Suppose $a,b,c,$ and $d$ are real numbers, $0 \lt a \lt b $, and $d \gt 0$. Prove that if $ac \ge bd$ then $c \gt d$

Question

Background: I'm currently working through the book "How To Prove It" by Daniel J. Velleman when I came across a problem in the simple proofs chapter I think I solved, but have no one to check over my proof. Here's the question:

Question : Suppose $a,b,c,$ and $d$ are real numbers, $0 \lt a \lt b $, and $d \gt 0$. Prove that if $ac \ge bd$ then $c \gt d$

My Approach : Proof by Contrapositive

So I decided to list my hypotheses:

1. $a,b,c,d \in \mathbb{R}$
2. $0 \lt a \lt b$
3. $d \gt 0$
4. $ac \ge bd$

and my conclusion:

1. $c \gt d$

After this I decided to attempt to prove the contrapositive so I wrote $$\lnot (c \gt d) \rightarrow \lnot(\text{Hypothesis})$$ $$d \ge c \rightarrow \lnot(\text{Hypothesis})$$

So then I worked out that $d \ge c$ and $b \gt a$ so I multiplied each of them by a,b, and c separately to get two that fit together: $$bd \ge bc $$ $$bc \gt ac $$ to make this: $$ bd \ge bc \gt ac \implies bd \gt ac$$ which is true if $d \ge c$ $$d \ge c \rightarrow bd\gt ac$$ $$\therefore \lnot (c \gt d) \rightarrow \lnot(ac \ge bd)$$ $$\therefore ac \ge bd \rightarrow c \gt d$$

While I reached the same answer as the book, it does not match the proof in the book so I don't know if I am wrong or correct and I don't want to be building proof writing habits that are incorrect. Thank you

Answer 1

Since $0 < a < b$. $c \geq bd/a > 1×d = d$ follows from $ac \geq bd.$
Your proof is excessively complicated.

Answer 2

I want to show you a fairly simple, direct way to prove this, and also how it could be written up.

By assumption, we have $a < b$ and $d > 0$. Multiplying our inequality by $d$, we obtain $ad < bd$. If, in addition, $bd \leq ac$, then $$ad < bd \leq ac,$$ hence $$ad < ac.$$ Dividing by $a$ gives us $d < c$, as desired.

(William Elliot's proposed proof is even simpler!)

The contrapositive is most often useful when your original statement is unwieldy, but the conclusion seems more manageable. Using the contrapositive here doesn't help much. For an example of when the contrapositive is useful, try proving this: If $n^2$ is odd, then $n$ is odd.