Can We Prove ac < bd Under Given Conditions?

[objectivesea]

Homework Statement

Prove that for all numbers a, b, c, d: if 0 \leq a &lt; b and 0 \leq c &lt; d then ac &lt; bd.

This is problem 5 from chapter 1 of Michael Spivak's "Calculus", 4th Edition. It is the text for my real analysis course.

I should also mention that this is not a homework problem, though similar content will be on my exam tomorrow and I'm hoping to have a better understanding by then.

Thanks for your help!

Homework Equations

Use only the following axioms. For all numbers,

P1. a+(b+c)

P2. a+0 = 0+a = a

P3. a+(-a) = (-a)+(a)=0

P4. a+b = b+a

P5. a \cdot (b \cdot c) = (a \cdot b) \cdot c

P6. a \cdot 1 = 1 \cdot a = a

P7. a \neq 0 \rightarrow \exists a^{-1}, a \cdot a^{-1} = 1

P8. a \cdot b = b \cdot a

P9. a \cdot (b + c) = a \cdot b+b \cdot c

Let P be be a collection such that a \in P \leftrightarrow a &gt; 0
Then for all numbers,
P10. Only one of the following is true:
<br /> \begin{align}<br /> &amp;a=0\\<br /> \textrm{or }&amp; &amp;a \in P\\<br /> \textrm{or }&amp; &amp;-a \in P<br /> \end{align}<br />
P11. a \in P \wedge b \in P \rightarrow (a+b) \in P
P12. a \in P \wedge b \in P \rightarrow (a \cdot b) \in P

"a &gt; b" is defined as the relation: : {(a,b): (a-b) \in P}
"a &lt; b" is defined as the relation: : {(a,b): b &gt; a}

The Attempt at a Solution

Only one of the following is true:
\begin{align}<br /> a&amp;=0 &amp;\wedge&amp; &amp;c &amp;= 0\\<br /> a&amp;&gt;0 &amp;\wedge&amp; &amp;c &amp;= 0\\<br /> a&amp;=0 &amp;\wedge&amp; &amp;c &amp;&gt; 0\\<br /> a&amp;&gt;0 &amp;\wedge&amp; &amp;c &amp;&gt; 0<br /> \end{align}

Suppose that anyone of (1), (2), (3) are true, then

ac = 0.

From P12, bd \in P. So By definition of P,
bd &gt; 0 =ac

Suppose that (4) is true, then
a \in P \wedge b \in P \wedge c \in P \wedge d \in P \wedge (b-a) \in P \wedge (d-c) \in P

I'm not sure where to go from here...

 

[objectivesea]

Figured it out!

(1) Since b \in P and (d - c) \in P, we get (bd - bc) \in P using P12.

(2) Since c \in P and (b - a) \in P, we get (bc - ac) \in P using P12.

Then from (1) and (2) we get (bd - bc + bc - ac) \in P using P11.

Using P3 and P2 we get (bd - ac) \in P, thus ac &lt; bd

 

[Odedude]

A bit late, but...

You want to show that bd - ac > 0. We have

bd - ac = b(d - c) + c(b - a) > 0 + 0 = 0. QED