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Problem 42: Show that V is not a vector space under the above defined operations.

Problem 42: Let R be the field of reals and V be the set of all ordered pairs$(x,y)$, where x, y are reals. Define.

$(x_1,y_1)+(x_2,y_2) = (x_1+x_2,y_1+y_2) \forall (x_1,y_1),(x_2,y_2) \in \V$ and

$\alpha(x,y)=(\alpha x,y) \forall \alpha \in \R$ and $(x,y) \in \V$.

Show that V is not a vector space under the above defined operations.

Solution: Given $\V = \{(x,y):x,y \in \R\}$

I. Properties under addition
i. Closure Property
Let \begin{align*} (x_1,y_1),(x_2,y_2) \in \V\\ \Rightarrow x_1,y_1,x_2,y_2 \in \R \Rightarrow x_1+x_2,y_1+y_2 \in \R\\ \therefore (x_1,y_1)+(x_2,y_2) = (x_1+x_2,y_1+y_2) \in \V \end{align*} $\Rightarrow \V$ is closed under addition.

ii. Associativity:
Let \begin{align*} (x_1,y_1),(x_2,y_2),(x_3,y_3)\in \V\\ \end{align*} Now, \begin{align*} \left[(x_1,y_1)+(x_2,y_2)\right]+(x_3,y_3) = \left[(x_1+x_2,y_1+y_2)\right]++(x_3,y_3)\\ = \left[(x_1+x_2)+x_3,(y_1+y_2)+y_3\right]\\ = \left[x_1+(x_2+x_3),y_1+(y_2+y_3)\right]\\ = (x_1,y_1)+\left[(x_2+x_3),(y_2+y_3)\right]\\ = (x_1,y_1)+\left[(x_2,y_2)+(x_3,y_3)\right]\\ \end{align*} $\Rightarrow$ Addition is associative in $\V$.

iii. Existence of additive identity:
For all $(x_1,y_1) \in \V, \exists (0,0) \in \V$
Now \begin{align*} (x_1,y_1)+(0,0)=(x_1+0,y_1+0) = (x_1,y_1)\\ (0,0)+(x_1,y_1)=(0+x_1,0+y_1) = (x_1,y_1)\\ \end{align*} $\Rightarrow (0,0)$ is the additive identity in $\V$.

iv. Existence of additive inverse:
Let $(x,y) \in \V$, be any element $(-x,-y)\in \V$
\begin{align*} \ \because x,y \in \R \Rightarrow -x,-y \in \R\\ \end{align*} Now \begin{align*} (x,y)+(-x,-y)= (x+(-x),y+(-y))=(0,0)\\ (-x,-y)+(x,y)= ((-x)+x,(-y)+y)=(0,0)\\ \end{align*} $\Rightarrow (-x,-y)$ is the additive inverse of $(x,y)$ for each $(x,y) \in \V$.

v. Commutative Law:
Let $ (x_1,y_1),(x_2,y_2) \in \V$, Now \begin{align*} (x_1,y_1)+(x_2,y_2)= (x_1+x_2,y_1+y_2) = (x_2+x_1,y_2+y_1) = (x_2,y_2)+(x_1,y_1) \end{align*} $\Rightarrow$ addition is commutative in $\V$.

II. Properties under Scalar Multiplication
vi. Closure under Scalar Multiplication
Let $\alpha\in \R$ and $(x,y) \in \V$ s.t. $x,y \in \R$ \begin{align*} \alpha (x,y) = (\alpha x,y) \in \V \therefore \alpha (x,y) \in \V \end{align*} $\Rightarrow \V$ is closed under Scalar Multiplication.

vii.Distributivity of Addition of Real Numbers
Let $\alpha, \beta\in \R$ and $(x,y) \in \V$ s.t. $x,y \in \R$ \begin{align*} (\alpha+\beta)(x,y)= ((\alpha+\beta) x,y) = (\alpha x + \beta x, y) \end{align*} And \begin{align*} \alpha(x,y)+\beta(x,y)= (\alpha x,y)+(\beta x,y) = (\alpha x+ \beta x, 2y) \end{align*} $\Rightarrow (\alpha+\beta)(x,y) \ne \alpha(x,y)+\beta(x,y)$ . $\Rightarrow $ Distributivity of Addition of Real Numbers does not exists.

Hence, $\V$ is not a vector space over reals$.

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