Solutions To Mathematics Textbooks/Algebra (9780817636777)/Exercises 26-50

a
If a and b had a nontrivial common factor k >= 2, then a = k*a' and b = k*b', so (ad - bc) = k(a'd - b'c) = ±1.

Alternatively, you must essentially show that a and b are coprime; that is the numerator and denominator share no common factor. Another way of saying this is to say that $$gcd(a, b) = 1$$.

Let $$g = gcd(a, b)$$. We can write $$ad-bc = \pm 1$$ as $$\left(\frac{a}{g} \cdot gd - \frac{b}{g} \cdot gc\right) = g \left(\frac{a}{g}d - \frac{b}{g}c \right) = \pm 1$$. Thus $$g$$ must be either -1 or 1, and thus a and b are coprime.