How can I prove:
If $0 < a$ and $0 < b$, then a < b if and only if $a^2
We are seeing this example in class but do not understand it.
These are the axioms we are using:
The axioms. The integers, which we denote by Z, is a set, together with a nonempty subset P ⊂ Z (which we call the positive integers), and two binary operations addition and multiplication, denoted by + and ·, satisfying the following properties:
- (Commutativity) For all integers a, b, we have
a + b = b + a and a · b = b · a.
- (Associativity) For all integers a, b, c, we have
a + (b + c) = (a + b) + c and a · (b · c) = (a · b) · c.
- (Distributivity) For all integers a, b, c, we have
(a + b) · c = a · c + b · c.
(Identity) There exist integers 0 and 1, such that for all integers a, we have a + 0 = a and a · 1 = a.
(Additive inverses) For any integer a, there exists an integer −a such that a+(−a) = 0.
(Closure for P) If a, b are positive integers, then a + b and a · b are positive integers.
(Trichotomy) For every integer a, exactly one of the following three possibilities hold: either
a is a positive integer, or a = 0, or −a is a positive integer.
- (Well-ordering) Every nonempty subset of the positive integers has a smallest element.
For inequalities:
- Trichotomy law.
- Transitive law.
- Compatibility of sum with order.
- Compatibility of product with order.
Would appreciate any help.
Answer
If $a,b\in\mathbb{R}^+$ with $b>a$, then $$b-a\in\mathbb{R}^+$$ and so $$b^2-a^2=(b-a)(b+a)\in\mathbb{R}^+$$ since $b+a$ is positive from our first assumption. Finally $$b^2-a^2\in \mathbb{R}^+\implies b^2-a^2>0\implies b^2>a^2$$
The other direction of your 'if and only if' statement is pretty much a disassembly of this proof put back together backwards.
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