Can We Prove ac < bd Under Given Conditions?

[objectivesea]

Homework Statement

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

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

"$$a > b$$" is defined as the relation: $$: {(a,b): (a-b) \in P}$$
"$$a < b$$" is defined as the relation: $$: {(a,b): b > a}$$

The Attempt at a Solution

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

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

$$ac = 0$$.

From P12, $$bd \in P$$. So By definition of $$P$$,
$$bd > 0 =ac$$

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 $$ac < bd$$

 

[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