How do we choose/find the rational sequence that is asymptotic to an irrational one?

Question

I read that to any irrational sequence we can assign at least one rational sequence which is asymptotic to it.
The fact that it is asymptotic I think implies that the rational sequence approximates the irrational one as much as we like.
But I am not clear how we find one such rational sequence?
E.g. if we have (just as an example) the expression $\sqrt{1 + \sqrt 2}$ a direct sequence to interpret it is:
$\sqrt{2.4}, \sqrt{2.41}, \sqrt{2.414}, ...$ and all the terms are irrationals.

But in general how do we choose/find the rational sequence to represent it instead which is asymptotic to the irrational one?
Would we pick just a rational approximation of each irrational term choosing up to an aribtrary number of digits after the decimal? E.g.
$1.549193338, 1.552417469 ...$?
These are the 9 first digits of $\sqrt{2.4}, \sqrt{2.41}$

Answer 1

If $(a_n)_{n\in\Bbb N}$ is a sequence of real numbers, then for each $n\in\Bbb N$, there is some rational number $q_n\in\left(a_n,a_n+\frac1n\right)$. So, $(q_n)_{n\in\Bbb N}$ is a sequence of rational numbers and $\lim_{n\to\infty}(a_n-q_n)=0$.

Answer 2

continued fractions give the best rational approximations to numbers.

Considering for example $0<x<1$, and let $T(x)=\frac{1}{x}-\lfloor \frac1x\rfloor$, where $\lfloor \cdot\rfloor$ is the floor function (a.k.a. maximum integer function).

Then $$x=\frac{1}{1/x}=\frac{1}{\lfloor \frac1x\rfloor + Tx}=\frac{1}{a_1 +Tx}$$

where $a_1(x)=\lfloor \frac1x\rfloor$. Continuing this way, and as long as $T^{n-1}x=T(T^{n-2}(x))\neq0$, let $a_n(x)=\lfloor \tfrac{1}{T^{n-1}x}\rfloor$. We obtain the sequences of rational numbers $$ x_n:=\frac{1}{a_1+\tfrac{1}{a_2 +\ddots\tfrac{1}{a_{n-1}+\tfrac{1}{a_n}}}}$$

It turns out that the rational approximations $x_n$ in some sense that can be made very precise:

0. If $x$ is rational, the sequence $x_n$ is final after a finite number of steps.

For irrational $x$, the sequence $x_n$ is infinite and

1. $x_n=\frac{p_n}{q_n}$ where \begin{align} p_n&=a_np_{n-1}+p_{n-2}\\ q_n&=a_nq_{n-1}+q_{n-2} \end{align} with $p_0=0$, $p_1=1$, $q_0=1$ and $q_1=a_1$.
2. It is relatively easy to check that $m.c.d(p_n,q_n)=1$ for all $n\geq0$.
3. With a little more effort we have the bounded $$\frac{1}{q_nq_{n+2}}<\Big|x-\frac{p_n}{q_n}\Big|<\frac{1}{q_nq_{n+1}}$$

That the rational $x_n$'s are the best rational approximations to $x$ is expressed in the following result

Theorem: If there is a rational number $a/b$ with $b>0$ such that $$\big|x-\frac{a}{b}\big|<\big|x-\frac{p_n}{q_n}\big|$$ for some $n>0$, then $b>q_n$.

A classic example is the golden mean $\phi:=\frac{\sqrt{5}-1}{2}$ which can me expressed as $$\phi:=\frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}$$

It is customary to use the notation $[a_1a_1\ldots]$ for the number $x$; other notations are $\frac{1}{a_1+}\frac{1}{a_2+}\cdots$.

A fun fact (easy to prove) is that if the continued fraction of a number is periodic, that is there is a pattern $a_{n_0}\cdots a_{n_0+k}$ that repeats (similar to periodic decimal expansions) then the irrational number it represents solves a quadratic question with integer coefficients, and vice versa.

Example: the golden mean $\phi$ satisfies $x=\frac{1}{1+x}$, or equivalently $x^2+x-1=0$

Example: $\sqrt{2}-1=\frac{1}{2+\frac{1}{2+\frac{1}{2+\ldots}}}$ satisfies $x=\frac{1}{2+x}$ , or equivalently $x^2+2x-1=0$

The number in your OP, $\beta=\sqrt{1+\sqrt{2}}$ does not solve a quadratic equation by a quartic: $x^4-2x^2-1=0$. Its continued fraction expansion is not periodic. This requires more finesse. There are very good algorithms nowadays that can produce large amounts of integers in the continued fraction of several algebraic numbers. Here is the first 10 elements of the continued fraction of $\beta-1=[1,1,4,6,1,2,2,2,1,1,6,\ldots]$.

---

Using also fine numerical methods to solve equations, can give you approximations for algebraic and wet known transcendental numbers in decimal expansion with different levels of accuracy.

---

The answer given in the different posting of course shoes that any irrational number can be approximated by rationals. This is a trivial fact due to the density of rationals in the real line (the kind of stuff cover in a course of Calculus or undergraduate Analysis). However, it is useless in practice for it does not tells you how to approximate numbers by irrationals, and how good the approximation is.

There are irrational numbers that are hard to approximate, the golden mean $\frac{\sqrt{5}-1}{2}$ being the hardest (in the sense of the theorem above). It is a delicate problem in fine numerical analysis, dynamical systems, and many applications in physics to have good approximations to irrational numbers, for systems may be extremely sensitive to errors.

I hope this motivates the OP to do some digging into continued fractions. Classic high school books such as Hall, H.S. and S.R. Knight, Higher Algebra, 1889 (Now edited by USA public domain selection), and Uspenski's Theory of equations dedicate a few chapters to this topics. The classic book on the subject is a book Perron, O., Die Lehre von den Kettenbrüchen, 1913. Another short and nice treatment appears in Vinogradov, I. Elements of Number Theory, English translation, Dover.

Answer 3

Consider the irrational sequence

$$\langle a_n \rangle ~: a_n = ~n \times \pi, ~n \in \Bbb{Z^+}.$$

In a manner very similar to that suggested in the original posting,
for $~n \in \Bbb{Z^+},~$
define the function $f(n)$ to be $\pi$ accurate to $(10)^n$ decimal places,
where the decimal expression of $\pi$ is truncated (rather than rounded)
after the $[(10)^n]$-th decimal place.

Now define

$$\langle b_n\rangle ~: ~b_n = n \times f(n), ~n \in \Bbb{Z^+}.$$

Then $~\displaystyle \langle b_n \rangle$ is a rational sequence that is asymptotic to the irrational sequence $~\displaystyle \langle a_n\rangle.$

---

Edit
Now, vary the above example. Instead of the irrational sequence $~\displaystyle \langle a_n \rangle~$ defined above, let $~\displaystyle \langle a_n \rangle~$ be any sequence of numbers, where each element of the sequence is irrational.

Then, in a manner very similar to the example at the start of my answer, you can construct the corresponding rational sequence $~\displaystyle \langle b_n \rangle~$, where $b_n$ represents the first $(10)^n$ decimal places (with truncation) of the irrational number $a_n$.

Then, as before, you will have that $~\displaystyle \langle b_n \rangle$ is a rational sequence that is asymptotic to the irrational sequence $~\displaystyle \langle a_n\rangle.$

Answer 4

There are many rational sequences converging to any particular irrational number you start with. There is no general rule for finding one.

Since you can always find some rational number in any interval, if you simply pick $a_n$ in the interval $(x, x+1/n)$ you will have a sequence of rationals converging to $x$.

For $\sqrt{2}$ the sequence $$ \frac{1}{1},\frac{3}{2},\frac{7}{5},\frac{17}{12},\frac{41}{29}, \ldots $$ is the best (in a particular rigorous sense). It's the continued fraction expansion.

You can always use continued fractions, as in other answers here. For quadratic irrationals they are particularly nice.

The successively more accurate initial segments of the decimal expansion will also do.

For roots of polynomials there's Newton's method.

Related: Computing irrational numbers

Answer 5

Your idea of using the decimal expansion is good and can be explicitly written as this $$a_n = \frac{1}{10^n}\left\lfloor10^n\ x_n\right\rfloor$$

It readily follows from the the properties of the floor function that $$x_n-\frac{1}{10^n}< a_n \le x_n$$ which proves that $a_n$ is asymptotic to $x_n$.

Notice that the sequence $10^n$ could be changed by any other sequence of integers tending to infinity. May be the simplest would be $$a_n = \frac{1}{n}\left\lfloor n\ x_n\right\rfloor$$

Answer 6

This is an expanded version of jjagmath’s excellent but terse answer, with a lot more explanation.

Consider the question’s example:

\begin{equation} \{\sqrt{2.4}, \sqrt{2.41}, \dots\} = \{1.549\dots, 1.552\dots, \dots\} \end{equation}

Using the question’s idea of choosing an arbitrary number of digits after the decimal point (where we need to use arbitrarily many digits as the sequence goes on to make the new sequence asymptotic to the original one), we can choose one digit for the first term, two digits for the second term, and so on. In general, we choose $n$ digits for the $n^\text{th}$ term $x_n$.

In other words, we drop all the digits after the first $n$ decimal places. Luckily, we have a well-known function that almost does this: the floor function $\lfloor \cdot \rfloor$ drops all the digits after the decimal point. Now, all we have to do is move the decimal point $n$ places to the right ($\times 10^n$), apply the floor function, and move the decimal point $n$ places to the left ($\div 10^n$). Putting this together, we have:

\begin{equation} a_n = \lfloor 10^n x_n \rfloor \div 10^n = \frac{1}{10^n} \lfloor 10^n x_n \rfloor. \end{equation}

Clearly, $a_n$ is rational. To prove that $\{a_n\}$ is asymptotic to $\{x_n\}$, we use a basic property of the floor function:

\begin{equation} x – 1 < \lfloor x \rfloor \leq x. \end{equation}

Applying this to our sequence, we have:

\begin{equation} 10^n x_n – 1 < \lfloor 10^n x_n \rfloor \leq 10^n x_n. \end{equation}

Dividing everything by $10^n$, we get:

\begin{align} x_n - \frac{1}{10^n} &< \frac{1}{10^n} \lfloor 10^n x_n \rfloor \leq x_n, \\ x_n - \frac{1}{10^n} &< a_n \leq x_n. \end{align}

Everything above still works if $\{10^n\}$ is replaced by any other rational sequence that tends to infinity, such as $\{n\}$:

\begin{equation} a_n = \frac{1}{n} \lfloor n x_n \rfloor. \end{equation}