Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 8

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Problem

Although 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:

  1. a = b,
  2. a < b,
  3. a > b.

(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.

Solution

(P10) (Trichotomy law) For every number a, one and only one of the following holds:

  1. a = 0,
  2. a is in the collection P,
  3. -a is in the collection P.

Let a be some arbitrary number. Let P be the set of all positive numbers. By P’10, assuming b = 0, one of the following is true about a:

  1. a = 0
  2. a < 0
  3. 0 < a

In the first case, a = 0.

In the second case, a < 0, so 0 < 0 - a = -a, so -a \in P.

The solution book uses a proof by contradiction to prove -a \in P. I think mine is just as valid.

In the third case, it’s clear that a \in P.

By P’10, these are the only three possible cases, therefore, for all a, a = 0, a \in P, or -a \in P.

(P11) (Closure under addition) If a and b are in P, then a+b is in P.

Let a and b be arbitrary numbers in P. It follows that 0 < a. By P’12, we know that 0 + b = b < a + b (b in P’12 is 0. c in P’12 is b here). Then, by P’11, we know that since 0 < b < a + b, it must be that 0 < a + b, which means a+b \in P.

Therefore, a, b \in P \to a+b \in P.

(P12) (Closure under multiplication) If a and b are in P, then a \cdot b is in P.

Let a and b be arbitrary numbers in P. It follows that 0 < a. By P’13, we know that 0 \cdot b = 0 < ab (b in P’13 is 0. c in P’13 is b here). So 0 < ab, which means ab \in P.

Therefore, a, b \in P \to ab \in P.

 

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