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Hi I am taking a number theory class and so far I have been proving modular congruences, modular arithmetic, and prime properties. There is this theorem that came up in the textbook and apparently it does not involve any modular arithmetic. The theorem is as follows

Suppose $f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ is a polynomial of degree $n>0$. Then there exists an integer $k$ such that if $x>k$ then $f(x)>0$.

I feel like this would come up in real analysis, but I have not come that far in my studies. I have an idea of applying induction and using the ceiling function somehow, but I have no idea how to start off this proof. Any help will do and thank you

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    $\begingroup$ You need $a_n>0$. $\endgroup$ – user10354138 May 19 at 6:53
  • $\begingroup$ the question does not state this in the book, or would that be part of the proof? $\endgroup$ – tytds May 19 at 6:55
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    $\begingroup$ @tytds, the highest coefficiet must be positive, because otherwise we have a counterexample to the statement: $n = 1, a_{1} = -1, a_{0} = 0$; i.e., $f(x) = -x$. $\endgroup$ – avs May 19 at 6:58
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One should note that the field of interest here (presumably) is $\mathbb{R}$, so that $f(x)$ is a real polynomial. Notice first that for $a_n<0$, this fails completely, because we can just choose $f(x)=-x$ and it does not have the property.

If we assume that $a_n>0$, then write $$ \frac{f(x)}{x^n}=a_n+a_{n-1}x^{-1}+\cdots +a_1 x^{1-n}+a_0 x^{-n}$$ so that $\lim_{x\to\infty} \frac{f(x)}{x^n}=a_n>0$. This implies that $\lim_{x\to\infty}f(x)=\infty$, so that $f(x)>0$ for $x\ge N$ for some $N\in \mathbb{R}$.

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  • $\begingroup$ One more thing. There is a similar theorem in the book that instead of $f(x)>0$, we must prove that for any number $M$, there exists an integer $k$ (entirely dependent on$M$), such that if $x>k$, then $f(x)>M$. Would the same idea apply? $\endgroup$ – tytds May 19 at 7:23
  • $\begingroup$ @tytds Consider the polynomial $g(x)=f(x)-M$ $\endgroup$ – Mark Bennet May 19 at 7:26
  • $\begingroup$ @MarkBennet then $f(x)$ would be a composite function? $\endgroup$ – tytds May 19 at 7:28
  • $\begingroup$ @tytds You have proved, for example that $x^2-x$ is eventually positive and want to prove that it eventually exceeds $M$. Consider $x^2-x-M$ which you prove to be eventually positive by the same method. This is saying the same as $x^2-x$ is eventually greater than $M$. $\endgroup$ – Mark Bennet May 19 at 8:06
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I don't know what you're allowed to assume. Anyway here's one way to see why (as someone mentioned in the comments the leading coefficient should be positive for the theorem to hold; also the coefficients should be real):

It is obvious that the polynomial cannot be written as a product of more than $n$ linear polynomials, possibly with complex coefficients (for otherwise, it should have a degree greater than $n$). Thus, $f(x)$ does not vanish at more than $n$ points on the real line. Hence, there is some real $x'$ so that for all $x>x',$ we must have $f(x)\ne 0.$ By continuity (and the fact that $f(x)\to+\infty$ as $x\to+\infty$ -- factor out $x^n$ to see this, recalling that $a_n>0$), it follows that $f(x)<0$ just before the last zero, so that just after this $f(x)>0.$ Taking $x'$ to be sufficiently close to, but after, this last zero, and recalling after $x'$ the polynomial retains constant sign, it follows that it must be positive for all $x>x'.$ Finally, by the principle of Archimedes you can find some positive integer greater than this $x'.$

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