Find Three Disjoint Bases for `RR^2` over `RR`.

Solution:
Reformulate the Given Exercise:
The task is to find three distinct sets , say `B_1 , B_2` , and `B_3` , such that:
- Each set is a basis for the vector space `RR^2` over the field `RR` (the real numbers).
- The sets are mutually disjoint , meaning the intersection of any two of them is the empty set:
`B_1 cap B_2 = emptyset , B_1 cap B_3 = emptyset , B_2 cap B_3 = emptyset`.

General Strategy for a Solution:
Recall the Dimension: The dimension of `RR^2` over `RR` is 2 , so every basis must contain exactly two vectors.
Define the Vector Pool: We need a total of `3 times 2 = 6` distinct vectors to form three disjoint bases.
Let `V = {v_1 , v_2 , v_3 , v_4 , v_5 , v_6}` be this set of six distinct vectors in `RR^2`.
Form the Bases: Partition the set V into three disjoint subsets:
`B_1 = {v_1 , v_2} , B_2 = {v_3 , v_4}` , and `B_3 = {v_5 , v_6}`.
Verify Linear Independence/Spanning: For each set `B_i` , choose the vectors such that they are linearly independent. Since the dimension of the space is 2 , any of 2 linearly independent vectors in `RR^2` automatically form a basis.
To ensure linear independence for a set `{a , b}` , we must check that `k_1 a + k_2 b = 0` implies `k_1 = k_2 = 0`. This is equivalent to checking that one vector is not a scalar multiple of the other.
Specific Theory Used: The solution relies on the fundamental consepts of Vector Space and Bases:
Vector Space: The space `(RR^2 , + , * , RR)` , consists of the set of all ordered pairs of real numbers with standard vector addition and scalar multiplication by real numbers.
Basis: A subset B of a vector space V is a basis if:
The set B spans V , meaning every vector in V can be written as a linear combination of vectors in B. And the set B is linearly independent , meaning no vector can be written as a linear combination of the others.
Dimension Theorem: For a finite-dimensional vector space V , all bases have the same number of vectors. This number is the dimension of V , denoted `dim(V)`. For `RR^2` over `RR` , dim`(RR^2) = 2`.
Criterion for a Basis in `RR^n`: Any set of n linearly independent vectors in `RR^n` is a basis for `RR^n`. In our case , any set of 2 linearly independent vectors in `RR^2` is a basis.