. (And obviously $18 \cdot a = a \cdot 18$ can't divide $n = 245$.)

*$d = 18$. This obviously doesn't work because $(18, 245) = 1$.

*$d > 18.$ Obviously, $d$ must be in form $d = a \cdot 18$ where $a | 24$. (So, $d$ cannot be for example $20$ because then $d$ would not divide $p$). But then $d$ cannot divide $n = 245$ because $(a, n) = 1$. Contradiction.

Basically, in each case $d$ must be in form $k \cdot l, l \in \{1, 2, 3, 4, 5, ..., a\}$ which is not possible to do because $kl | a$ but $kl = l \cdot k$ can't fit nicely into $n$ because $k$ can't in the first place - so can't any multiple of it.
It must be $d = 1$. Therefore $(ak, n) = 1$.
I used only lengths and visual arguments here (divisors as groups of lenghts or groups of groups of lenghts - this would be analogous to factorization in arithmetic).
EDIT. Wow, Euclid also thought about this. His proof:
https://mathcs.clarku.edu/~djoyce/java/elements/bookVII/propVII24.html.
