Is This Mathematical Proof Logically Sound?

[solakis1]

Is the following proof ,proving $$\forall A\forall B[ 0<AB\Longrightarrow (0<A\wedge 0<B)\vee(A<0\wedge B<0)$$,correct??

Proof:

Let, 0<ab

Let, ~(0<a& 0<b)...............1

Let , ~(a<0&b<0)...............2

But 0<ab => ~(ab=0) => ~(a=0) and ~(b=0) => ~(a=0)......3

For ,$$ 0<a \Longrightarrow\frac{1}{a}<0\Longrightarrow(ab)\frac{1}{a}<0\frac{1}{a}\Longrightarrow b<0$$,since 0<ab, $$\Longrightarrow 0<a\wedge0<b$$ a contradiction by using (2)

Hence ~(0<a)................4

In a similar way we prove : a<0 => (a<0&b<0) ,a contradiction by using (3)

Hence ~(a<0).................5

Thus from (4) and (5) we have :

~(0<a) and ~(a<0) => ~( 0<a or a<0) => a=o ,a contradictio by using (3)

Hence ~~(0<a & 0<b) => 0<a & 0<b => (0<a &0<b)or( a<0 & b<0)

 

[I like Serena]

Hey Solakis!

I have just noticed that you have 160 posts and 0 posts in which you have thanked anyone.
For the record, apparently your posts got thanked 11 times.

Is there any reason you expect that anyone wants to respond to your posts, considering that this is a site where people are only volunteering?

 

[solakis1]

Hey Solakis!

I have just noticed that you have 160 posts and 0 posts in which you have thanked anyone.
For the record, apparently your posts got thanked 11 times.

Is there any reason you expect that anyone wants to respond to your posts, considering that this is a site where people are only volunteering?

Thanks for reminding me