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

  1. #1
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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.

    Prove that the zero vector & additive inverse is unique - by contradiction?-0-vector-unique.jpg

    For the zero vector is unique proof:

    Is method 1 and method 2 both correct?

    If they are, which one is more preferable?
    =================

    Is the proof that the additive inverse is unique okay?
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  2. #2
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    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?
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    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
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    Re: Prove that the zero vector & additive inverse is unique - by contradiction?

    Quote Originally Posted by mathguy25 View Post
    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
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