Since i'm bored in class :

We know that the dimension of the linear vector space of polynomials on R of degree <= 2 (named after R[X]) is 3.

Let us prove that the linear vector space of polynomials on Q of degree <= 2 (named after Q[X]) is a subspace of R[X].

- The "0" of R[X] is the null polynomial, 0. 0 is in Q[X].

- If a is in Q[X] and b is in Q[X], then a+b is in Q[X].

- Let k be in R, and a be in Q[X]. k * a is in Q[X] (we don't add any roots, and all the roots stay the same)

Therefore, Q[X] is a subspace of R[X].

Hence, the dim(Q[X]) <= dim(R[X]) = 3

Now, we have the vectors x^2-5 , 3x^2 + 1, x+1. Let us prove that they are linearly independant.

Let a,b,c be so that for all x, a(x^2-5) + b(3x^2 + 1) + c(x+1) = 0.

This gives us :

a+3b = 0

c = 0

-5a+b+c = 0

Hence :

a+3b = 0

-15a +3b = 0

c = 0

And so :

c=0

a+3b = 0

16a = 0

Which gives us a = b = c = 0, hence the vectors are linearly independent.

Since we have 3 independent vectors and we know that the dimension of the space is at most 3, we have that the dimension of the space is 3 and hence that the 3 vectors form a basis (see

http://ltcconline.net/greenl/courses/203/Vectors/basisDimension.htm ).

CQFD.