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Question

Proposition: Every element in a vector space has a unique additive inverse. Proof: [...]

Answer

Suppose V is a vector space. Let v ∈ V. Suppose that w and w' are additive inverses of v. Then w' = w' + 0 = w' +(v + w) = (w'+v)+w = 0 + w = w. Thus w = w', as desired.

Question

Proposition: Every element in a vector space has a unique additive inverse. Proof: [...]

Answer

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Question

Proposition: Every element in a vector space has a unique additive inverse. Proof: [...]

Answer

Suppose V is a vector space. Let v ∈ V. Suppose that w and w' are additive inverses of v. Then w' = w' + 0 = w' +(v + w) = (w'+v)+w = 0 + w = w. Thus w = w', as desired.

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**Open it**

Proposition: Every element in a vector space has a unique additive inverse. Proof: Suppose V is a vector space. Let v ∈ V. Suppose that w and w' are additive inverses of v. Then w' = w' + 0 = w' +(v + w) = (w'+v)+w = 0 + w = w. Thus w = w', as desired.

#### Original toplevel document (pdf)

owner: eshi - (no access) - Sheldon_Axler_Linear_Algebra_Done_Right.pdf, p25

Proposition: Every element in a vector space has a unique additive inverse. Proof: Suppose V is a vector space. Let v ∈ V. Suppose that w and w' are additive inverses of v. Then w' = w' + 0 = w' +(v + w) = (w'+v)+w = 0 + w = w. Thus w = w', as desired.

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
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repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

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