Product of Two Real Negative Number is a Positive Number

We don't do homework, even on Christmas Day.

Go study.

This is more of a given definition than actually something to be proven. This does sound exactly like something that a student would be asked to do in a mathematics class. As such you should consult your textbooks for an answer and methodology relevant to your subject.

Actually there is proof something to do with field theory. These things aren't done just by definitions. I will try Sci.Math NG

Well there you go. You're hiding the added information that you're looking for a proof within the mathematical realm of field theory to prove that the product of two real negative numbers must be a positive number. Not only is this not strictly an engineering question . What you've posed here is a deceptive question. Shame on you. You cannot do any mathematics without the use of definitions.

When and if you ever find your answer, I defy to prove that 1+1=2 without the use of a definition.

Negative 5 minus negative 10 equals plus 5.

I even spelled it out for you.

If you really cared, you'd go take the test for him/her.

?????

Semantics? -5 x-10 = +50.

Note that the magnitude (the absolute value) of a product does not depend on the signs of the factors. That means we can abstract the signs-effect. In other words, we just have to show that -1 x -1 = 1. This should be obvious once we get a grip on the effect of 'adding' a minus sign in front of a quantity (like 1, e.g.). I hope you are familiar with the real number line: zero in the middle, minus infinity way off to the left, plus infinity way off to the right. What a minus sign in front of a number does is to "reflect the number in zero", i.e., it moves it to the other side of zero. If you do that twice, you end up where you started. A nice symbolic way to handle the multiplication is to require the same signs on both sides of any equation. That way you get -1 x -1 = -(-1): two minus signs on the left, two minus signs on the right. Now the right-hand side represents a double reflection of 1 in zero, which is 1 again.

OK, I'm obsessive.

There's an elementary-level discussion at

http://www.math.utoronto.ca/mathnet/questionCorner/minustimesaminus.html

It centers on additive inverses and the distributive law (a(b+c)=ab+ac). In plain-vanilla algebra:

For any real numbers a, b:

a + (-a) = 0 (additive inverse).

Multiply (both sides) by -b to get

-ab + (-a)(-b) = 0 (distributive law).

Add ab (to both sides) to get

(-a)(-b) = ab.

If all the necessary axioms were exposed, I expect they would not surprise.

One such axiom is rating your answer as off-topic whereas it deserves being the best answer.

Iron ringer from UofT?

O.K.,seems there is real interest and efforts done to find the answer so i took back my books and i copy just one demonstration: (-a)•(-b)=a•b ; first you should know these previous: i) -(-a)=a; ii) (a-b)•c=a•c-b•c (please try demonstrate that using distributive property); iii) Theorem: (-a)•(-b)=(0-a)•(0-b)=(0-a)•0-(0-a)•b=0-(0•b-a•b)=-(0-a•b)=a•b ;that's all, now you should try -•+=-;is interesting too,try later with order relationships like a>b and c<0 so a•c<b•c etc.-

Since we are all having fun here, I hope I'm not spoiling it with a word of caution:

for most of the above stuff it is necessary to first state that we are assuming a commutative system, say a×b = b×a (because there are systems where this does not hold!).

If so, the task is easy:

(-a)(-b) = (-1)×a × (-1)×b = (-1)^2 × ab = ab

Happy New Year *<[;o)))

I think i didn't use that property.In case we accept that property your demonstration is easier and seems not necesary the distr.prop.but you'll need trying to prove -1•a=-a which was oversupposed (E.I.B.M) in your post.P.D.:(E.I.B.M.)= "english invented by me".-

A _field_ (like the real number field or the complex number field, which includes the reals--in mathematics, where there is no end of things needing names, the term is overloaded, so "field" unqualified is ambiguous) is an abelian (_i.e._, commutative) group on which is defined a second operation. The latter is required (1) to be associative and (2) to distribute over the group operation. The group operation is conventionally called addition (no matter what it does), and the second operation is conventionally called multiplication. This structure fits the usual complex arithmetic to a T with the additional proviso that complex multiplication is commutative.

Subtraction can never be a basic field operation. It can't be the group operation because it's not commutative: a-b seldom equals b-a. It can't be the second operation because it's not associative: (a-b)-c seldom equals a-(b-c). What it _is_ is a convenient shorthand for an addition involving an additive inverse: a-b means, in terms of the basic field operations, a+(-b), where -b is the additive inverse of b.

A purported proof of a "field theorem" that uses subtraction must survive "deconstruction," _i.e._, translation into basic field operations. Let's translate the "proof" from Ferquiza's book:

(-a)•(-b)=(0-a)•(0-b)=(0-a)•0-(0-a)•b=0-(0•b-a•b)=-(0-a•b)=a•b

becomes

(-a)(-b) = (0+(-a)) · (0+(-b)) = ....

The logic fails here as the translation continues

= (0+(-a))·0 + (-((0+(-a))·b)) = ...,

while the logical continuation is clearly

= (0+(-a))·0 + (0+(-a))·(-b) = ...;

and similar errors follow. What's happening is that unjustified (not false, but effectively circular) inferential leaps are being made under cover of subtraction, which hides what's really going on. Expunging the errors reduces the "proof" to tautology: (-a)(-b) = (-a)(-b).

Bertrand Russell wrote that the germ of the _Principia Mathematica_ was his realization as a university student that the proofs being taught in his classes were not valid.

On another note, can we prove a(-b) = -ab as Ferquiza asks? With our newly resharpened insight, we're thinking additive inverses. What can we say? ab+(-ab)=0 for one thing, and a(-b)+ab=0 for another. Aha! Just add the inverse of ab to both sides of that (which comes from applying the distributive law to a((-b)+b)). Of course the mathematics isn't finished until it's been turned into mind-numbing, drawn-directly-from-the-vacuum student-repellent like this:

0 = 0

a(0) = ab + (-(ab))

a(b + (-b)) = ab + (-(ab))

ab + a(-b) = ab + (-(ab))

a(-b) = -(ab)

Q.E.D.

I think you should try prove ii) of post#18 first and your point is over, then see i asked first prove i) (same post #18) thats a necesary too.On the other hand i think i didn't ask to prove what you proved or tried to do.-

(-a)b = (0-a)b = 0·b - a·b = -ab.

Similarly a(-b) = a(0-b) = a·0 - a·b = -ab.

This shows that changing the sign of a factor changes the sign of the product. Hence from (-a)b = -ab we deduce (-a)(-b) = ab.

This, in essence, is the proof that appears in #18.

To avoid circularity neither a nor b can be negative. The restriction would be a wart on a real proof, but that's not what we have here. What we have is an illusion, a trick of notation, the kind of thing that leads reasonable people to ask an engineer for a real answer.

In #10, Slowlight points out that unary and binary minus are different. Here their similarity is turned into a shell game. Translating the initial lemma into basic field operations reveals the scam:

(-a)b = (0+(-a))·b = 0·b + (-a)b

a(-b) = a·(0+(-b)) = a·0 + a(-b)

Both equations are tautological. How could it be otherwise? The map is not the world. Writing -a as 0-a _adds no information_: nothing is happening but notation: nothing is changing but labels.

Reminds me of the famous joke about mathematicians attributed to Emmy Noether: "Let A be a set. Call it B."

Analyzing the trick in the first line above is instructive, because it shows there is a difference between algebra, properly understood, and the conventional rewriting rules we use to "do" algebra. I've never seen a better example.

Note that the desired transformation, from (-a)b to -ab, can be seen to require no more than the vanishing of the parentheses. Rewriting (-a)b as (0-a)b does two things: (1) it changes the unary to a binary minus and (2) it changes the role of the parentheses. In (-a)b the parentheses bind the unary minus to a. In (0-a)b the parentheses group the terms of the subtraction. When (0-a)b is expanded, the grouping parentheses are no longer needed. With the parentheses gone, the remaining binary is converted back to a unary minus. Presto!

If that were all, you might reasonably say, "So what?" But there's more, thanks to Emmy Noether and the others who have given us the theory of rings and fields. From them we know nothing happens in a field that is not expressible in terms of the field's elements and basic operations. So if the rewriting trick were OK, we could translate it, with no loss of meaning, into basic field operations.

We know that--except for the additive identity element, which is its own inverse--negation is represented by the pairing of additive inverses. We know that subtraction is represented by the addition of an additive inverse. Hence there's no problem executing the translation. But there is a problem with the result: the parentheses don't vanish. We conclude that the disappearance of the parentheses is an artifact deriving from the imperfect match between the rewriting rules used and the algebra they were supposed to represent.

In sum and in short, the "proof" is a misapplication of rewriting rules.

These proofs still depend on -1x-1=1 to be true.