proof concerning "Neighboring fractions"

Question

Brushing up on my Algebra skills with the book "Algebra" by I.M. Gelfand and reached a problem I am unable to solve.

Fractions $a/b$ and $c/d$ are called neighboring fractions if their difference $\frac{ad-bc}{bd}$ has numerator $\pm1$, that is, $ad-bc = \pm1$.

Prove (a) In this case neither fraction can be simplified (that is, neither has any common factors in numerator or denominator).

What is a correct answer to this problem? Thank you.

Answer 1

Assume one can be simplified, so for example $a=pk,b=qk$ with $k \gt 1$. Plug that in and conclude you cannot have $ad-bc=\pm 1$

Answer 2

Thank you Ross Millikan. This is a proof by contrapostive as I understand it. Is the following correct?

Assume $a = pk$ and $ b = qk$ with $k > 1$. Then,

$$ {a\over b} - {c\over d} = {pk\over qk} - {c\over d} = {{dpk - qkc} \over qkd} = {k({dp-cq}) \over bd} $$ ${k({dp-cq})} \ne \pm1$
Q.E.D.