MHB Proving Vector Subspace $U \notin R^3$

[bbelson01]

Hi All,

How do I prove that $U=\{(x,y,z)|x \mbox{ is an integer}\}$ is not a subspace of $R^3$?

I understand that I have to show $U$ is closed or not closed under vector addition and scalar multiplication but I'm unsure how I represent $x$ as an integer.

I would say:
let $V=\{v=(x,y,z) \in R^3\}$ a set of vectors in $R^3$.

$v_1 = (x_1, y_1, z_1)$ and $v_2=(x_2, y_2, z_2)$ then $v_1 + v_2=(x_1+x_2, y_1+y_2, z_1+z_2)$ a vector in $U$. $U$ is closed under vector addition.

Now let $c$ be an element of the set of real numbers. Then $cv=c(x,y,z) =(cx, cy, cz)$ which is an element of $U$. So $U$ is closed under scalar multiplication.

This is clearly not right as the question says to show it is NOT a subspace. Where am I going wrong?

Thanks

 

[Opalg]

Hi All,

How do I prove that U={(x,y,z)|x is an integer} is not a subspace of R^3?

I understand that I have to show U is closed or not closed under vector addition and scalar multiplication but I'm unsure how I represent x as an integer.

I would say:
let V={v=(x,y,z) \epsilon R^3} a set of vectors in R^3.

v1 = (x1, y1, z1) and v2=(x2, y2, z2) then v1 + v2=(x1+x2, y1+y2, z1+z2) a vector in U. U is closed under vector addition.

Now let c be an element of the set of real numbers. Then cv=c(x,y,z) =(cx, cy, cz) which is an element of U. So U is closed under scalar multiplication.

This is clearly not right as the question says to show it is NOT a subspace. Where am I going wrong?

Thanks

Think more carefully about the paragraph in red. For (cx,cy,cz) to be an element of U, its first coordinate cx must be an integer. Does that have to be the case?

 

[bbelson01]

Thank you for your response. That was pretty obvious wasn't it? Ooops.