Algebra Revisited 2018
Modern Algebra: A Logical Approach, Book Two,
circa 1966 [Allen, Pearson]
Chapter 6: Functions and Other Relations
Bounded Functions
p.325 (exercises 3, 5, 6) p.326 (exercise 8 proof)
(3)
M = {}
Since Thus M is bounded below. However, M is not bounded above.
(5)
S = {}
Since Thus S is bounded below. However, S is not bounded above.
(6)
T = {}
Since , the Absolute Maximum is
(8)
Given that F = {}, show that 1 is the absolute maximum of F.
We will need to use Theorem 8-3 and it's corollaries.
If , then ( a and b have like signs) and ( a and b have opposite signs).
Corollary 1: (1) If , then (2) If , then
Proof: If a = 0, then 00 = 0 If a , then the two factors of aa are the same and have the same sign. It follows from Theorem 8-3 that .
Corollary 2: 1 > 0 and -1 < 0 Proof that 1 > 0 follows from the fact that 1 = 1*1. Then the proof that -1 < 0 follows from the Corollary to Theorem 5-3. Theorem 5-3: If , then
Corollary to Theorem 5-3: A real number is negative if and only if its opposite is positive, and a real number is negative if and only if its opposite is positive. $$
Corollary 3: If and , then a and have like signs. $$
To prove that 1 is the absolute maximum of F = {}, we need to prove (i) is true when , and (ii) is true.
We know that if , and that is true if . This suggests a proof of the statement, "If , then "
(1) . ParseError: KaTeX parse error: Undefined control sequence: \$ at position 1: \̲$̲ REASONS: (…;;\mathbb{R};;;;$ Multiplication property of inequality.
NOTE: In step 6 we assumed that was positive. This can be proved as follows: ParseError: KaTeX parse error: Undefined control sequence: \$ at position 1: \̲$̲ (1);x \in \mathbb{R} \longrightarrow (2); x^2 \ge 0$, (3) .
REASONS: (1) Given, (2) Theorem 8-3 Corollary 1, (3) Theorem 8-3 Corollary 2, (4) Addition propert of inequality, (5) Theorem 8-3 Corollary 3.