Proof of surjectivity of a polynomial

Solution 1:

Your idea is sound but your details are a little confusing. This is easier. You know that $p$ can be arbitrarily large or small so, given $r\in R$ choose $a$ so that $p(a)\lt r$ and choose $b$ so that $p(b)\gt r$. The Intermediate Value theorem assures that $p$ takes all values between $p(a)$ and $p(b)$ so there exists $c\in (a,b)$ such that $p(c)=r$. Since this is true for all $r$, $p$ is surjective.

Solution 2:

You got the idea.

We know that

• $\forall M \ge 0, \exists H>0: \forall x > H \implies f(x) > M$.

• $\forall M \ge 0, \exists H>0: \forall x < -H \implies f(x) < -M$.

We can choose $H_1>0$ such that if $x> H_1$ then $f(x) > |y|$. Let $x_1 >H_1$.

We can choose $H_2>0$ such that if $x<-H_2$ then $f(x)<-|y|$. We choose $x_2<-H_2$.

Then we have

$$f(x_2)<-|y| \le y \le |y| < f(x_1)$$

Since $p$ being a polynomial is continuous, by intermediate value theorem, since $y \in $[f(x_2), f(x_1)] we conclude that we can find $x\in [x_2, x_1]$ such that $f(x)=y$.