Math help

[Shiny Noctowl]

How do I prove that (-1) * (-1) = 1 using only the following 12 postulates?

P1 Addition is associative:
For every a, b, c, a + (b + c) = (a + b) + c
P2 The number 0 is an additive identity:
For every a, a + 0 = 0 + a = a
P3 Additive inverses exist:
For every a, there exists a -a such that a + (-a) = (-a) + a = 0
P4 Addition is commutative:
For every a, b, a + b = b + a
P5 Multiplication is associative:
For every a, b, c, a * (b * c) = (a * b) * c
P6 The number 1 is a multiplicative identity:
For every a, a * 1 = 1 * a = a
P7 Multiplicative inverses exist:
For every a != 0, there exists an a^(-1) such that a * a^(-1) = a^(-1) * a = 1
P8 Multiplication is commutative:
For every a, b, a * b = b * a
P9 The distributive law:
For every a, b, c, a * (b + c) = a * b + a * c
P10 The trichotomy for P:
For every a, exactly one of the following holds:
a = 0
a is positive
-a is positive
P11 Closure under addition:
If a and b are in P, then so is a + b
P12 Closure under multiplication:
If a and b are in P, then so is a * b

 

[empoleon dynamite]

You're sending us your Homework?!

 

[Felly]

Blargh. Math. It's my worst subject, but I'll take a guess at this. Um... Multaplicative inverses, maybe? Or multaplicative identity. It's one of the multipication ones; I've gathered that much.

 

[Ben Cousins]

In the time you typed that, you could have figured out the answer.

 

[Shiny Noctowl]

In the time you typed that, you could have figured out the answer.

No, because it's so much harder than the other ones. The rest of the problems were easy things like "Prove that for every a != 0, a^2 > 0."

 

[Silktree]

It is first necessary to prove the following statement: For every a, a*0=0.

Proof: Using P2 and P9, we have: a*0=a*(0+0)=a*0+a*0.
Using P3, we can add the additive inverse of a*0 to both sides, yielding a*0=0. Q.E.D

Secondly, we will prove that for every b, the additive inverse -b is unique.

Proof: Let b be a real number. Assume that there exist numbers (not necessarily distinct; at least one exists by P3) b' and b'' such that: b+b'=b+b''=0.

Using P1 and P2, we have: b'=b'+0=b'+(b+b'')=(b'+b)+b''=0+b''=b''. In other words, the additive inverse of b is unique. Q.E.D

Finally, we shall prove a more general statement than required: For every c, (-1)*c=-c. Once we have proven that, it will follow that (-1)*(-1)=-(-1)=1.

Proof: Since c*0=0, we have c*(1+(-1))=0, from which we deduce using P6 and P9 that c+c*(-1)=c*1+c*(-1)=0.
It follows that c*(-1) is the unique additive inverse of c, or in other words, (-1)*c=-c. Q.E.D