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`.

Do the Exercise:
We will select three pairs of vectors , ensuring: The two vectors in each pair are not scalar multiples of each other (linear independence). And that all six vectors are distinct (disjoint property).

Basis 1 `(B_1)`. The Standard Basis: Let's choose the most straightforward independent vectors:

`B_1 = {([1] , [0]) , ([0] , [1])}`. We can check this:`([1] , [0])` is not a scalar multiple of `([0] , [1])`. `B_1` is a basis.

Basis 2 `(B_2)`. Non-Standard , Disjoint from `B_1`: We must choose two new , distinct vectors that are independent of each other:

`B_2 = {([1] , [1]) , ([1] , [-1])}`. We can check this:`([1] , [1])` is not a scalar multiple of `([1] , [-1])`. Also , neither

`([1] , [1])` nor `([1] , [-1])` is in `B_1`. `B_2` is a basis , and `B_1 cap B_2 = emptyset`.

Basis 3 `(B_3)`. A Third Disjoint Basis : We need two more vectors that are distinct from all vectors in `B_1` and `B_2` , and independent of each other. Let's try the following vectors:

`B_3 = {([2] , [1]) , ([3] , [1])}`. We have that `([2] , [1])` is not a scalar multiple of `([3], [1])`.

Let `a([2] , [1]) + b([3] , [1]) = ([0] , [0]) implies ([2a] , [a]) + ([3b] , [b]) = ([0] , [0]) implies ([2a + 3b] , [a + b]) = ([0] , [0]) implies a = -b`

and `2a + 3b = 0 implies 2 * (-b) + 3b = 0 implies b = 0 implies a = 0`. Or we can calculate the Determinant of

`[[2 , 3] , [1 , 1]]` , which will give us that `det(A) = (2 * 1) - (1 * 3) = -1 ne 0`. Since det(A) is not equal to zero , the two vectors are linearly independent.

Check Disjoint Property: `([2] , [1])` is not in `B_1` or `B_2` and `([3] , [1])` is not in `B_1` or `B_2`. So , `B_3` is a basis with `B_1 cap B_2 = emptyset and B_2 cap B_3 = emptyset`.

Conclusion: Three bases for `RR^2` over `RR` with no vector in common are:

`B_1 = {([1] , [0]) , ([0] , [1])} , B_2 = {([1] , [1]) , ([1] , [-1])} , B_3 = {([2] , [1]) , ([3] , [1])}`. No two of these bases have a vector in common. End