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I'm trying to parse the following excerpt from Pugh's Real Mathematical Analysis (in particular, the parts that I have bolded):

"One example is the field $\mathbb{R}(x)$ of rational functions with real coefficients. Each such function is of the form $$R(x) = \frac{p(x)}{q(x)}$$

where $p$ and $q$ are polynomials with real coefficients and $q$ is not the zero polynomial. (It does not matter that $q(x) = 0$ at a finite number of points.) Addition and multiplication are defined in the usual fashion of high school algebra, and it is easy to see that $\mathbb{R}(x)$ is a field. The order relation on $\mathbb{R}(x)$ is also easy to define. If $R(x) > 0$ for all sufficiently large $x$ then we say that $R$ is positive in $\mathbb{R}(x)$, and if $R − S$ is positive then we write $S<R$. Since a nonzero rational function vanishes (has value zero) at only finitely many $x ∈ \mathbb{R}$, we get trichotomy: either $R = S, R<S,$ or $S < R$. (To be rigorous, we need to prove that the values of a rational function do not change sign for $x$ large enough.) The other order properties are equally easy to check, and $\mathbb{R}(x)$ is an ordered field."

Firstly, I'm trying to understand what is meant by "since a nonzero rational function vanishes (has value zero) at only finitely many $x \in R...$" Is this just, essentially, a restatement of the Fundamental Theorem of Algebra (i.e., that any non-zero polynomial has a number of roots at most equal to its degree), but extended to rational functions somehow? Furthermore, I don't see how this implies trichotomy (which suggests I'm, indeed, misunderstanding the point he's trying to get across).

Secondly, I don't see why, in order to rigorously prove that the defined relation is an order on $\mathbb{R}(x)$ "we need to prove that the values of a rational function do not change sign for $x$ large enough." I understand (from basic calculus) that, for a rational function $f/g$, as $x$ approaches infinity, we need only to look at the the ratio of the leading coefficients (if $f$ and $g$ are of equal order), etc., but proving that (rigorously) is a different story, and more importantly why I need to prove it completely escapes me.

Any help unpacking this a bit would be greatly appreciated. Thanks.

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If a non-zero rational function $R$ vanishes, it implies that $p$ vanishes. By Fundamental Theorem of Algebra, $p$ only vanishes finite number of times. Hence $R$ vanishes only finite number of times.

We define $R$ to be positive if $R(x)>0$ for $x$ that is sufficiently large. We define $R(x)$ to be negative if $R(x) < 0$ for $x$ that is sufficiently large.

Are we able to compare two rational functions using the definition above? That is given $R$ and $S$, can we say whether $R=S$, $R>S$, or $R< S$? To answer that question, we have to check whether $R-S = 0 $, $R-S > 0 $ , or $R- S <0$. Suppose you have proven that $R-S$ remains a rational function, the question is now for each rational function $R$, can we always conclude that $R=0$, $R>0$ or $R<0$ using the definition of $R=0$, $R>0$ or $R<0$. We have to prove that rational function doesn't behave like sine or cosine where it switches signs forever. (well, one might know that a mathematical trick works, but not everyone knows why does it work, a proof helps in clearing this doubt).

Note that a polynomial can only changes sign at its root. Suppose $p \equiv 0$, then $R= 0$. Suppose not, then $p$ has finitely many roots $p_1< \ldots< p_k$. Also $q$ has finitely many roots $q_1< \ldots < q_m$. Hence polynomial doesn't changes sign after $\max(p_k, q_m)$. Hence either $R > 0$ or $R< 0$.

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You can define the order on $\mathbb{R}(x)$ by saying $\frac{f}{g}>0$ if the quotient of the dominant terms of $f$ and $g$ is $>0$. It's not that hard to show that this is an order relation. It turns out that $\frac{f}{g}>0$ in this order if and only if the values of $\frac{f}{g}$ are $>0$ for $x$ large enough. That is something not hard to show. It's based on this fact: the sign of $a_n x^n + a_{n-1} x_{n-1} + \cdot + a_0 = a_n x^n( 1 + \frac{a_{n-1}}{a_n x} + \cdot + \frac{a_0}{a_n x^n}) $ is the sign of $a_n$ if $x$ is large enough. The expression inside the brackets is positive. Why say it's not zero when you can say it's $>0$.

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  • $\begingroup$ You say that it's not hard to show that the above is an order relation, and yet here I am finding it quite difficult. For example, comparing two elements of the set, e.g., $\frac{f_1}{g_1} < \frac{f_2}{g_2}$, doesn't seem very simple. Sure, showing that any given element is positive or negative is not an issue, but say their leading coefficients are 1, 2, 3, and 4 (for $f_1, g_1, f_2, g_2$, respectively). Then, sure, both elements are positive, but how can one element be "greater than" the other? $\endgroup$ – thisisourconcerndude Sep 10 '17 at 19:50
  • $\begingroup$ @thisisourconcerndude: oh, you cannot just compare the ratios directly, you have to take the difference and see whether for this the ratio is $>0$ or not. It may go either way. $\endgroup$ – Orest Bucicovschi Sep 10 '17 at 19:54
  • $\begingroup$ Right, so I would want to assume that, for example, $\frac{f_1}{g_1} < \frac{f_2}{g_2}$, and then look at the leading coefficients of $\frac{f_2}{g_2} - \frac{f_1}{g_1} = \frac{f_2 g_1 - f_1 g_2}{g_2 g_1}$ and show that $>$ and $=$ can't also be the case for those two elements? $\endgroup$ – thisisourconcerndude Sep 10 '17 at 20:14
  • $\begingroup$ @thisisourconcerndude: I see. Well, first look at the difference, no need to assume something beforehand. One fact is true: if one fraction is larger than the other, then for large enough $x$ the value of first at $x$ > the value of the other at $x$. It is not that hard, but maybe Pugh makes it a bit too intricate. For fractions it's not that hard. $\endgroup$ – Orest Bucicovschi Sep 10 '17 at 20:20
  • $\begingroup$ Ok, so, when you say "fractions" do you mean, for example, elements of $\mathbb{Q}$? Because I know what it means when the symbol "<" is between two elements of $\mathbb{Q}.$ If, however, by "fractions" you mean elements of $\mathcal{R}$, I do not know what it means when the symbol "<" is between two elements of $\mathcal{R}$. I know when it means when "<" is between an element of $\mathcal{R}$ and the number $0$, but not two elements of $\mathcal{R}$. Similarly, symbols such as "+," "-," and "=" between two elements of $\mathcal{R}$ are a mystery to me. $\endgroup$ – thisisourconcerndude Sep 10 '17 at 20:43

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