# Did I skip a major step in this proof? + Theory of this proof

1. Sep 21, 2011

### flyingpig

1. The problem statement, all variables and given/known data

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

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

3. 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 similiar 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.

Last edited by a moderator: May 5, 2017
2. Sep 21, 2011

### 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.

3. Sep 22, 2011

### flyingpig

Any comment is fine...

4. Sep 22, 2011

### Punkyc7

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

5. Sep 22, 2011

### flyingpig

Yeah I took that for granted...

I don't know how to do that

6. Sep 22, 2011

### Punkyc7

1= 1+0= ?

does that help

7. Sep 22, 2011

### flyingpig

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

8. Sep 22, 2011

### Punkyc7

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

9. Sep 22, 2011

### flyingpig

I have one -1 though

10. Sep 22, 2011

### Punkyc7

pull out a negative one

11. Sep 22, 2011

### flyingpig

-1(-1 - 1 + 1)

12. Sep 22, 2011

### Punkyc7

now what do you notice...

13. Sep 22, 2011

### 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...

14. Sep 22, 2011

### 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

15. Sep 22, 2011

### 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

16. Sep 22, 2011

### 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 im 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

17. Sep 22, 2011

### flyingpig

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

18. Sep 22, 2011

### 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.

19. Sep 22, 2011

### flyingpig

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

20. Sep 22, 2011

### Punkyc7

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