# Help figuring out (simple) math proof

A few years ago I picked up the book An Introduction to Analysis (by James Kirkwood) at a textbook clearance sale. I finally got around to studying it recently during my commute time. For me, it’s mostly stuff that was touched on in college courses but I’ve never fully approached from the analysis angle. (I give this introduction so it doesn’t just look like I’m asking for help with homework.)

I’m in the very first section, where it’s setting up the real numbers. The problem I have a question about is this :

Prove that |a| <= |b| if and only if a[sup]2[/sup] <= b[sup]2[/sup].

Now, given what’s been in the text so far, the only way I can figure doing this is to consider every case and what results – though effective, it seems a tad inelegant. What really makes me think there is a better way is the ‘hint’ given at the back of the book :

|a| = sqrt( a[sup]2[/sup] )

Now, I’m not sure that proving that is any easier, but what I’m wondering is, where does that get me? The very next problem is to prove that x[sup]s[/sup] < y[sup]s[/sup] for positive x,y and s rational, so I’m wondering if this is supposed to be similar or not.

Anyone see what the book is getting at?
It’s difficult to give everything in the text to this point, but I’ll give the statements I have regarding order and absolute value:

F is an ordered field if :
There is a nonempty subset P of F which is closed under addition & multiplication.
For any a in F exactly one holds : a is in P, a = 0, or -a is in P.

The ordering operation :
For an ordered field F, and P as described, for a,b in F:
a < b if b - a is in P.

The absolute value :
For real a, |a| is defined as :
|a| = a, if a >= 0; -a, if a < 0.

The equality part is obvious, so let’s just look at the strict inequality. How about this:

Suppose |a| < |b|. Then |b| - |a| is in P.

|b| + |a| is also in P, so

b[sup]2[/sup] - a[sup]2[/sup] = (|b| - |a|)(|b| + |a|) is in P.

Therefore a[sup]2[/sup] < b[sup]2[/sup].

The converse should be straight-forward from here.

For the other direction, I might be tempted to use an indirect proof, and proceed in a manner similar to what Cabbage said.

Thanks for the help, guys. Normally I wouldn’t feel a need to just post a ‘thank you’, but I’m actually posting since I realized a mistake in my post that was nagging at me a bit. I wrote :

Obviously that can’t be proven true. S has to be positive as well (and the implied x < y must be true too). Maybe my statement could have been interpreted that way … but what it should be is x,y positive reals with x < y, and s a positive rational.

This new problem was somewhat separate from the one you solved – the lead-in for proving this one was the problem before, in which the same statement was proven for s (actually ‘n’) a positive integer. The induction of that one was the harder part; expanding it to rationals was fairly easy.