Why a double negative makes a positive. (Mathematics)

[Zhalentine]

Since we agreed that any positive number multiplied by negative 1 is the negative form, and that (-1)(-1) = -1, we can further simplify this:

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

So:

0 = -2???

If we place any other number (apart from +1) as the answer to the calculation (-1)(-1) = ?, the same thing happens. This is why two negatives make a positive in mathematics.

Reproduced for your enjoyment from http://mathforum.org/dr.math/faq/faq.negxneg.html

I don't think we did agree that (-1)(-1)= -1. Were you just trying to show that if (-1)(-1) = -1, then 0 = -2? You never mentioned you were doing that and when you said "For example, let's assume (-1)(-1) = +1." maybe you meant (-1)(-1) = -1.

Anyways, here is a math problem for you

"proof" that 0 = 1.
We compute the indefinite integral of (1/x dx) by parts. Set u = 1/x and
dv = dx, so that du = -1/x^2 dx and v = x.

We get:
indefinite integral of (1/x dx) = uv - indefinite integral of (v du)
= x/x + indefinite integral of (x/x^2 dx)
= 1 + indefinite integral of (1/x dx)

Subtracting indefinite integral of (1/x dx) from both sides, we get 0=1

[Tiberius]

I don't think we did agree that (-1)(-1)= -1. Were you just trying to show that if (-1)(-1) = -1, then 0 = -2? You never mentioned you were doing that and when you said "For example, let's assume (-1)(-1) = +1." maybe you meant (-1)(-1) = -1.

Yeah, corrected. Thanks for spotting the mistake!

Anyways, here is a math problem for you

"proof" that 0 = 1.
We compute the indefinite integral of (1/x dx) by parts. Set u = 1/x and
dv = dx, so that du = -1/x^2 dx and v = x.

We get:
indefinite integral of (1/x dx) = uv - indefinite integral of (v du)
= x/x + indefinite integral of (x/x^2 dx)
= 1 + indefinite integral of (1/x dx)

Subtracting indefinite integral of (1/x dx) from both sides, we get 0=1

Oh dear! Someone seems to be forgetting his constant!

You actually get:

0 = 1 - c, where c always equals 1.

Ergo you get: 0 = 1 - 1, which is correct

[Pope Alfred]

It works in mathematics, but in languages it is a silly old fallacy that a double negative adds up to an affirmative. My mother used to moan on about "It ain't no good" must logically mean "It is good".
That was true for Latin, and in those days English usage was forced by pedants into the mould of Latin.
In fact, in - for example - Middle English and Middle German, multiple negatives were stylish: Chaucer's He nevere yet no vileynye ne sayde contains three negators. The effect was to intensify the negative, not to cancel it out. The way negatives are expressed in correct modern French comes from precisely this; arguably 'je ne vais pas' is still a double negative today.
So... "It ain't no good" is perfectly OK in some common registers of English. It can be condemned for not conforming to the standard (if you like), but not for being illogical.

[Oldandeasilyconfused]

edited. It think my reply about logic was wrong.