Proof: By Euclid

Let $A$ be a medial (straight line).
I say that an infinite (series) of irrational (straight lines) can be created from $A$, and that none of them is the same as any of the preceding (straight lines).

fig115e

Let the rational (straight line) $B$ be laid down.
And let the (square) on $C$ be equal to the (rectangle contained) by $B$ and $A$.
Thus, $C$ is irrational [Def. 10.4] .
For an (area contained) by an irrational and a rational (straight line) is irrational [Prop. 10.20].
And ($C$ is) not the same as any of the preceding (straight lines).
For the (square) on none of the preceding (straight lines), applied to a rational (straight line), produces a medial (straight line) as breadth.
So, again, let the (square) on $D$ be equal to the (rectangle contained) by $B$ and $C$.
Thus, the (square) on $D$ is irrational [Prop. 10.20].
$D$ is thus irrational [Def. 10.4] .
And ($D$ is) not the same as any of the preceding (straight lines).
For the (square) on none of the preceding (straight lines), applied to a rational (straight line), produces $C$ as breadth.
So, similarly, this arrangement being advanced to infinity, it is clear that an infinite (series) of irrational (straight lines) can be created from a medial (straight line), and that none of them is the same as any of the preceding (straight lines).
(Which is) the very thing it was required to show.
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