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

• MHB
• bbelson01
In summary, the individual is asking for help in proving that the set U={(x,y,z)|x is an integer} is not a subspace of R^3. They understand that they need to show that U is not closed under vector addition and scalar multiplication, but they are unsure how to represent x as an integer. They provide a potential solution, but realize their mistake when it comes to showing that U is closed under scalar multiplication. They realize that for (cx,cy,cz) to be an element of U, cx must be an integer, but this may not always be the case.
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

Last edited by a moderator:
bbelson01 said:
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?

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

## 1. What is a vector subspace in $R^3$?

A vector subspace in $R^3$ is a subset of $R^3$ that satisfies three properties: closure under vector addition, closure under scalar multiplication, and contains the zero vector. In simpler terms, a vector subspace is a collection of vectors in $R^3$ that can be added and multiplied by scalars to produce other vectors within the subspace.

## 2. How do you prove that a set is not a vector subspace in $R^3$?

To prove that a set is not a vector subspace in $R^3$, you must show that it fails to satisfy at least one of the three properties mentioned above. This can be done by providing a counterexample, where two vectors within the set do not produce a vector within the set upon addition or scalar multiplication.

## 3. What is the importance of proving a set is not a vector subspace in $R^3$?

Proving that a set is not a vector subspace in $R^3$ is important because it allows us to identify which sets do not follow the properties of a vector subspace. This can help us to better understand the characteristics of vector subspaces and how they differ from other sets in $R^3$.

## 4. Can a set be a vector subspace in $R^3$ and $R^2$ at the same time?

No, a set cannot be a vector subspace in both $R^3$ and $R^2$ at the same time. This is because $R^3$ and $R^2$ have different dimensions, and the properties of a vector subspace in each dimension are different. A set must satisfy the properties of a vector subspace in its specific dimension in order to be considered a vector subspace.

## 5. What are some common examples of sets that are not vector subspaces in $R^3$?

Some common examples of sets that are not vector subspaces in $R^3$ include: a line that does not pass through the origin, a plane that does not contain the origin, and a circle that does not lie in the xy-plane. These sets do not satisfy the closure under vector addition property, as they do not contain the zero vector.

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