This question involves problem 5(viii) from chapter 1 of Spivak's Calculus, third edition.
I'm a layman trying to teach myself some more advanced mathematics, and although I've been making slow progress through the book, this problem has me stumped.
The question is as follows:
If $0 \leq a < b$ and $0 \leq c < d$, prove $ac < bd$.
This question has different answers, depending on whether $a$ and $b$ are equal to zero or not. I think the most difficult case is the one where $a, b \ne 0$, so that's the one I will use here.
My attempt:
In an earlier question, I've proved that if $a < b$ and $c > 0$ then $ac < bc$. This is done by multiplying: $(a < b)c = ac < bc$.
From now on I will use the notation that Spivak uses, where $a < b$ means that $b – a$ is in $P$. This leaves me with $b – a$ and $d – c$, both in $P$. Spivak has established that this means you can multiply the two together.
I will list the steps that I followed:
- $(b – a)(d – c) = bd – bc – ad + ac$
- $ bd – bc – ad + ac = bd – (bc + ad – ac)$
- $bd – (bc + ad – ac) = bd > (bc + ad – ac)$
- $bd > (bc + ad – ac) = bd > (bc + ad > ac)$
- $bd > (bc + ad > ac) = bd > bc + ad > ac$
This would appear to prove that $ac < bd$ (which is what the question asked) but the part $bd > bc + ad$ is quite obviously incorrect. Just a simple example proves this:
If $2 > 1$ and $3 > 2$, then $6 > 7 > 2$.
Intuition tells the me that the two seperate inequalities $bd > bc > ac$ and $bd > ad > ac$ are both correct, but I have no idea how I can formally get these inequalities to 'split', so to speak.
Help would be much appreciated!
…
Edited for spelling. And thanks for helping me with the problem guys!
Best Answer
If $c=0$, then $0 = ac < bd$ clearly. Otherwise, if $c>0$, then $ac < bc$ by what you've already proved, and also $bc < bd$ by what you've proved. Thus $ac < bc < bd$.