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bbelson01

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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

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

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