Spivak's calculus: is my proof wrong?

[johnnyies]

Homework Statement

If a, b > 0, and a² < b², then a < b.

Homework Equations

P is the collection of all positive numbers, n > 0

(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) If a and b are in P, then a + b are in P
(P12) If a and b are in P, then a x b is in P.

The Attempt at a Solution

If a² < b², then 0 < b² - a², hence b² - a² is in the collection P

0 < b² - a² which is the same as 0 < (b - a)(b + a)

by (P12), (b - a) and (b + a) are both in P,

if (b - a) is in P, this means b - a > 0, hence, b > a

this isn't the answer in the book, I will post that after dinner, but my proof seems sound but at the same time I can't tell if it's correct or wrong.

 

[Hurkyl]

by (P12), (b - a) and (b + a) are both in P,

That's not P12. The statement you used is

If a x b is in P, then a and b are in P​

which is the converse of P12, and is only true half the time.

 

[johnnyies]

I got it. If ab is in P then it's possible a and b are not in P, since it can be said: a < 0, b < 0 ,ab > 0 (product of two negatives). I missed that. Thanks.

 

[carlosbgois]

But can I say that if (b-a)(b+a) > 0, either (b-a) and (b+a) have to be positive, or either negative, and, as I know by hypothesis that (b+a) >= 0, (b-a) must be in P, hence (b-a) > 0, so b > a?

 

[micromass]

But can I say that if (b-a)(b+a) > 0, either (b-a) and (b+a) have to be positive, or either negative

You will need to prove this...
Otherwise it's correct.

and, as I know by hypothesis that (b+a) >= 0, (b-a) must be in P, hence (b-a) > 0, so b > a?

 

[carlosbgois]

Sure, but that is the easy part hehe. Thanks!