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objectivesea
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Homework Statement
Prove that for all numbers a, b, c, d: if [tex]0 \leq a < b[/tex] and [tex]0 \leq c < d[/tex] then [tex]ac < bd[/tex].
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. [tex]a+(b+c)[/tex]
P2. [tex]a+0 = 0+a = a[/tex]
P3. [tex]a+(-a) = (-a)+(a)=0[/tex]
P4. [tex]a+b = b+a[/tex]
P5. [tex]a \cdot (b \cdot c) = (a \cdot b) \cdot c[/tex]
P6. [tex]a \cdot 1 = 1 \cdot a = a[/tex]
P7. [tex]a \neq 0 \rightarrow \exists a^{-1}[/tex], [tex]a \cdot a^{-1} = 1[/tex]
P8. [tex]a \cdot b = b \cdot a[/tex]
P9. [tex]a \cdot (b + c) = a \cdot b+b \cdot c[/tex]
Let [tex]P[/tex] be be a collection such that [tex]a \in P \leftrightarrow a > 0[/tex]
Then for all numbers,
P10. Only one of the following is true:
[tex]
\begin{align}
&a=0\\
\textrm{or }& &a \in P\\
\textrm{or }& &-a \in P
\end{align}
[/tex]
P11. [tex] a \in P \wedge b \in P \rightarrow (a+b) \in P[/tex]
P12. [tex] a \in P \wedge b \in P \rightarrow (a \cdot b) \in P[/tex]
"[tex]a > b[/tex]" is defined as the relation: [tex]: {(a,b): (a-b) \in P}[/tex]
"[tex]a < b[/tex]" is defined as the relation: [tex]: {(a,b): b > a}[/tex]
The Attempt at a Solution
Only one of the following is true:
[tex] \begin{align}
a&=0 &\wedge& &c &= 0\\
a&>0 &\wedge& &c &= 0\\
a&=0 &\wedge& &c &> 0\\
a&>0 &\wedge& &c &> 0
\end{align}[/tex]
Suppose that anyone of (1), (2), (3) are true, then
[tex]ac = 0[/tex].
From P12, [tex]bd \in P[/tex]. So By definition of [tex] P [/tex],
[tex] bd > 0 =ac[/tex]
Suppose that (4) is true, then
[tex] 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[/tex]
I'm not sure where to go from here...
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