# Help with problem 1-5 from Calculus by Spivak (multiplying two inequalities together)

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:

1. $(b - a)(d - c) = bd - bc - ad + ac$
2. $bd - bc - ad + ac = bd - (bc + ad - ac)$
3. $bd - (bc + ad - ac) = bd > (bc + ad - ac)$
4. $bd > (bc + ad - ac) = bd > (bc + ad > ac)$
5. $bd > (bc + ad > ac) = bd > bc + ad > ac$

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!

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"Prove" is the verb, "proof" is the noun. – Pedro Tamaroff Aug 31 '14 at 22:33
We have $b - a = \epsilon > 0$ and $d - c = \delta > 0$. Then $bd = ac + a\delta + c\epsilon + \epsilon \delta > ac$. Or simply draw rectangles $[0, b] \times [0, d]$ and $[0, a] \times [0, c]$ and compare the areas. – akech Aug 31 '14 at 22:43
$b$ can never be $0$ because it is strictly greater than $a$, which is in turn at least $0$ – John Joy Sep 1 '14 at 0:47

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 $ac < bc < bd$.
The key thing here is to point out that $<$ is transitive. – Arthur Aug 31 '14 at 22:30