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How do I show that there exists a real number that equals its cube plus its square plus 1? I was thinking $x = x^3+x^2+1$ then solve for $x$?

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    @Zach: Thanks, I submitted a new edit proposal. I'm here slowly trying to learn the extra syntax rules the math editor adds.2011-04-07

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To make things as simple as possible: you want $x$ such that $x = x^3 + x^2 + 1$, that is $x^3 + x^2 - x + 1 = 0$. And every cubic with real coefficients has a real root (because it has different signs at $x$ and $-x$ for large enough $x$).

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Let $f(x) := x^3 + x^2 -x + 1$. Consider the following,

$\mathrm{sgn}\left(\lim_{x \to \infty} \quad f(x) \right),$

and,

$\mathrm{sgn}\left(\lim_{x \to -\infty} \quad f(x) \right).$

What can you conclude?

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    there exists a real root2011-04-07
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Another way is to show that it is continuous, and then find a value of the curve $y=x^3+x^2+1$ below the $y=x$ line and another point above the $y=x$ line.

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    and use$\mathbb{R}$$\mathbb{R}$2011-04-07
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If a polynomial has real coefficients then its solutions will be either real or come in complex conjugate pairs. That implies that every polynomial of odd degree with real coefficients has at least one real root.

The reason that the non-real solutions must come in conjugate pairs is because they must have the same symmetry as the polynomial: The polynomial does not change when you switch i with -i so the solution set cannot change when you do that either.


A polynomial (with real coefficients) is a statement in the language of rings that defines a set of points $X$ in $\mathbb C$. These sets are called $\mathbb R$ definable and the symmetries of $\mathbb C$ that fix $\mathbb R$ are relevant, let $\sigma$ generate them then $x \in X \iff x \in \sigma(X)$.

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    @Bill: I mean that the definition / construction of $\mathbb{R}$ is partially of an analytic / topological nature rather than being purely algebraic. The same applies to $\mathbb{R}[\sqrt{-1}]$. Any proof of the algebraic closure of $\mathbb{C}$ has to use analysis or topology. Conversely, the *least* amount of such that I have ever seen is in Artin's proof which requires *only* that every odd degree polynomial over $\mathbb{R}$ has a root. Proving this by recourse to the fundamental theorem of algebra need not be circular (depending on the proof), but it's using a lot to prove a little.2011-04-08