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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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    That answer depends on what course it is. You are on the right track.2011-04-07
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    Note that you don't have to *solve* that equation, only to prove it has a real solution.2011-04-07
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    If the discriminant equals or is larger than zero can i claim a proof?But suppose i don't remember how to solve 3° equations what could i do?2011-04-07
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    I believe *continuity* is the word you're supposed to think about (I hope I'm not revealing too much :)2011-04-07
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    @Bill your edit left an extra +1 in the question.2011-04-07
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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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    That it at some point it crosses the x axis so we can conclude that2011-04-07
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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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    As every polynomial is continuous on $R$ and use the Intermediate value theorem?2011-04-07
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    Yup you got it.2011-04-07
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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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    This is certainly true, but its justification seems to require some analytic reasoning with (e.g.) the Intermediate Value Theorem. Thus the latter answer seems more fundamental. On the other hand, your answer is accessible to a precalculus student and the OP has not specified her background, so there is a good chance this is exactly what her instructor is expecting to hear.2011-04-07
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    @Pete, I have tried to express it clearer I don't think it needs any analysis.2011-04-07
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    Hmm. First, a polynomial with real coefficients is not a statement in the language of rings for two reasons: (i) the variable $x$ is unbound by quantitifers, so it is a rather a *formula*, (ii) but it is not a formula either, because the names of the individual real numbers are not part of this language. Second and more to the point: you seem to be assuming that $\mathbb{R}[\sqrt{-1}]$ is algebraically closed. This is an analytic fact; one famous proof starts by showing that every odd degree real polynomial has a root and then applying Sylow theory.2011-04-07
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    (And by the way, why are you bringing model-theoretic language into the question? This question lies somewhere in the closed interval from precalculus to calculus.)2011-04-07
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    @Pete, I was going to say we don't need algebraic closure (which is true).. but we do need existence of solutions! Algebraic closure is probably the best way to get it... This arguments depends on much stronger things than it's proving. Thanks for helping me realize that.2011-04-07
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    @Pete: What precisely do you mean when you say "$\mathbb R[\sqrt{-1}]$ is algebraically closed" is an "analytic fact"?2011-04-07
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    Fundamental Theorem of Analysis2011-04-07
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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