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We want to prove the following:

For any $n+1$ distinct real numbers $a_0, a_1, ..., a_n$ and any $n+1$ real numbers $b_0, b_1, ..., b_n$, there exists a polynomial of degree at most $n$ taking the value $b_i$ at $a_i$ for all $i=0, 1, ..., n$.

This was discussed in a class as a follow-on from the Chinese Remainder Theorem, but I can't see how so.

Real numbers have unique prime factorisations, so I thought about considering ideals of the form $p_i \mathbb{Z}$ where $p_i$ is the $i$th prime.

Then let $p_n$ be the greatest prime that features in the prime factorisations of the $a_i$.

Then applying the CRT, but I'm not sure if this approach is misguided.

I'd be interested to see an answer/guidance on how this can be done.

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  • $\begingroup$ Do you mean real numbers, rational numbers, or integers? I don't see what the CRT would have to do with real numbers. $\endgroup$ Commented Jun 10, 2013 at 22:34
  • $\begingroup$ I mean real numbers. $\endgroup$
    – Mathmo
    Commented Jun 10, 2013 at 22:36

1 Answer 1

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Hint: apply CRT to the system $\ f(x) \equiv b_i \pmod{x-a_i},\,$ using that the $\,x-a_i\,$ are pairwise comaximal, since the $\,a_i\,$ are distinct.

It deserves to be much better known that Lagrange interpolation is a special case of CRT.

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  • $\begingroup$ Thanks, I see how this works now. $\endgroup$
    – Mathmo
    Commented Jun 10, 2013 at 22:41

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