(-a)(-b)=ab prove it if you can. Iam sure many can't?

Question

you can use numbers instead of a and b too.

Answer 1

Consider that ∀x, 0*x=0. Also, per the definition of -a, a+(-a) = 0. Therefore, (a+(-a))*(-b) = 0, so by the distributive property, a*(-b)+(-a)*(-b) = 0, which means that (-a)*(-b) = -(a*(-b)). However, we also have that 0 = a*(b+(-b)) = a*b + a*(-b), so a*b = -(a*(-b)). Therefore, by transitivity, (-a)*(-b) = a*b. Q.E.D.

Answer 2

Are we allowed to assume that a and b belong to a subset of the reals? It's not specified, and I'm not sure off the top of my head whether the statement is true in all rings. Certainly in a ring without a unit (which do exist), my proof wouldn't hold. Anyway, here's the proof, which holds for all reals.

Assume that a and b are real numbers. Then (-a) = (-1)*a and (-b) = (-1)*b. So (-a)(-b) = (-1)(a)(-1)(b). By the commutative property of the reals, (-1)(a)(-1)(b) = (-1)(-1)(a)(b).

Now we have reduced the problem to showing that (-1)(-1)=1. Since the real numbers are a field, every real number has a multiplicative inverse. So there exists a number x such that (-1)*x = 1. Take the absolute value of both sides. |(-1)*x| = |1|. Since the expression in the absolute value is a product, we can distribute the absolute value operation to get |-1|*|x|=|1|.

This implies that x=1 or x=-1. Try them both in the equation (-1)*x = 1. We see that x = 1 doesn't work because 1 is the multiplicative identity, so (-1)*1 = -1. Therefore the only possible multiplicative inverse for (-1) is (-1).

Thus (-1)*(-1) = 1, and this directly implies that (-1)(-1)(a)(b) = (a)(b). QED.

Answer 3

Using specific values doesn't constitute a valid proof. You can't "prove by example".

If you REALLY wanted to be strict and do a formal proof SOLELY from core axioms, you could do this:

-a is defined as the inverse of a, such that a + -a = 0. Here's the proof that -a = -1*a:
a = a*1 (multiplicative identity)
a + -1*a = a*1 + -1*a
a + -1*a = a*1 + a*(-1) (commutative)
a + -1*a = a(1 + -1) (distributive)
a + -1*a = a(0) (additive identity)
a + -1*a = 0 (the zero identity)
a + -1*a = a + -a (additive identity)
-1*a = -a

From here we can write
(-a)(-b) = (-a)(-1*b) = (-1 * - a)(b) = (-(-a))(b) = ab

Answer 4

I cant comprehend why somebody along with your medical subject residing interior the comparable climate as Jamaica etc the factor of Jamaica could desire to flow to there completely. the only place with lots of well-being center and medical centre strategies available 24-7 is in Kingston. look no extra, as you could desire to make your well-being a concern. you're youthful additionally, so Kingston would be the place it is at for all styles of activities AND in case you will get an enduring interest. Please make confident that each and all and sundry your docs are so as, artwork enable and all. Have a blast in Jamaica!!

Answer 5

It's not really provable. It's pretty much a fact. But here are my two shots at it anyway:

1) (-1)(-2) = 1 * 2

2 = 2 [tah-DAH]

2) Divide both sides by (-a)...

-b = ab/-a

-b = -b

Answer 6

It is very simple. We all know that 2 negative numbers multiplied together causes a postive number. So take this for example. (-2)(-4)=8
Simple as that.

Answer 7

this rule just states that multiplying two negative numbers is teh same as multiplying two positive numbers.
example:
(-3)(-2) = 6
(3)(2) = 6
so, we can say that (-3)(-2) = (3)(2)
this proves that (-a)(-b) = (a)(b)
hope this helped...not really sure though cuz we skipped chapter in our class

Answer 8

a negative number multiplied by negative is positive, letters next to each other means multiply therefore a negative "a" times negative "b" is the same as positive "a" times positive "b".

Answer 9

if a + b = ab, and in numerical maths -2*-3= 6 and () mean multiply then (-a)(-b) =ab.

Answer 10

-3(-2) = 6