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I'm 9th grade student, and my teacher said that $|x|$ is not rational expression ( expression like $\frac{p(x)}{q(x)}$ s.t $p(x)$ and $q(x)\neq 0$ are polynomial) but he didn't have convincing reason. One of my friends said that it is provable by use of differential, but I don't know about calculus. Is there any proof for this fact, without use of calculus?

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    $\begingroup$ You just need to understand the definitions. A rational function is, as you said, a ratio of polynomials. The absolute value function is not a polynomial of any kind, nor is it the ratio of polynomials. $\endgroup$ – Git Gud Apr 14 '16 at 20:12
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    $\begingroup$ @GitGud good, but why? have you any proof? $\endgroup$ – Arsalan Apr 14 '16 at 20:13
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    $\begingroup$ You can split it into cases according to the degrees of $p(x)$ and $q(x)$. If $\deg q(x)\ge \deg p(x)$, then you can show that for large enough values of $x$ (large in comparison to the coefficients of the two polynomials), the ratio will become too small. If $\deg p(x)\ge \deg q(x)+2$, then the ratio becomes too large. In the remaining case $\deg p(x)=\deg q(x)+1$ the ratio will become negative for either positive or negative values of $x$ far away from zero. $\endgroup$ – Jyrki Lahtonen Apr 14 '16 at 20:19
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    $\begingroup$ @GitGud the fact that it is not written as a quotient of polynomials does not a priori out rule the possibility that admits an expression as a rational function. $\endgroup$ – neth Apr 14 '16 at 20:19
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    $\begingroup$ @GitGud It is clear that in this context we would say a function "is" a polynomial if there is a polynomial function with which it agrees at every point of the domain. $\endgroup$ – neth Apr 14 '16 at 20:27
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If there were polynomials $p$ and $q$ such that $|x|=\frac{p(x)}{q(x)}$ for all $x\in\Bbb R$, then $\frac{p(x)}{q(x)}$ would be defined for all $x\in\Bbb R$, and $q(x)$ could never be $0$. Moreover, we’d have $p(x)=|x|q(x)$ for all $x\in\Bbb R$. If $x\ge 0$, this means that $p(x)=xq(x)$, and if $x<0$ it means that $p(x)=-xq(x)$.

Let $r(x)=p(x)-xq(x)$; this is certainly a polynomial, and

$$r(x)=\begin{cases} 0,&\text{if }x\ge 0\\ -2xq(x),&\text{if }x<0\;. \end{cases}$$

I’m going to assume that you know that a polynomial of degree $n\ge 1$ has at most $n$ real roots, though you’ve probably never seen a proof. Our supposed polynomial $r(x)$ evidently has infinitely many real roots, since each non-negative real is a root, so it must be the constant function $r(x)\equiv 0$. But then $-2xq(x)=0$ for each $x<0$, and it follows that $q(x)=0$ for each $x<0$. We saw at the beginning that this is impossible: $q(x)$ can never be $0$. This contradiction shows that in fact no such polynomials $p$ and $q$ exist, and $|x|$ is not a rational function.

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    $\begingroup$ Thanks for your complete answer, I enjoyed it. $\endgroup$ – Arsalan Apr 14 '16 at 20:30
  • $\begingroup$ @Arsalan: You’re welcome. $\endgroup$ – Brian M. Scott Apr 14 '16 at 20:30
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If the value of a polynomial function is specified at infinitely many points, then the polynomial is determined completely. So, if we just look at $x \geq 0$ then we must have $p(x) = x q(x)$. Similarly, we must have $p(x) = -x q(x)$. But $x q(x) \neq -x q(x)$ unless $q(x) = 0$.

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  • $\begingroup$ Where do you get p)x)=-xq(x) for x>0? $\endgroup$ – DanielWainfleet Apr 14 '16 at 20:25
  • $\begingroup$ @user254665 I get that for $x \leq 0$. $\endgroup$ – Seven Apr 14 '16 at 20:27
  • $\begingroup$ thanks for your simple explanation. $\endgroup$ – Arsalan Apr 14 '16 at 20:35
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For instance, because a rational function has a derivative at each point of its domain, and $\lvert x\rvert$ doesn't have one at $0$.

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    $\begingroup$ This is a nice simple answer, but the OP asked for a proof without the use of calculus. $\endgroup$ – nukeguy Apr 14 '16 at 20:18
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    $\begingroup$ I don't know differential calculus. $\endgroup$ – Arsalan Apr 14 '16 at 20:19
  • $\begingroup$ You don't even know the derivative? $\endgroup$ – Bernard Apr 14 '16 at 20:20
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    $\begingroup$ I am grade 9, we have no any lesson in calculus. $\endgroup$ – Arsalan Apr 14 '16 at 20:22
  • $\begingroup$ I don't what grade 9 is, except it corresponds to some level in high school. Do you know about limits? $\endgroup$ – Bernard Apr 14 '16 at 20:23

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