Construction of Geometric Sequence in Lowest Terms/Proof 1

Theorem
It is possible to find a geometric sequence of integers $G_n$ of length $n+1$ with a given common ratio such that $G_n$ is in its lowest terms.

Proof
Let $r = \dfrac a b$ be the given common ratio.

Let the required geometric sequence have a length of $4$.

Let $a^2 = c$.

Let $a b = d$.

Let $b^2 = e$.

Let:
 * $a c = f$
 * $a d = g$
 * $a e = h$

and let:
 * $b e = k$

As:
 * $a^2 = c$
 * $a b = d$

it follows from that:
 * $\dfrac a b = \dfrac c d$

As:
 * $a b = d$
 * $b^2 = e$

it follows from that:
 * $\dfrac a b = \dfrac d e$

As:
 * $a c = f$
 * $a d = g$

it follows from that:
 * $\dfrac c d = \dfrac f g$

As:
 * $a d = g$
 * $a e = h$

it follows from that:
 * $\dfrac d e = \dfrac g h$

As:
 * $a e = h$
 * $b e = k$

it follows from that:
 * $\dfrac a b = \dfrac h k$

Putting the above together:
 * $c, d, e$ are in geometric sequence with common ratio $\dfrac a b$
 * $f, g, h, k$ are in geometric sequence with common ratio $\dfrac a b$

We have that $a$ and $b$ are the smallest numbers with the same ratio.

So by :
 * $a \perp b$

where $\perp$ denotes coprimality.

We also have that:
 * $a^2 = c, b^2 = e$

and:
 * $a c = e, b e = k$

so by :
 * $c \perp e$
 * $f \perp k$

But from, these are the least of those with the same common ratio.

Hence the result.