Can someone help me with part b Let b0 b1 bn and a0 a1 an be

Can someone help me with part b?

Let {b_0, b_1, ...,b_n} and {a_0, a_1, ...,a_n} be subsets of a given field F. We wish to construct a polynomial p of degree at most n such that p(a_i) = b_i for each i = 0, 1,..., n. Define q_i(x) = (x - a_0) ... (x - a_i-1)(x - a_i + 1) ... (x - a_n). Explain why q_i(a_j) = 0 for j notequalto i, but q_i(a_i) notequalto 0. Write q_i(a_i) = c_i and show that the polynomial p(x) = b_0/c_0 q_0(x) + b_1/c_1 q_1(x) + ... + b_n/c_n q_n(x) is a polynomial of degree at most n for which p(a_i) = b_i, for each i = 0,1,..., n. Prove that there is one and only one polynomial p(x) of degree at most n such that p(a_i) = b_i for each i = 0, 1,..., n.

Solution

First, note that if ai = aj , then bi better equal bj—else no polynomial can equal both bi and bj when evaluated at ai = aj

For a contradiction, suppose there are two distinct polynomials P(x) and Q(x) of degree at most n such that for all i,

P(ai) = Q(ai) = bi

Then consider the polynomial R(x) = P(x) Q(x). It has degree at most n, since it is the difference of two polynomials of degree at most n. Moreover,

R(ai) = P(ai) Q(ai) = 0 for all the n + 1 settings of i = 0, 1, . . . , n.

Once again, R is a polynomial of degree at most n, with n + 1 roots. By the contrapositive of the theorm which states any non-zero polynomial of degree at most n has at most n roots, R(x) must be the zero polynomial. And hence P(x) = Q(x), which gives us the contradiction.