Proving 0v=0 using only the 10 Axioms

[torquerotates]

Well, I'm supposed to prove 0v=0

It is stated that I'm only allowed to use the following axioms.

let a,b,c be vectors and V is a vector space, then
1)a&b is in V then a+b is in V
2)a+b=b+a
3)a+(b+c)=(a+b)+c
4)0+a=a+0=a
5)a+(-a)=(-a)+a=0
6)a is in V implies ka is in V
7)k(a+b)=ka+kb
8)(k+m)a=ka+ma
9)k(ma)=(km)a
10) 1a=a

The book does it like this, and i think its wrong

0v=(0+0)v=0v +0v { axioms 4&8}

now subtract 0v from both sides { axioms ?}
we get 0=0v

you see the problem here? there's no justification for the subtraction step because there is no axiom allowing the step. Logically I assume that I'm only allowed to use the 10 axioms to prove this theorem.

 

[torquerotates]

Actually I figured it out

0v=(0+0)v=0v+0v

Now I just use another equation that has nothing to do with the first.
namely, 0v+(-0v)=0
well, this implies 0v+(-0v)=0=(0v+0v)+(-0v)
implies 0=0v+(0v+-0v)=0v+0=0v

 

[torquerotates]

But I was working on another problem in which the author uses incorrect reasoning.

authors' proof:
Prove: if a+c=b+c, then a=b
(a+c)+(-c)=(b+c)+(-c) add (-c) to both sides
a+(c-c)=b+(c-c)
a+0=b+0
a=b

you see how in this example, (-c) was actually added to both sides. There is no way to justify this according to the 10 axioms.

 

[mrandersdk]

you don't need a axiom to add an element to both sides of a equallity, that is how it works.

That is if

a = b then a+c=b+c

this is not a axiom, this is how equality works, he proves that

a = b if and only if a+c=b+c

 

[HallsofIvy]

You will do a lot better in math if you stop assuming anytime you don't understand something, the author is wrong.

You don't have to have an axiom that says "if a= b then a+ c= b+ c", that's part of the definition of "binary operation" and is assumed whenever you have a binary operation.

 

[torquerotates]

Its not a+c=b+c iff a=b. Its a+c=b+c implies a=b. If it is just a binary operation,(I presume you guys to mean it is a definition), then why can it be proved using axioms only? No assumptions are needed except the hypothesis. Having the need to create a definition merely indicates that the statement cannot be proven.proof: (a+c)+(-(a+c))=0 ax.5
(a+c)+(-(b+c))=0 hypothesis
(a+c)+((-b)+(-c)))=0 ax.7
(a+c)+((-c)+(-b))=0 ax2
((a+c)+(-c))+(-b)=0 ax3
(a+(c+(-c)))+(-b)=0 ax3
(a+0)+ (-b) =0 ax5
(a)+(-b)=0 ax.4
a=b ax.5

 

[mrandersdk]

in your last step from

(a)+(-b)=0 ax.4

to

a=b ax.5

what do you use there?

yep, you are right you use what you are trying to prove (a+c=b+c implies a=b), just with
a -> a+(-b)
b-> 0
c-> b

so you haven't proved anything

 

[mrandersdk]

ax. 5 says

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

not

a+(-b)= 0 => b=a

 

[torquerotates]

I see. Well how about this:

a=a+0 ax4
a=a+(c+(-c)) ax5
a=(a+c)+(-c) ax3
a=(b+c)+(-c) hypothesis
a=b+(c+(-c)) ax3
a=b+0 ax5
a=b ax4

Well? Haven't I disproven that the binary operation is "merely" a definition?

 

[torquerotates]

Well, in the defense of the last step my first proof,
a+(-b)=0

b+0=b=(a+(-b))+(b)=a+(b+(-b))=a+0=a ax.4,3,5&4 in order form left to right.
therefore a=b