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There is a slight part of the following proof in my textbook which I don't quite get.

THEOREM

If $x_0, x_1,...,x_n$ are distinct real numbers, then for arbitrary values $y_0, y_1,...,y_n$, there is a unique polynomial $p_n$ of degree at most $n$ such that

$$p_n(x_i) = y_i \quad (0 \leq i \leq n)$$

PROOF

Let us prove the uniqueness or unicity first. Suppose there were two such polynomials, $p_n$ and $q_n$. Then the polynomial $p_n - q_n$ would have the property $(p_n - q_n)(x_i) = 0$ for $0 \leq i \leq n$. Since the degree of $p_n - q_n$ can be at most $n$, this polynomial can have at most $n$ zeros if it is not the $0$ polynomial. Since the $x_i$ are distinct, $p_n - q_n$ has $n+1$ zeros; it must therefore be $0$. Hence, $p_n = q_n$.

OK, so the only thing I don't quite here is the part:

"Since the $x_i$ are distinct, $p_n - q_n$ has $n+1$ zeros"

How does it follow that this must have exactly $n+1$ zeros. I have pondered this for some time, but just can't see why this is obvious. I'm probably overlooking something simple here, but if anyone can help me out, I would greatly appreciate it!

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  • $\begingroup$ Do you agree with the statement that $(p_n - q_n)(x_i)$ would be equal to $0$ for $i = 0,1,\ldots,n$? $\endgroup$
    – littleO
    Commented Nov 19, 2012 at 21:03
  • $\begingroup$ Yes! See below. I understand it now. Thanks! $\endgroup$
    – Kristian
    Commented Nov 19, 2012 at 21:09

2 Answers 2

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You don't know that it has exactly $n+1$ zeros, just that it has at least that many because you have found them: $x_0, x_1, \ldots, x_n$. But an $n^{\text{th}}$ degree polynomial can only have $n$ zeros, unless it is identically zero...

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  • $\begingroup$ Ah, thanks. I get it now :). Really appreciate your answer! $\endgroup$
    – Kristian
    Commented Nov 19, 2012 at 21:08
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The phrase "Since the $x_i$ are distinct, $p_n - q_n$ has $n+1$ zeros" is intended to mean that there are at least $n+1$ zeros. It would have been a little better to say "at least," but being absolutely precise can be a bit tedious.

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  • $\begingroup$ Thanks a lot! Yes, I get it now :) $\endgroup$
    – Kristian
    Commented Nov 19, 2012 at 21:09

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