You left out the case that b= 0.
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!