Question

Question: Show that ${\mathit{Q}}^{\mathbf{2}}$is a vector space over $\mathit{Q}$.

Short answer

Answer:

${\mathit{Q}}^{\mathbf{2}}$is a vector space over$\mathit{Q}$.

Step-by-step solution

Definition of space.

A space consisting of vector, together with the associative and commutative operations of addition of vectors and the associative and distributive operation of multiplication of vectors by scalars.

Showing that Q2 is vector space over Q.

Let the set ${\mathit{Q}}^{\mathbf{2}}$over a field $\mathit{Q}$of rational numbers. As ${\mathit{Q}}^{\mathbf{2}}$is a group under addition and $\left(0,0\right)\mathbf{\in }{\mathit{Q}}^{\mathbf{2}}$is its additive identity.

Let

$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\in }{\mathit{Q}}^{\mathbf{2}}$. So,

$$\begin{gathered} \left(x,y\right)\mathbf{+}\left(-x,-y\right)\mathbf{=}\left(x-x\right)\mathbf{,}\left(y.y\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(0,0\right) \end{gathered}$$

Therefore the negative of $\left(x,y\right)$is $\mathbf{(}\mathbf{-}\mathbf{x}\mathbf{,}\mathbf{-}\mathbf{y}\mathbf{)}$.

Let $\mathit{a}\mathbf{\in }\mathit{Q}$and $\left(x,y\right)\mathbf{\in }{\mathit{Q}}^{\mathbf{2}}$.

Then scaler multiplication defined as

$\mathit{a}\left(x,y\right)\mathbf{=}\left(ax,ay\right)$

Hence set ${\mathit{Q}}^{\mathbf{2}}$is a vector space over$\mathit{Q}$.