I have posted an exercise below about basic proof writing with fractions and I would appreciate if you could check my work/comment on the logical structure of my proofs.
Problem
Fractions $\frac{a}{b}$,$\frac{c}{d}$ are called neighbor fractions if their difference $\frac{ad-bc}{bd}$ has numerator $±1$, that is, $ad-bc=±1$.
Prove that:
(a) in this case neither fractions can be simplified.
(b) if $\frac{a}{b}$,$\frac{c}{d}$ are neighboring fractions, then $\frac{a+c}{b+d}$ is between them and is a neighboring fraction for both $\frac{a}{b}$ and $\frac{c}{d}$.
(c) no fraction $\frac{e}{f}$ with positive integer $e$ and $f$ such that $f<b+d$ is between $\frac{a}{b}$ and $\frac{c}{d}$.
Proof of (a):
Assume that $\frac{a}{b}$ can be simplified. Then there exist $k,p,q\not= 0\inℤ$ s.t. $a=kp$ and $b=kq$. Substituting in for $a$ and $b$ we obtain, $kpd-kqc=±1$. Factoring out $k$ we get $k(pd-qc)=±1$. Now, $k\not=0$, and the product with $k\not=0$, so $(pd-qc)\not=0$. Therefore, k=$±1/(pd-qc)$ with $(pd-qc)\in ℤ$. So $k\notinℤ$, a contradiction. An analogous argument shows for $\frac{c}{d}$. $\blacksquare$
Proof of (b):
Without loss of generality, assume $\frac{a}{b}<\frac{c}{d}$. It follows that $ad<bc$. Adding $dc$ to both sides we get $ad+dc<bc+dc$, and then factoring out $d$ from LHS and $c$ from RHS we get $d(a+c)<c(b+d)$ and so $\frac{a+c}{b+d}<\frac{c}{d}$. Likewise, adding $ab$ to both sides we get $ad+ab<bc+ab$, and then factoring out $a$ from LHS and $b$ from RHS we get $a(b+d)<b(c+a)$ and so $\frac{a}{b}<\frac{a+c}{b+d}$.
By the definition of neighboring fractions, we need to show $a(b+d)-b(a+c)=±1$ and $d(a+c)-c(b+d)=±1$. Because $\frac{a}{b}$,$\frac{c}{d}$ are neighboring fractions, we get $ad-bc=±1$. It follows that $-bc=±1-ad$ and $ad=±1+bc$.
$a(b+d)-b(a+c)=ab+ad-ba-bc=ad-bc=±1+bc-bc=±1$
$d(a+c)-c(b+d)=da+dc-cb-cd=da-cb=da±1-ad=±1$ $\blacksquare$
Proof of (c):
How would you show part (c)?
Best Answer
Your proof of (a): I'll ignore the typos for now, and focus on the maths. You need to assume $k\neq \pm 1$ as well (otherwise, you're not simplifying at all). Then, (WLOG that $ad-bc = 1$ and $k=1$) the contradiction comes from that $k$ is forced to be an integer between $0$ and $1$.
Your proof of (b): Both parts look fine.
For part c: Are you assuming that $\frac{a}{b}$ and $\frac{c}{d}$ are neighbor fractions? Otherwise, there are counterexamples such as $\frac12 < \frac35 < \frac34$, for example.
Good work on writing this up. You've asked a good question.