Check my proof: Showing how $\frac{c-a}{d-b}$ compares to $\min(\frac{a}{b},\frac{c}{d})$ and $\max(\frac{a}{b},\frac{c}{d})$

Question

This is a follow-up inquiry to a (now-deleted) MSE question.

Here is my initial query:

Where is the fraction $\dfrac{c-a}{d-b}$ located with respect to $$\text{min}\left(\dfrac{a}{b},\dfrac{c}{d}\right)$$ and $$\text{max}\left(\dfrac{a}{b},\dfrac{c}{d}\right),$$ assuming $a$, $b$, $c$, and $d$ are all positive integers?

MY ATTEMPT

WLOG, we may assume that $b \neq 0$, $d \neq 0$ and $b \neq d$. Also, WLOG we can assume that $$\dfrac{a}{b} \leq \dfrac{c}{d}.$$

IF the equation $$\dfrac{a}{b} = \dfrac{c}{d}$$ holds, then we obtain $$\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{a - c}{b - d}$$ since the last relationship holds if and only if $$\Bigg(a(b - d) = b(a - c)\Bigg) \land \Bigg(c(b - d) = d(a - c)\Bigg)$$ which is true if and only if $$\Bigg(ad = bc\Bigg) \land \Bigg(ad = bc\Bigg),$$ something which obviously holds.

Now, I claim that IF the inequality $$\dfrac{a}{b} < \dfrac{c}{d}$$ holds, then we have the inequality $$\text{min}\left(\dfrac{a}{b},\dfrac{c}{d}\right) = \dfrac{a}{b} < \dfrac{c - a}{d - b} < \dfrac{c}{d} = \text{max}\left(\dfrac{a}{b},\dfrac{c}{d}\right).$$

PROOF

Since $a$ and $b$ are positive (integers), then $$0 < \dfrac{a}{b} < \dfrac{c - a}{d - b} < \dfrac{c}{d}.$$

Subtracting and simplifying, we obtain $$\dfrac{a}{b} - \dfrac{c - a}{d - b} = \dfrac{ad - ab - bc + ab}{b(d - b)} = \dfrac{ad - bc}{b(d - b)} < 0,$$ which means that $b < d$. This implies that $a < c$.

We may now cross-multiply (without worrying about the signs): $$\dfrac{a}{b} < \dfrac{c - a}{d - b} \iff ad - ab = a(d - b) < b(c - a) = bc - ab \iff ad < bc$$ which is obviously true; and $$\dfrac{c - a}{d - b} < \dfrac{c}{d} \iff cd - ad = d(c - a) < c(d - b) = cd - bc \iff bc < ad,$$ which is false.

CONCLUSION

Hence, if the inequality $$\dfrac{a}{b} < \dfrac{c}{d}$$ holds, then what we actually have is the inequality $$\text{min}\left(\dfrac{a}{b},\dfrac{c}{d}\right) = \dfrac{a}{b} < \dfrac{c}{d} = \text{max}\left(\dfrac{a}{b},\dfrac{c}{d}\right) < \dfrac{c - a}{d - b}.$$

Here is my final query:

Is my argument above logically sound? If not, how can it be mended so as to produce a valid proof?

Answer 1

See my comments for elaborations on the 2 errors in your writeup.

1. Phrasing of the contradiction.
2. Not considering all possible cases to be eliminated.

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A better way to present your work is (This is a sketch where you have to fill in the missing details)

1. WLOG $ \dfrac{a}{b} \leq \dfrac{c}{d} $. So $ bc - ad \geq 0 $
2. Observe that $ \dfrac{a}{b} - \dfrac{ c-a}{ d - b} = \dfrac{ad - bc } { b(d-b) } $, so

• If $ bc = ad$, then $\dfrac{a}{b} = \dfrac{c-a}{d-b} $.
• Else If $ d < b $, then $\dfrac{a}{b} > \dfrac{c-a}{d-b}$.
• Else (IE $ b < d $), then $\dfrac{a}{b} < \dfrac{c-a}{d-b}$.

3. Observe that $ \dfrac{c}{d} - \dfrac{c-a}{d-b} = \dfrac{ ad-bc} { d(d-b) } $, so

• If $ bc = ad$, then $\dfrac{c}{d} = \dfrac{c-a}{d-b} $.
• Else If $ d < b $, then $\dfrac{c}{d} > \dfrac{c-a}{d-b} $.
• Else (IE $ b < d $), then $\dfrac{c}{d} < \dfrac{c-a}{d-b}$.

4. Hence, what is the relevant conclusion? (Depends on how strict you want to be with $ < , \leq, = $ signs in the cases.)

• If $bc = ad$, then $\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{c-a}{d-b}$.
• Else (If $bc \neq ad$ and $d < b$), then $\dfrac{c-a}{d-b} < \dfrac{a}{b} < \dfrac{c}{d}$.
• Else (If $bc \neq ad$ and $b < d$), then $\dfrac{a}{b} < \dfrac{c}{d} < \dfrac{c-a}{d-b}$.

Note:

• It is possible to have $ b < d $ and $ a > c$, which would lead to $ \frac{c-a}{ d-b } < 0 $.