Can any one prove that "Product of two negative numbers is positive???"?

Answer 1

Let a and b be positive numbers.

We shall determine the value of -a x -b, the product of two negative numbers.

If we add a + (-a), we get 0.

So (a + (-a)) x -b = 0 x -b = 0

By the distributive property, we have:

a x -b + (-a) x -b = 0

We know (or can derive) that a x -b = -ab

So we have:
-ab + (-a) x -b = 0

Adding ab to each side of the equation, we have:

-a x -b = ab
showing that a negative times a negative equals a positive (since ab, the product of two positive values, is positive).

Answer 2

This will be rather lengthy. Before we prove anything, we should clearly understand what we accept as true (axioms, definitions and theorems), to use as the basis for proving it.

In this case, let us start with natural numbers. (These are 1,2,3...). We know their properites under addition amd multiplication.

The number 1 (the multiplicative identity)has the interesting property that:
1xn = nx1 = n for all natural numbers n. But there is no such identity for addition.

So, let us invent the additive identity 0, and define it by the property
0 + n = n + 0 = n for all natural numbers and
0 + 0 = 0

Now, we have the new set of numbers, natural numbers extended with 0.

Starting from addition again, we can define subtraction as:

a - b = c if and only if a = b + c.

We can immediately notice that this is rather restricted, because we cannot do the subtraction for arbitrary pairs. 1 - 3 cannot be any number, because you cannot find any c that makes 1 = 3 + c!
Now, we define negative numbers: suppose for every number n, we define -n as
n + (-n) = 0. -n will be called the additive inverse of n. Clearly -0 = 0. Now we can extend the set of natural numbers to include 0 and the negative numbers to form the set of integers. In this extended set of numbers, we can do subtraction for any pair.

With this introduction, how do we prove
-a x -b = ab?

or that
-a x b = -(ab)?

The second one actually means that the left hand multiplication gives the additive inevrse of (ab), so we must prove:

ab + -axb = 0.

You must invoke the distribution property over multiplication to see that
ab + -axb = (a+(-a))b = 0xb = 0
(NOTE: We did not discuss above that somewhere we should have proved that 0xn = 0 for all n! And many other properties!)

After this we can prove the first one:
ab + (-ab) = ab + -a(b) = (a + -a)b = 0xb = 0,

proving -ab is the additive inverse of ab.

The whole development of the theory of numbers from natural numbers to intergers to rational numbers to real numbers is a great story, in which is founded on a set of axioms called (Peano's axioms) that chracterises the set of natural numbers. Everything else can be defined and proved!

If you can get hold of it please see the book "Foundations of Analysis: E. Landau".

Answer 3

okay good question to start with, here is a solution that will set you well on your way. Note that this proves that -1 x -1 = 1 which implies that the product for all negative numbers is positive (but i have not fully shown the last part, I'll leave that to you!):

Proof:

Obvious:

1 = 1

Taking 1 away from both sides:

1 - 1 = 0

Multiplying boths sides by -1:

-1 x (1 - 1) = 0

Multiplying out the brackets:

(-1 x 1) + (-1 x -1) = 0

Evaluating the first bracket:

-1 + (-1 x -1) = 0

Adding 1 to both sides:

-1 x -1 = 1

Done.

Now if you wish to be more elaborate you can extend this proof quite easily. But this is the main idea. If your stuck let me know ( I shall check the question post later.)

Answer 4

If I borrow two rupees from you ( that is a negative # for you ).
You borrow two rupees from me ( " " " " " " " me ).

So we are both rich. The rusult of two negatives is momentarily positive.

If you look at co-ordinate geometry in terms of x and y co-ordinates, you can darw a rectangle that is 2 units length in X direction and 2 units high in the Y direction. So this rectangle will have an area of 4 units.

Now you can also create a rectangle by drawing the X and Y lines completely opposite by -2 units in the X direction and -2 units in the Y direction. The result will be still another rectangle of 4 units area that is diametrically opposite. But it does does exist and is real.

Answer 5

-1*-1 = 1

this can be simply proved as beow
we know 1 -1 = 0
multiply by -1 on both sides
-1(1-1) = 0
-1 *1 -1(-1) = 0
-1 - 1(-1) = 0 as 1 is multiplicative identity -1 *1 = -1
add 1 on both sides
-1(-1) = 1
I have not done any tricks above (simple maths)
Additionally going by the statement in http://www.timecube.com
-1 is opposite.
so -1(-1) is opposite of opposite.
is opposite of opposite not same
based on that -1 * -1 = 1
muliply by positive a * b to get (-a)*(-b) = ab

so product of 2 -ve numbers = positive

Answer 6

this is is not a difficult question
suppose if we have a<0 and b<0
and then there will be three possibilities according to trichotonmy rule either
a*b <0
a*b=0
or
a*b>0
now we will examin them one by one
suppose a*b<0
then (a*b)/b>0/b (inequality changed as we divided by a -ive number i.e b)
we have a>0
which is a contradiction as a<0
now in case two
if we have
a*b=0
then a*b/b =0/b
a=0
agian contradiction
so there is only one possibility left that a*b>0
i.e product of two negitive number is always negitive.hope that gives the answer to the question

Answer 7

NICE QUESTION . LET'S HAVE A LOOK.
Suppose x & y are two negative nos .
hence x < 0 , y < 0
Now as we know , if a < b then -a > -b . Using this let's multiply both sides of Eq. x < 0 by y as it is less than 0 . As a result of this inequality sign will be reversed .
hence we get x * y > 0 .

Answer 8

The product of two negative nos is always +ve as,
minus sign x minus sign is always +ve
Eg. -5 x -5=25

Answer 9

A man's son in college visits him 4 times a month. Each time he visits, he takes Rs. 500 from his dad. In a month, the Dad loses Rs.2,000.00 to his son this way. Now if the son were to visit only once during a month, and assuming he still asks for only Rs. 500, then, the number of visits (from norm) is -3

Man saves -3 x -500 = Rs. 1,500.00

This is the simplest I could make it !

Answer 10

hey dats simple .... use complex polar geometry..
1st convert negative(assuming numbers are real bcos imaginary nos. cant b classified as positive/negative) nos. into polar magnitude-exponential form.
2nd both magnitudes being positive and product of two positive nos. is +ve(pls. dont ask me 2 prove this also ok?)..the resultant mag. is +ve.
3rd..now during multiplication exponents r added. exponents of both -ve nos. r "j*pi rad."Hence their addition is "j2xpi rad"
4th..now combine 2 parts of product..it has +ve magnitude and exponent 'j*2*pi'
5th..now convert this to coordinate plane...+ve mag*1..bcos angle in cmplx.plane corresponds to rotation which will be 2pi rad i.e. complete circle end hence along positive direction of x-axis..hence product lies on positve direction of x axis (mag*direction)..which is REAL POSITIVE NUMBER..giving ya da requisite proof...so how wuz dat.. rigorous proof aint it?