Proof that all real numbers great than 0

Webmonkey

Hi There,

Just working on a few things here for an exam - I need to proof:

For all real numbers a, a^2 >= 0

I have half of proof where:

One must hold: a >= 0 or a <= 0
So consider these cases separately.

Case 1
a >= 0 --> 0 <= a

So by the Axiom (if a<=b and 0<=c then ac<=bc) we have
if we have 0<=a and 0<=a (i know same) then we have
0a <= aa = 0 <= a^2. IE: a^2 >= 0
(Taking a = 0, b = a, c = a in the axiom)

So now how do I prove the other case?
I mean when I do the other case I end up with a^2 <= 0. So my two cases are a^2 <=0 or a^2 >= 0.

Obviously a^2<=0 is incorrect but how do I say that - can't assume anything here!

Thanks.

Fremen

1 + (- 1) = 0

multiply both sides by -1

(1)(-1) + (-1)(-1) = 0 by associativity and the fact that a.0 = 0

-1 + (-1)(-1) = 0 by a property of 1

add 1 to both sides:

(-1)(-1) = 1 > 0

Now it's your turn, generalise the argument for A, and not just 1.

Webmonkey

Thanks for that,

Unfort I can't assume properties of 1, since they weren't proved before this proof, I guess he'd accept it but id rather not assume 1 > 0 (Though obviously it is).

The proof for the second case above, his solution was:

This is just a special case of the theorem where we take b=a. QED

Unfort that doesn't help me much - I could write that down i suppose and get marks but rather understand what he meant here!

Thanks once again for that.

Fremen

Well, if you start the proof with "let a > 0"
then a + (-a) = 0
a(-a) + (-a)(-a) = 0

-1(a)(a) + (-a)(-a) = 0

(-a)(-a) = (a)(a) > 0 by your proof above.

I've assumed -1(a)(a) = (-a)(a), or equivalently -1(a) = -a

If you need to prove this too, add a to both sides

-1(a) + a = 0

then -1(a) is the additive inverse of a, which is -a by definition.

Beyond that, I don't think I can help you without a list of the axioms you're allowed to use. I'm kind of going on guesswork here.

This is just a special case of the theorem where we take b=a. QED

did he prove a theorem where ab > 0 if (a > 0 and b > 0) of if (a < 0 and b < 0)?

Webmonkey

Fremen wrote:

did he prove a theorem where ab > 0 if (a > 0 and b > 0) of if (a < 0 and b < 0)?

Yes something similar, the lemma right before this as a matter of fact.

Let a,b be real numbers, and suppose a<=0 and b<=0, then ab>=0

Thanks once again.

Edit - It looks clear now what he means, if a <= 0 and a <=0 (Taking b=a for second one), then ab>=0, ie aa>=0 so a^2>=0
Looks like the solution - cheers mate.