# Real Euclidean Space

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Consider the following subset B of R^3 (real Euclidean space):

B={(x,y,z) in R^3 such that x=y+z}

With the usual "component wise" addition and scalar multiplication in R^3, is B a vector space? Prove or disprove your answer.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Proposition: The subset B <= R^3 with B = {(x, y, z)|x = y + z|} is a vector subspace of R^3.

Proof: We will show that B is closed under both scalar multiplication and subtraction.

Let (x, y, z) <- B, (u, v, w) <- B and r <- R be arbitrary. Then

x - y - z = 0 and u - v - w = 0. Hence,

(x - u) - (y - v) - (z - w) = (x - y - z) - (u - v - w) = 0

Therefore, (x, y, z) - (u, v, w) = (x - u, y - v, z - w) <- B. Moreover,

rx - ry - rz = r(x - y - z) = r . 0 = 0

Therefore, (rx, ry, rz) <- B. Hence, B is a subspace of R^3.

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