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Prove that the zero vector & additive inverse is unique - by contradiction?

I know this can be done by a direct proof, but I'm interested in doing it by contradiction.

I have done two methods for proving the zero vector in a vector space is unique.

Attachment 28097

For the zero vector is unique proof:

Is method 1 and method 2 both correct?

If they are, which one is more preferable?

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Is the proof that the additive inverse is unique okay?

Re: Prove that the zero vector & additive inverse is unique - by contradiction?

I'm not sure why you want to prove this by contradiction - your proof is exactly the same as the direct proof, so it adds nothing in terms of understanding. In general, it is inadvisable to use proof by contradiction if a direct proof is just as easily obtained. This is because if you make any mistakes in your working, passing that mistake off as a contradiction gives an invalid proof, even though the conclusion is correct.

For the uniqueness of additive inverse proof, there is a lot of things that may not be well defined. What is (-v)? Is it a third additive inverse? Is it w? Is it u? If (-v) is just an arbitrary additive inverse, does the symbol (-v) represent the same element on both sides? Without knowing uniqueness (which is what you want to prove), how can you add (-v) on each side of the equality without showing they are the same?

Re: Prove that the zero vector & additive inverse is unique - by contradiction?

There is nothing wrong with proving the method of contradiction, this is essentially the method you use when showing uniqueness. You basically assume two objects possess a certain characteristic and then show that these two objects must in fact be the same. Same reasoning as assuming they are different and finding a contradiction.

However, in your proof (1) you assumed u and w to be the zero vectors and used that assumption to conclude they were the same zero vector, seems a little too easy and too good to be true.

(2) is a good proof except use v^(-1) instead of -v unless you can prove that v^(-1) = -v

Proof of the uniqueness of inverses:

Let v be a vector in vector space V. Let v' and v'' be the inverse of v. Then by inverse additive axiom, v + v' = 0 and v + v'' = 0. Then v + v' = v + v''. Then add either v' or v'' to both sides. Then v' + v + v' = v' + v + v''. Then (v' + v) + v' = (v' + v) + v''. Then 0 + v' = 0 + v''. Then v' = v''. Thus, inverses are unique.

Proof that v^(-1) = -v.

v + v^(-1) = 0

-v + v + v^(-1) = -v + 0

(-1 + 1)v + v^(-1) = -v

0v + v^(-1) = -v

0 + v^(-1) = -v

v^(-1) = -v

Re: Prove that the zero vector & additive inverse is unique - by contradiction?

Quote:

Originally Posted by

**mathguy25** There is nothing wrong with proving the method of contradiction, this is essentially the method you use when showing uniqueness. You basically assume two objects possess a certain characteristic and then show that these two objects must in fact be the same. Same reasoning as assuming they are different and finding a contradiction.

However, in your proof (1) you assumed u and w to be the zero vectors and used that assumption to conclude they were the same zero vector, seems a little too easy and too good to be true.

(2) is a good proof except use v^(-1) instead of -v unless you can prove that v^(-1) = -v

Proof of the uniqueness of inverses:

Let v be a vector in vector space V. Let v' and v'' be the inverse of v. Then by inverse additive axiom, v + v' = 0 and v + v'' = 0. Then v + v' = v + v''. Then add either v' or v'' to both sides. Then v' + v + v' = v' + v + v''. Then (v' + v) + v' = (v' + v) + v''. Then 0 + v' = 0 + v''. Then v' = v''. Thus, inverses are unique.

Proof that v^(-1) = -v.

v + v^(-1) = 0

-v + v + v^(-1) = -v + 0

(-1 + 1)v + v^(-1) = -v

0v + v^(-1) = -v

0 + v^(-1) = -v

v^(-1) = -v

By v^(-1), do you mean that this is the negative of the multiplicative identity?

So, from multiplicative identity axiom we have: 1v = v

-1v = -(1v) = -v