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Can $$ x^{2/3} + y^{2/3} $$ be expressed as a fraction of two polynomials in x and y?

How can we see this easily?

(It is the curve swept by a stick sliding down a wall)

KReiser
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Yan King Yin
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    If you evaluate a rational function using rational number inputs, then the result will be a rational number (easy to see this). In the case of your function, if you plug in $x = y = 2,$ then you will NOT get a rational number. Therefore $\ldots$ – Dave L. Renfro Mar 17 '24 at 21:14
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    @DaveL.Renfro Doesn't it depend on what the coefficients of the polynomial are allowed to be? If the coefficients might be any real numbers... – Mark Mar 17 '24 at 21:34
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    @Mark: Yes, the numerical coefficients definitely matter! I forgot about that. Incidentally, this is an algebraic function with rational-poly (hence, integer) coefficients. This can be shown analogous to the rationalizing calculations in Answer 1 and Answer 2 and Answer 3. Letting $u=x^{2/3}+y^{2/3},$ we have $u-x^{2/3}-y^{2/3}=0.$ Using $(a-b-c)(a^2+b^2+c^2+ab-bc+ac)=a^3-b^3-c^3-3abc$ with $a=u$ and $b=x^{2/3}$ and $c=y^{2/3},$ (continued) – Dave L. Renfro Mar 18 '24 at 06:36
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    we can multiply both sides of $u-x^{2/3}-y^{2/3}=0$ by (the $u,$ $x,$ $y$ equivalent of) $a^2+b^2+c^2+ab-bc+ac$ to get $u^3-(x^{2/3})^3-(y^{2/3})^3-3ux^{2/3}y^{2/3}=0,$ or $u^3-x^2-y^2=3ux^{2/3}y^{2/3}.$ Now cube both sides to get $(u^3-x^2-y^2)^3=27u^3x^2y^2.$ It's now clear that this last equation, when expanded and terms transposed so that one side is equal to $0,$ gives a $9$'th degree polynomial in the variable $u$ whose coefficients are integer-coefficient polynomials in $x$ and $y.$ @Wang YeFei (in case interested) – Dave L. Renfro Mar 18 '24 at 06:36
  • @DaveL.Renfro What you did seems to be an implicitization of the function $u$ in polynomial form, but that does not imply $u$ is a polynomial function. Interesting. – Yan King Yin Mar 18 '24 at 08:32
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    does not imply $u$ is a polynomial function --- I was showing that $u$ is an algebraic function (note that I began with "this is an algebraic function"), a question that was raised in the comments to Wang YeFei's answer. In fact, it's an explicit algebraic function (see comments to this MSE question) -- every explicit algebraic function is an algebraic function (this requires some work) but not every algebraic function is an explicit algebraic function. – Dave L. Renfro Mar 18 '24 at 08:57

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Your $f(x,y) = x^{\frac{2}{3}}+y^{\frac{2}{3}}$ is not a rational function. For if it were, the it can be written as a ratio of two polynomial in $x,y$. That is $f = \dfrac{P}{Q} \implies P = fQ$. But $fQ$ is then no longer a polynomial since it has terms with non-integer exponents. This contradicts $P$ being a polynomial. Another way to see this is by plugging some numbers, say $x = 3 = y$ into the expression and the output is an irrational number which is a contradiction since it is supposed to be a rational number.

Wang YeFei
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  • Thanks, what are these functions called? Are they transcendental? – Yan King Yin Mar 17 '24 at 21:24
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    I think it is still called algebraic function. Transcendental ones are like: $\sin(\pi\cdot x) + x^2 - \ln(x+7)$ for example. – Wang YeFei Mar 17 '24 at 21:26
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    @WangYeFei so would an algebraic function be one which maps the algebraics to themselves? – Malady Mar 17 '24 at 21:47
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    Of course it would but that preserving property wouldn't be helpful here as it doesn't keep the rationality of a real. – Wang YeFei Mar 17 '24 at 22:25
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    then no longer a polynomial since it has terms with non-integer exponents --- To not be a polynomial requires more than this observation, since maybe the expression could be rewritten so that all terms have integer exponents. Remember that polynomial, rational function (or rational number), etc. each mean that something CAN be expressed in a certain way, not that it IS expressed in a certain way. So for non-rational, you have to show that it CAN'T be written in a certain way, not that it ISN'T written in a certain way. – Dave L. Renfro Mar 18 '24 at 06:45