Verifying that the product of two negatives is positive

[flyingpig]

Homework Statement

[PLAIN]http://img717.imageshack.us/img717/2285/unledxig.png

The problem is to verify (v), that is for some x and y in R, we have (-x)(-y) = xy

The Attempt at a Solution

(-x)(-y) = (-1)x * (-1)y by Prop 2.7 iv

= x*(-1) * (-1)y by M2
= x*(-1*-1) * y by M1
= x*(+1)*y
= xy

Q.E.D

Not sure if I did it right.

Now I am going to throw in another question.

My prof did a lot of proofs in class in the similar manner as me (except for being wrong...). The thing is that they all look so silly. Like why do I have to go throw all this trouble just to show that (-1)(-1) = +1? That took like 4 lines. This is like math made unnecessarily hard.

 

[flyingpig]

Actually wait, I didn't need 4 lines at alll...

(-x)(-y) = (-1)x(-1)y By Prop.2.7 iv

=(-1)(-1)xy byM2

= 1 * xy By M3

= xy

Q.E.D.

 

[flyingpig]

Any comment is fine...

 

[Punkyc7]

Couldnt agree with you more that this is math made unnecessarily hard, but I think you have to show -1*-1=1

 

[flyingpig]

Couldnt agree with you more that this is math made unnecessarily hard, but I think you have to show -1*-1=1

Yeah I took that for granted...

I don't know how to do that

 

[Punkyc7]

1= 1+0= ?

does that help

 

[flyingpig]

1= 1+0= ?

does that help

1 = 1 + 0 = 1 + 1 - 1?

But this is addition.

 

[Punkyc7]

yes...but you can distribute a -1 by prop 4 and m4D

 

[flyingpig]

yes...but you can distribute a -1 by prop 4 and m4D

I have one -1 though

 

[Punkyc7]

pull out a negative one

 

[flyingpig]

pull out a negative one

-1(-1 - 1 + 1)

 

[Punkyc7]

now what do you notice...

 

[flyingpig]

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

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

Is that it? Because this seem really really unnecessary...

 

[Punkyc7]

yep... that is it, and I could not agree more, but pure math people seem to love this kind of stuff because now they now for sure that -1*-1=1 and they are not relying on what others have said before

 

[flyingpig]

How do I include that in my original proof?

Assume (-x)(-y) = xy

Conditions [ =(-x)(-y) = (-1)x * (-1)y by Prop 2.7 iv
= x*(-1) * (-1)y by M2
= x*(-1*-1) * y by M1
= stuck here. How do I introduce 1 = 1 + 0 unawkwardly?]

Result [...]

Q.E.D

 

[Punkyc7]

just call it a lemma and put it before the proof.. of course if you have proven it before you do not need to include but I am guessing you havent.

lemma 1 = -1(-1)

pf/
1 = 1 + 0 = 1 + 1 - 1 = -1(-1 - 1 + 1)

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

pf/
(-x)(-y) = (-1)x * (-1)y by Prop 2.7 iv

= x*(-1) * (-1)y by M2
= x*(-1*-1) * y by M1
= x*(+1)*y
= xy

Q.E.D

 

[flyingpig]

Lemma: (-1)(-1) = 1 or do I have to write out that whole step we did...?

 

[Punkyc7]

All the step. Or how is someone who is reading your proof going to know that -1*-1=1 if it hasnt been shown to them.

 

[flyingpig]

All the step. Or how is someone who is reading your proof going to know that -1*-1=1 if it hasnt been shown to them.

But I thought Lemma was the results, not the proof. Isn't it just showing the results that you should believe in?

 

[Punkyc7]

you can't believe in anything anymore it must be proven to be true.

 

[flyingpig]

Lemma: (-1)(-1) = 1

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

Proof: (-x)(-y) = xy= x*(-1) * (-1)y by M2
= x*(-1*-1) * y by M1
= x*(+1)*y
= xy

Q.E.D

Did I need to include "by ..." in my Lemma? Why is it that I didn't need to show that 1 - 1 = 0?

 

[Punkyc7]

That looks good, I don't think you need to include the by lemma but if you want to be extra clear you can put it in, also you don't need to because of field properties