Find a basis for `RR(sqrt 2)` over `RR`.

Reformulate the Given Exercise: The task is to find a set of elements B that forms a basis for the field extension `RR(sqrt 2)` when it is viewed as a vector space over the base field `RR` (the real numbers).

General Strategy for a Solution:

1. Understand the Field Extension `RR(sqrt 2)`: Determine the form of the elements in this field extension and relate it the base field `RR`.
2. Determine Linear Dependence/Independence: Check if the the element `sqrt 2` is "new" or algebraic over the field `RR` , or if it is already an element of `RR`.
3. Find the Dimension: The dimension of `RR(sqrt 2)` over `RR` is the number of elements in the basis.
4. Construct the basis: Use the determined dimension to construct the basis B.

Specific Theory Used:

Field Extension `RR(sqrt 2)`: The smallest field containing both the base field `RR` and the element `sqrt 2`. Since `sqrt 2 = 1.414...` , which is a real number , `sqrt 2` is already an element of `RR`.
Vector Space Property: If `alpha` is an element of the base field F , then the field extension `F(alpha)` is simply equal to F itself , meaning `bb{F(alpha) = F}`.
Dimension of Trivial Extension: If V is a vector field over itself (i.e. V = F) , then the dimension of V over F is always 1: `bb{[F(alpha) : F] = 1}` if `bb{alpha in F}`.
Basis for a Field Over Itself: A basis for any field F as a vector space over F is the set containing only the unity element 1 of the field.

Do the Exercise:

Step 1: Determine the Relationship between `sqrt 2` and `RR`:
1. The base field is `F = RR` (the real numbers).
2. The element is `alpha = sqrt 2`.
3. Since `sqrt 2` is a real number , it is already contained within the base field : `sqrt 2 in RR`.
4. Therefore , the field extension is trivial: `RR(sqrt 2) = RR`.

Step 2: Determine the Dimension and Basis:
1. We are finding the basis for the vector space `RR` over the field `RR`.
2. The dimension of any vector space V over itself is 1: `bb[[RR(sqrt 2):RR] =[RR:RR] = 1}`.
3. A basis for a 1-dimensional space is any non-zero element. By convention , the simplest basis is the unity element 1.

Verification:

Spanning: Any element `y in RR(sqrt 2)` is a real number `(y in RR)`. It can be written as a linear combination of the single basis element 1 using a scalar `bb{y in RR}`: `bb{y = y * 1}`.
Linear Independence: The set `bb{{1}}` is linearly independent because `c * 1 = 0 implies c = 0` , where `c in RR`.

Conclusion: Basis for `RR(sqrt 2)` over `RR` is `bb{{1}}`. END