Question

proof:

Answer 1

if a < b and c < 0, then ac > bc
proof: since c < 0
0 = c + (-c) < 0 + (-c) = -c
so -c is positive, hence by (x), we get -ac < -bc,
-ac + (bc + ac) < -bc + (bc + ac)
hence bc < ac, and this is equivalent to ac > bc

I think there's some application of the following axioms:
transitive law: if a < b and b < c, then a < c

law of trichotomy: for every real number a, one and only of the following alternatives holds: a = 0, or a < 0, or 0 < a;

if a < b, then a + c < b + c;

if a < b and 0 < c, then ac < bc

Answer 2

Did you get that from my notes?

Answer 3

I recall doing something similar.

Answer 4

I don't get this part: 0 = c + (-c) < 0 + (-c) = -c
No, it's from a review of real numbers ineq. and absolute values.

Answer 5

Oh okay.. Let's try to break it up:
0 = c + (-c) right?

Answer 6

yeah

Answer 7

So we can simplify it down to:
0 < 0 + (-c), because:
We are given that c < 0, meaning it's negative. Therefore -c > 0 (double negative is a positive), so this line is simply telling you that:
0 < -c, or
-c > 0

Answer 8

Did you follow?

Answer 9

If not I think I can make it even easier.

Answer 10

yeah by assuming that -c is positive, but that would mean we're blowing away the fact that -1 (-c) = c > 0

Answer 11

OHHHH I GET IT NOW

Answer 12

That Eureka moment in math... ahhhh..

Answer 13

ya! thanks. that -c is an insertion or expansion using the existence of subtraction axiom.
-a such that a + (-a) = 0