# Is following subset of $\mathbb{Q^3}$ a subspace?

Given a set of vectors:

$$M = \{(a, b, c)\ |\ a^2= b^2 = c^2 \vee 3a = b+c \}$$

Is M a subspace of $$\mathbb{Q^3}$$?

I've already done the multiplication with a scalar, but I'm stuck at the addition of two vectors:

So if we have two vectors $$(a_1,b_1,c_1)$$ and $$(a_2,b_2,c_2)$$ which fulfill these conditions, for their sum we have:

$$a_1^2 + 2a_1a_2 + a_2^2 = b_1^2 + 2b_1b_2 + b_2^2$$, and thus

$$a_1a_2 = b_1b_2$$

For both $$a_1$$ and $$a_2$$ we have $$a^2-b^2=0 \Rightarrow a = b \vee a = -b$$

If $$a_1=b_1$$ then $$a_2 = b_2 \vee a_2 = -b_2$$, so I guess $$M$$ is not a subspace, because the equality doesn't hold always. Am I right?

• $\Bbb Q_3=\Bbb Q^3$? $x=a$? $y=b$? $z=c$? – Lord Shark the Unknown Dec 21 '18 at 21:04

Well, the triples $$(1,1,1)$$ and $$(-1,1,1)$$ are elements of $$M$$, however their sum $$(1,1,1) + (-1,1,1) = (0,2,2)$$ is not, so $$M$$ cannot be a linear subspace.