Thread: Prove ab > 0 (basic proof)

Suppose a,b are elements of the Real numbers and a*b > 0

Then either a>0 and b>0 OR a<0 and b<0


I figured the easiest way to do this proof is using trichotomy, but it seems like there should be a way to compress what I have into fewer steps (but maybe not).

So this is what I have:

assume b>0

if a = 0 then 0*b > 0 is false since 0 = 0

if a < 0 then a*b > 0 is false since (ab < 0) the inequality would flip when multiplying by a value less than 0

if a > 0 then a*b > 0 is true

assume b < 0

if a = 0 then 0*b > 0 is false since 0=0

if a < 0 then a*b > 0 is true

if a > 0 then a*b > 0 is false since (ab < 0) the inequality would flip when multiplying by a value less than 0

Thus we see when b > 0 and a > 0 then ab >0, and when b < 0 and a < 0 then ab > 0

Any criticism on my proof would be greatly appreciated

Thanks!

You left out the case that b= 0.