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How can I prove:

If $0 < a$ and $0 < b$, then a < b if and only if $a^2<b^2$

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.

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2 Answers 2

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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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One way is easy. If a< b and a> 0 then a^2< ab. If a< b and b> 0 then ab< b^2. Therefore a^2< b^2.

The other way: If a^2< b^2 then b^2- a^2= (b- a)(b+ a)> 0. The product of a positive and a negative number is always negative so b- a and b+ a must have the same sign- either both negatives or both positive.

case 1: b- a> 0 and b+ a> 0. From b- a> 0 we have immediately that b> a.

case 2: b- a< 0 and b+ a< 0. Now we have both b< a and b< -a. Since a is positive -a is negative and b< -a means that b is negative. Since we are given that b is positive, this case is impossible.

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