Question

Prove the following \[\forall a,b \in \mathbb{r}, a < 0, b < 0 \implies a * b > 0\] Prove the following \[\forall a,b \in \mathbb{r}, a < 0, b < 0 \implies a * b > 0\] @Mathematics

Answer 1

because minus * minus is plus result that this is true allways

Answer 2

I need a stronger proof

Answer 3

right ???

Answer 4

suppose that not is true and check the result so by reductio ad abbsurdum prove style,
methode

Answer 5

the book says you have to begin by first proving the following statement\[(-a) * b = -(a * b)\]

Answer 6

sorry

Answer 7

(-a) * b + a*b = b(-a + a)
= b*0
= 0, which we proved in the previous question
hence, (-a)*b + a*b -(a*b) = -(a*b)
therefore, (-a)*b = -(a*b)

Answer 8

a<0, b<0
this implies -a> and -b>o
this implies (-a)(-b)>0
this implies ab>0(because (-a)(-b)=ab)

Answer 9

ok, this is my attempt at this:
you already proved that (-a)*b + ab = 0
therefore:
-b*( (-a)*b + a*b ) = 0
(-b)*(-a)*b + (-b)*a*b = 0
( (-b)*(-a) + (-b)*a )*b = 0
(-b)*(-a) + (-b)*a = 0

you also proved (-a)*b = -(a*b)
therefore:
(-b)*a = -(a*b)

plug this into above to get:
(-b)*(-a) + -(a*b) = 0
therefore:
(-b)*(-a) = (a*b)

Answer 10

@asnaseer that was how the book did it

Answer 11

Consider a number \(c\) defined as \(c=ab+(-a)(b)+(-a)(-b)=ab+(-a)[b+(-b)]\).
\(b+(-b)=0\). Therefor \(c=ab+(-a)[0] \implies c=ab\).
We can also write \(c=b(a+(-a))+(-a)(-b) \implies c=b(0)+(-a)(-b)=(-a)(-b).\)
We have \(c=ab\) and \(c=(-a)(-b)\), which implies that \(ab=(-a)(-b)\).