Why is $\mathbb{R}^2$ not a subspace of $\mathbb{R}^3$? 
I cannot understand why $\mathbb{R}^2$ is not a subspace of $\mathbb{R}^3$.

My reasoning is as follows:
Choose any elements $v_1$ and $v_2$ from $\mathbb{R}^2$, add them together you get an element of $\mathbb{R}^2$ and same for scalar multiplication and $0$ vector is an element of $\mathbb{R}^2$. So where did I go wrong?
 A: The space $\mathbb{R}^2$ is isomorphic to the subset $(a,b,0)$ of $\mathbb{R}^3,$ but it's also isomorphic to infinitely many other 2-dimensional subspaces of $\mathbb{R}^3.$ Therefore, there's no canonical embedding, and you don't usually think of $\mathbb{R}^2$ as being contained in $\mathbb{R}^3.$ 
Another explanation is the vector (a,b) is not the same as the vector (a,b,0). We have a vector with 2 components & a vector with 3 components, so they are different objects.
Before you can discuss whether $\mathbb{R}^2$ is a subspace of $\mathbb{R}^3$ you need to embed $\mathbb{R}^2$ in $\mathbb{R}^3$ by defining an isomorphism between a subset of $\mathbb{R}^3$ & all of $\mathbb{R}^2.$ An obvious one is
\begin{equation*}
(a,b)\in\mathbb{R}^2\leftrightarrow (a,b,0)\in\mathbb{R}^3.
\end{equation*}
As I mentioned above there are infinite ways to do it. For example, another isomorphism to a subspace of $\mathbb{R}^3$ is 
\begin{equation*}
(a,b)\in\mathbb{R}^2\leftrightarrow (a,0,b)\in\mathbb{R}^3
\end{equation*}
In order to discuss whether elements of $\mathbb{R}^2$ are closed under addition in $\mathbb{R}^3,$ you first need to know how you map an element of $\mathbb{R}^2$ into $\mathbb{R}^3$ (the isomorphism). If you've done that, you should be able to show (using the subspace criteria), that $\mathbb{R}^2$ is isomorphic to a subspace of $\mathbb{R}^3.$ 
However you cannot say $\mathbb{R}^2$ is a subspace of $\mathbb{R}^3.$ 
What if you choose to embed $\mathbb{R}^2$ in $\mathbb{R}^3$ by 
\begin{equation*}
(a,b)\in\mathbb{R}^2\leftrightarrow (a,b,1)\in\mathbb{R}^3?
\end{equation*}
Clearly the zero vector is not in the embedded $\mathbb{R}^2,$ so it is not a subspace of $\mathbb{R}^3.$
Does that help? 
