Proof: By Euclid
(related to Proposition: 7.29: Prime not Divisor implies Coprime)
 For if $B$ and $A$ are not prime to one another then some number will measure them.
 Let $C$ measure (them).
 Since $C$ measures $B$, and $A$ does not measure $B$, $C$ is thus not the same as $A$.
 And since $C$ measures $B$ and $A$, it thus also measures $A$, which is prime, (despite) not being the same as it.
 The very thing is impossible.
 Thus, some number cannot measure (both) $B$ and $A$.
 Thus, $A$ and $B$ are prime to one another.
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"