In the paper Safe Prime Generation with a Combined Sieve by Michael J. Wiener, the author states:
For any small odd prime $r$, we can eliminate candidates for $q$ that are congruent to $(r − 1)/2\mod r$ because they lead to $p$ being divisible by $r$.
(where $p = 2q + 1$)
But he did not explain why for the uninitiated. As such I would like to know why $(p-1)/2 \equiv (r-1)/2 \mod r$ means that $p \equiv 0 \mod r$.
Or to put it another way, why does $q \equiv (r-1)/2 \mod r$ mean that $2q \equiv -1 \mod r$?