Proof using only field axioms Prove if $x, y ∈ R$ and $xy > 0$ then
either $x > 0$ and $y > 0$, or, $x < 0$ and $y < 0$ using only the field axioms.
These include the Field axioms for addition, multiplication, distribution, order axioms (trichotomy and transitivity), ordered field axioms.
I'm stuck on this and not sure how to approach it. Would proof by contrapositive be a good approach in this case?
 A: Yes, or it boils down to a few cases to consider:


*

*$x=0$: Then $xy=0\not>0$

*$y=0$: Again $xy=0\not>0$

*$x,y<0$: Nothing needs to be shown

*$x,y>0$: Nothing needs to be shown

*$x<0<y$: Then $xy+(-x)y=0y=0$ and $(-x)y>0$ as product of two positives, so $xy<0$ (why?)

*$y<0<x$: Similar.

A: Prove smaller props along the way.
If x > 0 and y > 0 then xy > 0 is an ordered field axiom.
0x = 0 is a proposition to be proven:
0x +0x = (0+0)x=0x (distributive) so 0 = 0x - 0x = 0x +0x - 0x = 0x.
If x > 0 then -x < 0 and vice versa can be proven:
x>0 then by ordered field axiom 0=x-x>0-x =x.(same for <)
If x > 0 and y < 0 then xy < 0.
Proof:y < 0 so -y > 0 so x (-y) > 0 so x (-y) +xy > 0 + xy = xy.  So x (-y + y) = x0 =0 < xy.
If x <0 and y < 0 then xy > 0.
Proof:x < 0 so -x>0 so (-x)y < 0 so 0= (x-x)y = xy + (-x)y < xy + 0 = xy.
By the exclusive property of order and by the commutativity of multiplication these are the only possibilities to consider.  xy > 0 only when x>0 and y>0 or x <0 and y <0.
