Extending basis of a subspace (From Gilbert Strang's Linear Algebra(4e). Section 3.5 . Problem 22)

Question: Suppose S is a 5-dimensional subspace of $R^6$. True or false.
a) Every basis for S can be extended to a basis for $R^6$ by adding one more vector
b) Every basis for $R^6$ can be reduced to a basis for S by removing one vector

For part a), the basis for S contains 5-dimensional vectors. How can adding another 5-dimensional vector help create a 6-dimensional basis for $R^6$ ? Isn't there a dimension mismatch ? Similarly for part b) ?
 A: You seem to think that if $v$ is a vector in a $5$ dimensional vector space, then $v$ has $5$ components, i.e., that it looks something like $v=(x_1,\cdots ,x_5)$. This is not the case. In fact, vectors in general need not look anything like a list of numbers. For instance, the space of all continuous functions $f:\mathbb R \to \mathbb R$ is a linear space. The dimension of a vector space is the number of vectors in a basis (after proving all bases of a vector space have the same size). 
Further hints:
a) Any linearly independent set of vectors in a vector space can be extended to a basis.
b) think of $\mathbb R^5$ sitting inside $\mathbb R^6$. Can you find a basis of $\mathbb R^6$ such that none of the basis vectors is in $\mathbb R^5$?
A: Since S is 5-D subspace of$ R^6$ ,so S consists of 5 Linearly Independent Six dimensional vectors.Now if you add one more Six dimensional vector((not 5-dimensional)  This is where u went wrong,) then you will have 6 Linearly indep. vectors which will span six dimensional space i.e $R^6$    
For part (B)converse is true
A: Personally, I think you need a strict definition of that vector space with some conditions, e.g. relate to orthogonality to other vector subspaces. For example, if that vector space is linear function of other vector spaces, your questions would be false.
Please leave your comments.
