Proof concerning "neighbor fractions"

Hello, I am working with Israel Gelfand's book "Algebra" and have another question about one of the exercises!

**Fractions $\displaystyle \frac{a}{b}$ and $\displaystyle \frac{c}{d}$ are called neighbor fractions if their difference $\displaystyle \frac{ad-bc}{bd}$ has numerator ±1, that is, ad–bc = ±1. Prove that**

**(a)** in this case neither fraction can be simplified (that is, neither has any common factors in numerator and denominator);

**(b)** if $\displaystyle \frac{a}{b}$ and $\displaystyle \frac{c}{d}$ are neighbor fractions, then $\displaystyle \frac{a+b}{c+d}$ is between them and is a neighbor fraction for both $\displaystyle \frac{a}{b}$ and $\displaystyle \frac{c}{d}$; moreover,

**(c)** no fraction $\displaystyle \frac{e}{f}$ with positive integer $\displaystyle e$ and $\displaystyle f$ such that f < b+d is between $\displaystyle \frac{a}{b}$ and $\displaystyle \frac{c}{d}$.

I understand that this is a three-part proof and so I'm not expecting any complete answers, but just some comments or tips in the right direction would help immensely.

For (a), I understand how it is true and I can sort of describe why ("multiples of factors greater than one will either be identical or will be separated by a distance greater than one, so if x is the factor then a(x)d–b(x)c will either equal 0 or have an absolute value greater than the absolute value of one"), but I'm not sure how to write this up or if it's totally accurate.

For (b) and (c) I am confused because from the examples I tried it seems like $\displaystyle \frac{c+d}{a+b}$ should be the one in between $\displaystyle \frac{a}{b}$ and $\displaystyle \frac{c}{d}$ and not $\displaystyle \frac{a+b}{c+d}$. Also it seems like (c) should be a stipulation for (b) (and not just an addition to it), since there are pairs like $\displaystyle \frac{4}{9}$ and $\displaystyle \frac{3}{7}$ which are neighbor fractions but for which neither (b) nor (c) are true.

Any help is greatly appreciated!