Give a basis for the vector space `QQ(sqrt 2)` over the field `QQ`.

Reformulate the given exercise:
The task is to find a set of elements B in the field extension `QQ(sqrt 2)` such that:
- The set B is a basis for `QQ(sqrt 2)` when viewed a a vector space over the field `QQ` (the rational numbers).
- Specifically , we need to find the set B and determine the dimension `[QQ(sqrt 2):QQ]`.

General Strategy for a Solution:

1. Understand the elements: Recall the definition of the field `QQ(sqrt 2)`. It consists of all numbers that can be formed by rational numbers and `sqrt 2`.
2. Find the minimal Polynomial: Identify the non-zero polynomial p(x) with coefficients in `QQ` of the lowest possible degree such that `sqrt 2` is a root.
3. Relate Degree to Dimension : The degree of the minimal polynomial of `sqrt 2` over `QQ` is equal to the dimension of the vector space `QQ(sqrt 2)` over `QQ`.
4. Construct the Basis: If the dimension is n , a basis is given by the set `bb{{1 , alpha , alpha^2 , ... , alpha^{n-1}}}` , where `alpha = sqrt 2`.

Specific Theory used:

Field Extension `QQ(sqrt 2)`: This is the smallest field containing both `QQ` and `sqrt 2` , and the elements of `QQ(sqrt 2)` are of the form `bb{a + b sqrt 2}`, where `a , b in QQ`.
Minimal Polynomial: This is the unique monic polynomial `p(x) in QQ[x]` of the smallest possible degree such that `p(sqrt 2) = 0`. For `sqrt 2` , the minimal polynomial is `bb{x^2 - 2}` , because `sqrt 2` is a root and it is irreducible over `QQ` (by the Rational Root Theorem or Eisenstein's Criterion).
Dimension of a Field Extension (Tower Law): The degree of the field extension , denoted by `[QQ(sqrt 2):QQ]` , is equal to the degree of the minimal polynomial of the element `sqrt 2` over `QQ`: If `bb[deg(p(x)) = n]` , then the set `B = {1 , alpha , alpha^2 , ... , alpha^{n-1}}` is a basis for `QQ(alpha)` as a vector space over `QQ`.

Do the Exercise:

Step 1 : Determine the Minimal Polynomial and Dimension:
1. The algebraic number is `alpha = sqrt 2`.
2. We look for `bb{p(x) in QQ}` such that `p(sqrt 2) = 0`. We start with `x = sqrt 2 implies x^2 = 2 implies x^2 - 2 = 0`.
3. The minimal polynomial for `sqrt 2` over `QQ` is `bb {p(x) = x^2 - 2}`.
4. The degree of this polynomial is n = 2.
5. Therefore , the dimension of the vector space is `bb{[QQ(sqrt 2):QQ] = 2}`.

Step 2 : Construct the Basis:
1. Since the dimension is 2 , the basis set B will contain n elements.
2. Using the form `B = {1 , alpha , alpha^2 , ... , alpha^{n-1}}` with `alpha = sqrt 2` and n = 2 , we get `bb{B = {1 , sqrt 2}}`.

Verification:

Spanning: Any element in `QQ(sqrt 2)` has the form `a + b sqrt 2` , where `a , b in QQ`. This is clearly a linear combination of the vectors `{1 , sqrt 2}` with scalars a and b from the field `QQ`: `bb{a * 1 + b * sqrt 2}`.
Linear Independence: We must show that `c_1 * 1 + c_2 * sqrt 2 = 0` , where `c_ , c_2 in QQ` , implies `c_1 = 0` and `c_2 = 0`. Assume `c_2 ne 0`, then `sqrt 2 = -c_1/c_2`. Since `c_1 , c_2 in QQ` , their ratio is also in `QQ`. This means `sqrt 2` is a rational number , which is a contradiction. Therefore we must have that `c_2 = 0`. Substituting `c_2 = 0` back into the equation gives `c_1 * 1 + 0 * sqrt 2 = 0` , so `c_1 = 0`. The set `{1 , sqrt 2}` is linearly independent.

Conclusion: Since `bb{B = {1 , sqrt 2}}` is linearly independent and spans `QQ(sqrt 2)` over `QQ` , it is a basis. END