From Chapter 1, page 9 of Spivak's Calculus, 3rd edition:
In fact, it is convenient to consider the collection of all positive numbers, denoted by $P$, as the basic concept, and state all properties in terms of $P$:
(P10) (Trichotomy law) For every number $a$, one and only one of the following holds:
(i) $a=0$, (ii) $a$ is in the collection $P$, (iii) $-a$ is in the collection $P$.(P11) (Closure under addition) If $a$ and $b$ are in $P$, then $a+b$ is in $P$.
(P12) (Closure under multiplication) If $a$ and $b$ are in $P$, then $a\cdot b$ is in $P$.
This is question 8 on page 15:
Although all the basic properties of inequalities were stated in terms of the collection $P$ of all positive numbers, and < was defined in terms of $P$, this procedure can be reversed. Suppose that P10-P12 are replaced by
(P'10) For any numbers $a$ and $b$ one, and only one, of the following holds: (i) $a=b$, (ii) $a<b$, (iii) $b<a$.
(P'11) For any numbers $a$, $b$, and $c$, if $a<b$ and $b<c$, then $a<c$.
(P'12) For any numbers $a$, $b$, and $c$, if $a<b$, then $a+c<b+c$.
(P'13) For any numbers $a$, $b$, and $c$, if $a<b$ and $0<c$, then $ac<bc$.
Show that P10-P12 can then be deduced as theorems.
And here's the solution as in the manual:
Two applications of P'12 show that if $a<b$ and $c<d$, then $a+c<b+c<b+d$, so $a+c<b+d$ by P'11. In particular, if $0<b$ and $0<d$, then $0<b+d$, which proves P11. It follows, in addition, that if $a<0$, then $-a>0$; for if $-a<0$ were true, then $0=a+(-a)<0$, contradicting P'10. Consequently, any number $a$ satisfies precisely one of the conditions $a=0$, $a>0$, $a<0$, the last being equivalent to $-a>0$. This proves P10. Finally, P'13 shows that if $0<a$ and $0<c$, then $0<ac$, which proves P12.
My question is, why doesn't P10 immediately follow from P'10, by substituting $0$ for $b$ ?
Best Answer
Because then what we would have would be:
It is not obvious that you can jump from this to:
You will need some connection between $<$ and the arithmetic operations in $\Bbb R$.