Why do two negatives equal a positive in multiplication?

[Hurkyl]

It's good for a heuristic proof, but if you want a really formal, rigorous proof, these things can be tricky to get right, because we're so used to assuming all of the niceties of arithmetic!

(Of course, it also depends on from what axioms and definitions you start! You could start with the definition -b := (-1)b, but then you'd have to prove that b + (-b) = 0)

 

[Hurkyl]

I guess the real question is if Gale is happy with the responses!

 

[eNathan]

One of my teachers put it this way:

The Friend of my Friend is my Friend | + * + = +
The Friend of my Enemy is my Enemy | + * - = -
The Enemy of my Friend is my Enemy | - * + = -
The Enemy of my Enemy is my Friend | - * - = +

While not exactly a mathematical proof, it makes sense.

Well put, motai! This example is excellent for people to grasp the concept of a negative

 

[Werg22]

Hurkyl if we keep it that way we will eventually have to prove that we exist.

 

[Gale]

I guess the real question is if Gale is happy with the responses!

am i ever really happy? i was basically satisfied after the first link... but you guys are really having fun with it... so let's prove it in at least... 3 more different ways.

 

[HallsofIvy]

Hurkyl if we keep it that way we will eventually have to prove that we exist.

I can prove (to my satisfication) that I exist. I'm not so sure about the rest of you.

 

[roger]

I can prove (to my satisfication) that I exist. I'm not so sure about the rest of you.

Hallsoivy, please could you tell me how to prove the existence of the uniqueness of addition ?

In summary , what is the minimum number of definitions or axioms one needs to prove that -a*-b = a*b ?

 

[HallsofIvy]

First, I didn't say I could prove the existence of anything but myself!

Second, you don't mean "prove the existence of the uniqueness of addition", you mean simply "prove that the result of addition is unique".

Exactly now you do that depends on what number system you are using.

The simplest is the natural numbers where you take as axioms, the existence of a number "1" and the existence of the "successor" of any number as a one-to-one, onto function from the natural numbers to the natural number without "1" and take the "principle of induction" as an axiom (If a set X contains the number 1 and, whenever it contains the number k, it also contains the successor of k, then X is the set of all natural numbers"
We then define addition by "n+ 1 is the successor of n. If m is not 1, then it is the successor of some number, say p. We define n+ m as "the successor of n+ p".
Now we can prove that m+ n is unique:
Let n be a counting number and let X be the set of all m such that n+ m is unique. n+ 1 is, by definition, the successor of n. That is unique since "successor" is a function. Suppose k is in X. Then n+ successor of k is, by definition,the successor of n+ k. n+ k is unique since k is in X. The successor of n+ k is then unique because "successor" is a function. Therefore the succesor of k is in X. Therefore X is all natural numbers: For every natural number, n+ m is unique for all natural numbers m.

Once we have that we can define the integers by: Let NxN be the set of all pairs of natural numbers. We say that two such pairs, (a,b) and (c,d) are equivalent if and only if a+d= b+ c. It's easy to show that that is an equivalence relation and so partitions the set of pairs into equivalence classes. An "integer" is such an equivalence relation.
We define addition of integers by: If x and y are two integers, that is, two such equivalence classes, choose two "representative" pairs, (a,b) from x, and (c,d) from y. "x+ y" is the equivalence class containing the pair (a+c, b+d). In order to show uniqueness here, we must show that choosing different pairs from x and y would give the same equivalence class for x+ y.
That is, suppose (a,b) and (a',b') are both in equivalence class x and (c,d) and (c',d') are both in equivalence class y. That is, a+b'= b+a' and c+d'= d+c'. Adding those two equations, a+b'+c+d'= b+a'+d+c' so (a+c)+(b'+d')= (b+d)+ (a'+c') showing that (a+c, b+d) is equivalent to (a'+c', b'+ d').

 

[Robokapp]

I got the easiest answer. My math teacher showed this.

(-1)*(-1)=1 Given
1(-1)*1(-1)=1 1 multiplied to any factor equls that factor.
(1*1)*[(-1)*(-1)]=1 Association property

1*1=1 so substituting that above...

1*[(-1)*(-1)]=1

as stated in second line, 1 times any number = that number, so we can write

(-1)*(-1)=1 and that has to be true due to last form of the expression. Imagine the (-1)*(-1)=x

solve 1*x=1

there you go!

 

[jcsd]

-a*-b = a*b folows on form the follwing properties:

additve identity, additive inverse, additive associativty and distributivity

so it is true in any ring.

 

[Hurkyl]

In summary , what is the minimum number of definitions or axioms one needs to prove that -a*-b = a*b ?

Just 1: take as an axiom that -a*-b=a*b.

It may seem like a cheap trick, but these sorts of silly things are very useful.

In any case, I think you meant to ask about trying to weaken the axioms of arithmetic as much as possible -- I just wanted to make this explicit.