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Suppose we have rational functions $f$, $g$, and $h$ defined for all natural numbers $n$ such that $f \leq g \leq h$ for all $n \in \mathbb{N}$.

How can we prove that there is no rational functions $f_1$ and $g_1$ such that $f \leq f_1 \leq g \leq g_1\leq h$ ?

Knowing that the field $\mathbb{Q}$ is dense, that is for any $q_1$ and $q_2$ in $\mathbb{Q}$, there always exists $q \in \mathbb{Q}$ such that $q_1< q< q_2$. Is this already sufficient to show that my previous question's implication is false; that is there are always rational functions $f_1$ and $g_1$.

Now suppose a function $g$ not defined by any elementary operation but the range $R$ of $g$ is a subset of $\mathbb{Q}$, did my previous question now possible?

Thanks

Keneth Adrian
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  • Sorry, not for all natural $n$ rather, but for all defined $n$. thanks – Keneth Adrian Aug 29 '13 at 03:51
  • @Kenneth could you say more? What kind of defined n? And are you referring to the rational functions or the more general functions? – Betty Mock Aug 29 '13 at 04:24
  • @BettyMock, since rational functions have variables in the denominator, that is in this case its $n$.so when the polynomial in the denominator becomes zero for that particular $n$ that means its not part of the domain of $f$, thats why its for all defined n of the functions – Keneth Adrian Aug 29 '13 at 04:52

2 Answers2

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Perhaps I am misinterpreting what you mean by rational function (usually this refers to a function that is the ratio of two polynomials).

Consider $f(n) = n-2, g(n) = n,$ and $h(n) = n+2$. These are defined for all natural numbers $n \in \mathbb{N}$ and the desired inequality holds.

Now use $f_1(n) = n-1$ and $g_1(n) = n+1$ to obtain the inequalities you were trying to avoid. (!)

Moreover, from the phrasing of your question: why not just take $f_1 = f$ and $g_1 = g$?

Benjamin Dickman
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  • $\mathbb{Z}$ is not dense that is there is no integer between 1 and 2. so there exists a way to order it, unlike in $\mathbb{Q}$.would the case be different to rational functions? – Keneth Adrian Aug 29 '13 at 04:55
  • @KenethAdrian Can you write down your definition of "rational function"? – Benjamin Dickman Aug 29 '13 at 05:20
  • @ Benjamin Dicknam: Rational functions are functions where two polynomials say P and Q where $Q \neq 0$ such that $\frac{P}{Q}$ – Keneth Adrian Aug 29 '13 at 05:28
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Let $f$, $g$, and $h$ be rational functions of $n \in \mathbb{N}$, such that

$$f(n) \leq g(n) \leq h(n).$$

In general, there is always a family of rational functions $f_1, g_1$ that satisfies $$f(n) \leq f_1(n) \leq g(n) \leq g_1(n) \leq h(n);$$ we can simply take $$f_1(n) = \frac{f(n) + g(n)}{2},$$ and $$g_1(n) = \frac{g(n) + h(n)}{2}.$$

Jose Arnaldo Bebita Dris
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  • @ Dris: Suppose we have a collection of rationals, say $S$ such that $S \subset \mathbb{Q}$? where $S$ cannot be expressed as ration of two polynomials...and this set represents the function $g$. how can we show then that $f_1$ and $g_1$ still exists..? – Keneth Adrian Aug 29 '13 at 05:46
  • @KenethAdrian, if $f, g, h \in \mathbb{Q}$, then

    $$f_1(n) = \frac{f(n) + g(n)}{2} \in \mathbb{Q},$$

    and

    $$g_1(n) = \frac{g(n) + h(n)}{2} \in \mathbb{Q}.$$

     – Jose Arnaldo Bebita Dris Aug 29 '13 at 05:50
  • Suppose there exists rational functions $f$ and $h$ such that it bounds $S$ the range of $g$. if rationals $f_1$ and $g_1$ exists such that its the only rational functions that bound $g$ and we say they are the "tight" bounds of $g$. How can we show such situation? since we are concluding that there is no other way? – Keneth Adrian Aug 29 '13 at 05:53
  • Hold on. What is true is that $f(n) \leq g(n) \leq h(n)$, right? This is true for each $n$. I guess you have to be more careful when you say "Suppose there exist functions $f$ and $h$ such that [they] bound $S = g(n) \subset \mathbb{Q}$." – Jose Arnaldo Bebita Dris Aug 29 '13 at 05:58
  • Additionally, I am not quite sure if "tight bounds" correspond to "best-possible bounds", but I will assume that they do. In this case, if you want to show that $f_1$ and $g_1$ are tight bounds for $g(n)$, you can do this by proving that, if you assume the negation of your original hypotheses, then you either get $g(n) < f_1$ or $g(n) > g_1$. – Jose Arnaldo Bebita Dris Aug 29 '13 at 06:03
  • @ Dris: Yes..:) – Keneth Adrian Aug 29 '13 at 06:04
  • So contradiction is the approach?. Thank you @Dris – Keneth Adrian Aug 29 '13 at 06:08