- #1

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like proving that (-1)

__u__=-

__u__in a vector space.

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- Thread starter sbo
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- #1

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like proving that (-1)

- #2

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- #3

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- #4

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thanx now i know that i dont need to prove axioms, they are given. thanks to that.

- #5

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Im still worried about this vector thing and ill try to prove that the negative of a vector in V is unique, thaks all :)

- #6

Fredrik

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- #7

HallsofIvy

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[itex](-1)\vec{u}[/itex] is the is the additive inverse of the multiplicative identity in the

To show that they are the same you use the basic properties (axioms) of vector spaces: specifically that -1 and 1 are addivitive inverses in the field of scalars, that [itex]1\vec{u}= \vec{u}[/itex], that [itex]0\vec{u}= \vec{0}[/itex], and the distributive law [itex](a+ b)\vec{u}= a\vec{u}+ b\vec{u}[/itex].

[itex](1+ -1)\vec{u}= 0\vec{u}= \vec{0}[/itex]

and [itex](1+ -1)\vec{u}= 1\vec{u}+ (-1)\vec{u}= \vec{u}+ (-1)\vec{u}[/itex].

Since those are both equal to [itex](1+ -1)\vec{u}[/itex] they are equal to each other and [itex]\vec{u}+ (-1)\vec{u}= \vec{0}[/itex], which is precisely the

- #8

Fredrik

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This step requres proof as well. We have[itex] 0\vec{u}= \vec{0}[/itex]

[tex]0\vec u=(0+0)\vec u=0\vec u+0\vec u[/tex]

which implies that

[tex]0\vec u+\vec 0=0\vec u+0\vec u[/tex]

and now the result [itex]0\vec u=\vec 0[/itex] follows from the theorem I mentioned in #6.

- #9

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can you say that proving that condition is equivalent to proving that -(-

- #10

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can you say that proving that condition is equivalent to proving that -(-

- #11

Fredrik

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