Why is negative x negative = positive?

[murrmac]

IMO, the multiplication of negative (and positive numbers as well for that matter) is best illustrated by visual representation.

To start with, let us consider the simple case of 10 x 10.

Now, 10 can also be expressed as (1+9), (2+8) , (3+7) …(9+1)

For argument’s sake we will consider the case of (8+2)

So 10 X 10 can also be expressed as (8+2) x (8+2)

This can be expanded by multiplying each of the numbers in the second bracket by each of the numbers in the first

therefore (8+2)(8+2) =64+16+16+4, and this can be illustrated visually as in the diagram below.

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Now , using the same diagram (just because it would take me too long to do a new one) let us consider the case where a number is expressed not as the sum of two integers, but as the difference of two integers, and just because it is convenient we will consider the number 8, which can be expressed as (10-2)

So , 8 x 8 can also be expressed as (10-2) x (10-2)

Expanding this gives us 100-20-20 …plus or minus whatever (-2) X (-2) might be.

Using the same diagram, let us do it step by step.

the diagram as it stands shows 10 x10 =100

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The next step is to show 100-20, so …

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Now we need to subtract another 20, so

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But this leaves us with 56, and we know that 8 x 8 =64, so where do we find the missing four ? The answer is that the last term , which is (-2) x(-2) must be equal to 4. Therefore multiplying two negative numbers yields a positive number.

I realize that this little demo isn’t exactly on the same level as proving Fermat’s Last Theorem, but hopefully it will be helpful in enabling some people to visualize
numerical problems graphically.

[Animastryfe]

@ultrafilter

There are certain properties that we would really like our arithmetic to have. One of them is that whenever you have a number x, there’s another number -x such that x + -x = 0. Since -x is a number as well, there must be some number -(-x) such that -x + -(-x) = 0. We can then say that x + -x = -x + -(-x), and conclude that x = -(-x).

One of the other properties we’d like our arithmetic to have is that (x + y)z = xz + yz. If that’s the case, then (-x)y + (-x)(-y) = -x(y + -y). We know that y + -y = 0 and x0 = 0, so it must be the case that (-x)y + (-x)(-y) = 0. However, (-x)y + xy = (-x + x)y, and by the same logic this is also equal to zero. So by the same reasoning as above, we can conclude that (-x)(-y) = xy.

I like your explanation very much. Despite being a physics major, I keep forgetting the proof for why a negative number times a negative number equals a positive number.

[Indistinguishable]

One more presentation of ultrafilter’s (and most everyone else’s) argument, in case it is of use to anyone:

Suppose A + A’ = 0 and B + B’ = 0 (A’ and B’ may be called negations of A and B, respectively).

Then consider AB + A’B + A’B’. On the one hand, this is (A + A’)B + A’B’ = 0B + A’B’ = A’B’. On the other hand, this is AB + A’(B + B’) = AB + A’0 = AB. Thus, we must have that A’B’ = AB.

Accordingly, negating both the factors in a product leaves the product unchanged.

[In math jargon, this shows that, given any bilinear operator (multiplication) on a monoid (addition), taking similarly one-sided additive inverses to both factors of a multiplication leaves the multiplication unchanged. Note that we needn’t assume anything (e.g., commutativity of + or *, general inverses or cancellativity for + or *, associativity of *, …) beyond this.]

[Indistinguishable]

But the more I think about it, the more I think the best ten-second answer is simply this:

A) Each time you negate one of the factors in a multiplication, the overall product negates
B) So if you negate both the factors, the overall product double-negates (which means it stays the same)
C) Thus, (-3) * (-7) = -(3 * (-7)) = --(3 * 7) = 3 * 7, and so on…

One can give more argument for A) and for B), should someone demand it, but I suspect this answer as is will satisfy many people…