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## Homework Statement

Let X=ℝ

^{3}and let V={(a,b,c) such that a

^{2}+b

^{2}=c

^{2}}. Is V a subspace of X? If so, what dimensions?

## Homework Equations

A vector space V exists over a field F if V is an abelian group under addition, and if for each a ∈ F and v ∈ V, there is an element av ∈ V such that all of the following conditions apply for all a,b ∈F and all u,v ∈V -- a(u + v) = au + av, (a + b)v = av + bv, a(bv)= (ab)v, and 1 × v = v.

## The Attempt at a Solution

a

^{2}+b

^{2}=c

^{2}is abelian under addition because real numbers are commutative under that operation. I believe we would not have to concern ourselves with multiplication because we are not multiplying, and we would not have to concern ourselves with associativity under multiplication for the same reason. If we multiply this my the multiplicative identity, we get 1 × (a

^{2}+b

^{2}=c

^{2})=a

^{2}+b

^{2}=c

^{2}. So, we are good for a vector space.

V is a subspace of X because it is nonempty, and closed under linear operations -- our friend Pythagoras showed that is true several millennia ago. Addition and multiplication are closed under ℝ, so we are good to go. Also, it passes through the origin, 0^2 + 0^2 = 0^2, obviously enough. Therefore, I have is as a subspace existing in three dimensions.

Thank you all for your time,

SY