# How to prove that $n^{\frac{1}{3}}$ is not a polynomial?

I'm reading Barbeau's Polynomials, there's an exercise:

How to prove that $n^{\frac{1}{3}}$ is not a polynomial?

I've made this question and with the first answer as an example, I guess I should assume that:

$$n^{\frac{1}{3}}=a_pn^p+a_{p-1}n^{p-1}+\cdots+ a_0n^0$$

And then I should make some kind of operation in both sides, the resultant difference would be the proof. But I have no idea on what operation I should do in order to prove that.

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Presumably on the right you want $t$, not $n$. –  André Nicolas Nov 30 '12 at 3:55
The question should be more clearly stated. $n^\frac13$ is not a polynomial in $n$ because it is not of the right form (a linear combination of monomials in $n$), period. By the same token $\exp(2\ln x)$ is not a polynomial in $x$, even though for $x>0$ one has $\exp(2\ln x)=x^2$. The question should be to show that $n\mapsto n^\frac13$ is not a polynomial function of $n$, where the domain of the function should be specified (for instance the name $n$ suggests integers only, in which case differentiation arguments cannot be used). –  Marc van Leeuwen Nov 30 '12 at 8:29
@MarcvanLeeuwen I just copied as I've seen in the book. –  Vÿska Nov 30 '12 at 17:11

Cube both sides, then consider whether the difference could be identically $0$.

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Lol. I just realized I had the same answer. Bleh. I'll leave it there. –  Patrick Da Silva Nov 30 '12 at 8:06

If $t^{1/3}$ were a polynomial, then its degree would be at least one (because it is not constant). This would imply $$\lim_{t\to\infty}\frac{t^{1/3}}t\ne0.$$ But, precisely, the limit above is indeed zero. So $t^{1/3}$ cannot be a polynomial.

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If $t^{1/3}$ is a polynomial, then its derivative $\frac{1}{3}t^{-2/3}$ also is. But it's clearly impossible (for example it would not be continuous at $0$).

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Assuming $\sqrt[3]{t} = a_n t^n +a_{n-1}t^{n-1}+\cdots +a_0$ and using $\sqrt[3]{8t}=2\sqrt[3]{t}$, we have $$8^na_n t^{n}+8^{n-1}a_{n-1}t^{n-1}+\cdots + a_0 = 2a_n t^n+2a_{n-1}t^{n-1}+\cdots+2a_0.$$ Use uniqueness of coefficients for polynomial functions on $\mathbb R$ to reach a contradiction.

Alternatively, using $\sqrt[3]{t^3}=t$, we have $$t=a_n t^{3n} +a_{n-1}t^{3n-3}+\cdots +a_0,$$ leading to a similar contradiction.

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If $t^{1/3}$ is a polynomial, it is non-zero and $$t^{1/3} = a_n t^n + \dots + a_0$$ for $n \ge 0$ with $a_n \neq 0$, but we also have $$t = (a_n t^n + \dots + a_0)^3 = a_n^3 t^{3n} + \dots + a_0.$$ Therefore, $a_n^3 = 0$, so $a_n = 0$, which is contradicting the assumption that $a_n \neq 0$.

This proof is different from the others in the sense that it is from an algebraic point of view. It does not only say that $t^{1/3}$ is not a polynomial function. It also says that there is no polynomial whose cube is $t$, which makes sense as a statement even if $t$ is not a real number or a function. In algebra, $t$ would be called "an indeterminate". Indeterminates are useful mostly when you want to plug in different kinds of things rather than just real numbers. Their main use is in algebraic theories though.

Hope that helps,

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Okay seriously. That downvote was completely irrelevant. –  Patrick Da Silva Nov 30 '12 at 21:41

To begin with, the way the question is posed in a somewhat unclear fashion. From an elementary point of view, one may say that $X^{1/3}$ is not a polynomial because polynomials are different sort of expressions.

Maybe the question is whether the function $X\mapsto f(X)=X^{1/3}$ is polynomial in the sense that there is a polynomial $P(X)$ such that $f(x)=P(x)$. There's a number of ways to prove this false, like the fact--borrowing from Joel Cohen's answer--that polynomial functions admit derivatives continuous everywhere, which is not the case for the function $f(X)$.

In my opinion the correct way to rephrase the question is whether there may be a polynomial $P(X)$ such that $$P(X)^3=X.$$ This is impossible by considering degrees: the degree of the left hand polynomial is a multiple of $3$ whereas the degree of $X$ is $1$, not a multiple of $3$.

Yet, we learn from algebra courses that one may consider polynomials with coefficients in more general algebraic structures (commutative rings). If the coefficient ring is a domain, in particular a field, the very same degree argument applies. But in more exotic situations the degree argument is not sufficient, like the example $$(1+2X)^2=1\qquad\text{in (\Bbb Z/4\Bbb Z)[X]}$$ shows. Thus, a different sort of argument is required. But I stop here since I believe that this is actually beyond the scopes of the question.

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Related: The degree argument wouldn't work in general even when the coefficient ring is a field if the question is about whether a polynomial $P$ exists such that $f(x)=P(x)$ for all $x$, as you suggested. But it does work for an infinite field like $\mathbb R$, which is probably what is intended in the question. –  Jonas Meyer Nov 30 '12 at 14:43
@JonasMeyer : the degree argument doesn't work for fields if you are talking about polynomial functions as the notorious $X^p-X$ example in characteristic $p$ shows, but it certainly works for polynomials since $X^p-X$ is NOT the zero polynomial –  Andrea Mori Nov 30 '12 at 18:28
Agreed.${}{}{}{}{}$ –  Jonas Meyer Nov 30 '12 at 19:23