Question

hey cow this might sound stupid but is there a way to prove a negative times a negative is positive?

Answer 1

let put this formally:
Let a and b be positive real numbers. Then -a(-b)=ab.

Answer 2

assume opposite and prove it false by counterexample i guess:
assume (-a)(-b) = -ab
counter-example: a=2,b=3
(-2)(-3) = 6
thus premise is false and neg times neg is positive

Answer 3

i wish i could think of a proof for this

Answer 4

multiplication is just a repeated sum, so the reason for negXneg=pos i think comes from why -(-3) = 3 or 2-(-1)=2+1=3

Answer 5

right but i want to prove -(-3)=3

Answer 6

we know\[-a=ae^{i(2n+1) \pi}\] and \[-b=be^{i(2n+1) \pi}\]
so (-a)(-b)=\[ae^{i(2n+1) \pi}be^{i(2n+1) \pi}=abe^{i2(2n+1)\pi}=ab\]

Answer 7

i think u gao my approach

Answer 8

omg seriously wow i think this is good

Answer 9

yeah that will work :)

Answer 10

:) yes!

Answer 11

thanks..

Answer 12

A>0
B>0
(A)+(-A)=0
B(A)+B(-A)=0
so this proves a negative times a positive is negative since a positvex +negativex=0
(A)+(-A)=0
-B(A)+(-B)(-A)=0
in order for this to be zero one of terms need to be negative and one needs to be positive
-B(A) is negative
so
(-B)(-A)=BA
what about this too?

Answer 13

its also right...but here u considered that +ve*+ve is +ve