Problem 54: Examples of NOT Vector Spaces

Problem 54: Let V the set of all pairs $(x,y)$ of real numbers, and let $\F$ be the field of real numbers. Examine in each of the following cases whether V is a vector space over the field of real numbers or not?

$(i). (x,y)+(x_1,y_1) = (x+x_1,y+y_1)$
$c(x,y) = (|c|x,|c|y)$
$(ii). (x,y)+(x_1,y_1) = (x+x_1,y+y_1)$
$c(x,y) = (0,cy)$
$(iii). (x,y)+(x_1,y_1) = (x+x_1,y+y_1)$
$c(x,y) = (c^2x,c^2y)$

Solution:
$(i)$. We shall show that in this case the postulate
$(a+b)u = au+bu \;\;\forall a,b \in \F$ and $u \in \V$ fails.
Let $u=(x,y)$ and $a,b \in \R$. we have
$(a+b)u = (a+b)(x,y) = (|a+b|x,|a+b|y) $
Also
$au+bu = a(x,y)+b(x,y) = (|a|x,|a|y)+(|b|x,|b|y)$
$= (|a|x+|b|x,|a|y+|b|y) = ((|a|+|b|)x,(|a|+|b|)y)$
$\therefore (a+b)u \ne au+bu$.
Hence $V(\R)$ is NOT a vector space.

$(ii)$. We shall show that in this case the postulate
$1u = u \;\;\forall u \in \V$ fails.
Let $u=(x,y)$ and $x,y \in \R$. we have
$1u = 1(x,y) = (0,1y) = (0,y) \ne (x,y)$
$\therefore 1u \ne u$.
Hence $V(\R)$ is NOT a vector space.

$(iii)$. We shall show that in this case the postulate
$(a+b)u=au+bu \;\;\forall a,b \in \R$ and $u \in \V$ fails.
Let $u=(x,y)$ and $a,b \in \R$. we have
$(a+b)u = (a+b)(x,y) = ((a+b)^2x,(a+b)^2y) $
Also
$au+bu = a(x,y)+b(x,y) = (a^2x,a^2y)+(b^2x,b^2y) = (a^2x+b^2x,a^2y+b^2y) = ((a^2+b^2)x,(a^2+b^2)y)$
$(a+b)^2 \ne a^2+b^2$
$\therefore (a+b)u \ne au+bu$.
Hence $V(\R)$ is NOT a vector space.