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This is my first post of math.stackexchange so bear with me if the formatting is weird, I learnt simple mathjax just to type this up

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:

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 $c \gt d$

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 $$\lnot (c \gt d) \rightarrow \lnot(hypothesis)$$ $$d \ge c \rightarrow \lnot(hypothesis)$$

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

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  • $\begingroup$ I'm dubious about your assertion that $bc>ac$. I can see this will be true if $c>0$, but that is not one of your hypotheses. $\endgroup$ – Lord Shark the Unknown May 11 '18 at 3:08
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    $\begingroup$ @LordSharktheUnknown I got $c \gt 0$ from $ac \ge bd$ because for that to be true and both a and b be positive, shouldn't c have to be greater than 0? $\endgroup$ – Nragis May 11 '18 at 3:11
  • $\begingroup$ @Nragis The answer to your question is yes. $\endgroup$ – Benedict Voltaire May 11 '18 at 3:23
  • $\begingroup$ @Nragis Your third inequality $bd > ad$ is the same as the first inequality. Was this supposed to be something else? $\endgroup$ – Benedict Voltaire May 11 '18 at 3:26
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    $\begingroup$ @Nragis -- I think the chain $bd>bc>ac$ should be $bd\geq bc>ac$ . But the consequent is the same, $bd>ac$ . $\endgroup$ – mr_e_man May 11 '18 at 3:30
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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.

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  • $\begingroup$ Thanks! The book said to use a contrapositive in the hint so I blindly followed it without checking for a much simpler method such as this one. $\endgroup$ – Nragis May 11 '18 at 4:32
  • $\begingroup$ Not a wasted effort! It's good to try the contrapositive, and excellent practice to prove things more than one way. $\endgroup$ – rwbogl May 11 '18 at 4:36
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Since $0 < a < b$. $c \geq bd/a > 1×d = d$ follows from $ac \geq bd.$
Your proof is excessively complicated.

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  • $\begingroup$ Yeah, I see that now, because $\frac{b}{a}$ is always greater than 1. Thanks, was trying to be fancy and use new techniques I'd learned and completely overlooked the obvious I guess $\endgroup$ – Nragis May 11 '18 at 3:38
  • $\begingroup$ @mr_e_man Yes I do, first day learning set theory and my big lesson is perfect notation is everything. and check everything a million times $\endgroup$ – Nragis May 11 '18 at 3:42

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