Can I make an assumption about arbitrary numbers in a proof? Assume  $ x_1, x_2... x_{10} $ are different numbers and $ y_1, y_2... y_{10} $ are some arbitrary numbers. Prove that there exists some unique polynomial of degree not exceeding 9 such that:
$P(x_1)=y_1,P(x_2)=y_2...P(x_{10})=y_{10}$
Proof to the problem rests on the assumption that there is some polynomial Q(x) of degree not exceeding 9 such that:
$ Q(x_1)=y_1,Q(x_2)=y_2...Q(x_{10})=y_{10}$. 
Is it a valid assumption in a proof of this sort?
 A: Your assumption is incorrect because proving the existence of such a polynomial is an important part of the problem you are working on.
Here is a hint: Suppose you were given the problem with only two pairs $(x_1,y_1)$ and $(x_2,y_2)$. Then how would you find such a polynomial? Think in terms of a straight line. You are given two distinct (since $x_1 \neq x_2$) points so there always exists a straight line that passes through these points and the equation of that line will be polynomial of the required degree. 
If you were given $3$ points then a parabola passes through it and so on. 
If you are still unable to make progress then look for Lagrange Interpolation.
A: You need to prove two things: that there is such a polynomial and that there is only one.  To prove that there is one, you can show a way to construct it.  Can you find a way to construct a polynomial $Q_1(x)$ of degree not exceeding $9$ such that $Q_1(x_1)=y_1,Q_1(x_2)=0...Q_1(x_{10})=0$?  Now if you construct nine other polynomials with similar conditions and add them together, you get the $Q(x)$ you want.  Linearity is wonderful.  Now you need to prove it unique.  Assume there is another polynomial $R(x)$ of degree at most $9$ that satisfies the requirement.  How many roots does $R(x)-Q(x)$ have?
