How can we measure how "irrational" a number is?

Question

I wonder, can we, for instance, say that $\pi$ is more irrational than $e$? Or that $e$ is more irrational than $\sqrt{2}$? What kind of irrationality measurement can we use to say that $A$ is (much) more irrational than $B$? (in particularly rigorous sense)

Answer 1

Well, there is a sort of extension of the idea of irrationality called "transcendental". You can think of "irrational" as meaning "There is no way to relate this number to $1$ by thinking only about addition and subtraction". That is, a number $q$ is rational if we can write something like: $$q+q+q-1-1-1-1=0$$ or some other similar equation involving only $q,\,+,$ and $1$. The above represents the number $\frac{4}3$ and more generally, you can, for $q=\frac{a}b$ write: $$\underbrace{q+\ldots+q}_{b \text{ repetitions}}-\underbrace{1-\ldots-1}_{a \text{ repetitions}}=0.$$ or appropriately change signs for negative $a$.

The next logical step would be to consider the idea of allowing multiplication too - and this extends the rational numbers to algebraic numbers, which are (potentially) irrational, but can be related to the rationals via multiplication and addition. For instance, $x=\sqrt{2}$ satisfies $$x^2-2=0$$ and we can come up with similar expressions (the set of such equations would be polynomials to be precise) for any expression built with radicals and addition and such - and we could say that the degree of an algebraic number (i.e. how many multiplications we need) represents how irrational it is - that is $\sqrt{2}$ can be related to integers by squaring it, but $\sqrt[3]{2}$ requires cubing to get to an irrational - and numbers like $\sqrt{2}+\sqrt{3}$ require fourth powers. We might say that the degree represents a sort of "distance" from the rational numbers.

However, this only extends to a narrow class of numbers - $\pi$ and $e$ are both transcendental, meaning "not algebraic". We have much less understanding of these, since we can't relate them to the rationals through arithmetic - so we could be justified in saying that they are less well behaved then algebraic numbers. One way we can talk about their irrationality measure which essentially tells us how far from rational numbers a given irrational is, relating the growth of the denominator of the best rational approximations to how close they are - this represents how these numbers are comprehended by looking at sequences of rational approximations, rather than algebraic properties. However, this isn't terribly useful to compare actual numbers, because we hardly know how to calculate any. We do know that, from its infinite series, that $e$ acts very similarly to an algebraic number (looking at its series definition) and its irrationality measure is $2$. We don't know the irrationality measure of $\pi$, though we might suspect that it greater than $2$ - but it's hard to compare numbers this way, given our limited knowledge thereof, and all it means to have a small irrationality measure is that it's "far" from any rationals with small denominators.

Answer 2

A first thing to note is that there is a dichotomy among irrational numbers, some are roots of non-zero polynomials with rational coefficients, they are called algebraic numbers.

The others are called transcendental.

One has that $\sqrt{2}$ is algebraic, while $e$ and $\pi$ are not. In that sense $e$ and $\pi$ are perhaps more irrational.

There is a lesser known notion of periods, roughly things that can be expressed as integrals using only rational parameters. It is easy to see that $\pi$ is a period while it is unknown and doubtful that $e$ is a period.

In that sense $e$ is perhaps more irrational than $\pi$.

There is also such a thing as a measure of irrationality. It is known that the irrationality measure of every rational is $1$, of every non-rational algebraic number it is $2$, and it is at least two for transcendental numbers.

It is known that this measure is $2$ for $e$ while this is not known for $\pi$, though it might well be the case it is also $2$.

Long story short, yes, there are ways to further classify how irrational some number is, the details can however be complex and there are different ways to approach the subject.

Answer 3

There is another take on this that doesn't use the concepts of trancendentality or algebraity.

For any given irrational $a$, it's not hard to see that there exists a unique sequence of rationals $\{\frac{a_1}{1}, \frac{a_2}{2}, \frac{a_3}{3}, \ldots, \frac{a_n}{n}, \ldots\}$ such that each term is the closest rational to $a$ with that denominator.

Of course, we can then take a subsequence of terms $\sigma(k)$ such that $|a - \frac{a_{\sigma(k)}}{\sigma(k)}|$ is monotonically decreasing (arbitrarily, take the one that is biggest, in the sense of set inclusion).

We could say that irrational number $b$ is`more irrational' than $a$ if its corresponding monotonic sequence $\{|b - \frac{b_{\tau(k)}}{\tau(k)}|\}_{k \in \mathbb{N}}$ converges slower than $\{|a - \frac{a_{\sigma(k)}}{\sigma(k)}|\}_{k \in \mathbb{N}}$.

This is very related continued fractions - I haven't looked at the link APGreaves posted, but I presume it's similar.

Answer 4

As noted by @quid and @meelo the irrationality measure is a tool that we can use, but it's very difficult to calculate and does not capture the idea of "how different is a real number by a rational number in an algebraic sense". I don't know if such an idea has a well defined mathematical meaning, but I think that it's interesting and I'm tempting to capture such idea by means of the "amplitude" of the field extension of the rational field that we need to "catch" the given real number.

From $\mathbb{Q}$, adding the operation $\sqrt{\cdot}$ to the classical arithmetic operations, we can buid the quadratic closure of $\mathbb{Q}$ that coincide with the filed of Constructible numbers ($\mathbb{G}$ for Geometric), then we can buid the field of algebraic numbers $\mathbb{A}$ and we have $\mathbb{Q}\subset \mathbb{G} \subset \mathbb{A}$. So that we can say that a number in $\mathbb{A}/\mathbb{G}$ is "more irrational" than a number in $\mathbb{G}$.

Now we can build an "exponential" field adding to the operations in $\mathbb{A}$ the exp operation $a^b \quad a,b \in \mathbb{A}$ and taking the field closure, so finding a new field $\mathbb{E}$ such that $\mathbb{A} \subset \mathbb{E} $ (the construction of $\mathbb{E}$ can be made as that). In this field there are many transcendental numbers but, as far as I know, we don't know if $e,\pi \in \mathbb{E}$.

We can further enlarge $\mathbb{E}$ adding the series of the form $$ \sum_{n=1}^\infty f_{(a,b,c\cdots)}(n) \qquad a,b,c, \in \mathbb{E} $$ where $a,b,c,\cdots$ is a finite set of constants and $f$ is an expression constructed with the operations just defined in $\mathbb{E}$ that is the same for all terms of the series, as an exemple : $$ f(n)=\dfrac{a^{n^2-1}}{n!-b^{n+1}} $$ If we call $\mathbb{S}$ (for Series) the field closure of such set, we have $ \mathbb{Q} \subset \mathbb{G} \subset \mathbb{A} \subset \mathbb{E} \subset \mathbb{S}$.

Here we begin to have some trouble, because I don't know if $\mathbb{S}$ is bigger or smaller then the ring of periods ($\mathbb{P}$) nor what is his relation with computable numbers ($\mathbb{T}$ for Turing). But if we hypothesize that
$$ \mathbb{Q} \subset \mathbb{G} \subset \mathbb{A} \subset \mathbb{E} \subset \mathbb{S} \subset \mathbb{P} \subset \mathbb{T} \subset \mathbb{U}\subset \mathbb{R} $$ where $ \mathbb{U}$ is the set of uncomputable numbers, we have a chain of fields that can give a sort of measure of how away is a number from a rational. Note that in this chain only $\mathbb{U}$ ( and obviously $\mathbb{R}$) are uncountable sets.

I see that all this is more a suggestion (with many open questions) than an answer but I hope it can be useful. Anyway, others classical approaches to this hard question are here.

Answer 5

One way to view things would be to make the distinction between:

• algebraic numbers, which are irrational but not too far from rational numbers, in the sense that they are roots of polynomial equations with rational (or even integer) coefficients. For instance $\sqrt 2$ satisfies the equation $x^2=2$. The set of all algebraic numbers is countable, just as the rational numbers.
• transcendental numbers which are not roots of any polynomial with rational coefficients. Among these numbers $\pi$ and $\mathrm e$ are transcendental. It was proved only towards the end of the 19th century, although it was known since the mid-18th century that $\pi$ is irrational (Johann Lambert, 1761). It is not known if the Euler-Mascheroni constant is rational od irrational. Transcendental numbers are much numerous than algebraic numbers – actually as many as real numbers.

The fact that $\pi$ is not an algebraic number provides a negative answer to the problem of squaring the circle, a problem that dated back to the Ancient Greek.

Answer 6

No rational number is irrational, but in the same way we can measure how "black" a shade of grey is, given that every irrational number has an infinite continued fraction, a fair measure of how irrational a rational number is, is by the length of its continued fraction.

As part of my studies of the Collatz conjecture I discovered (proof thanks to Torsten Schoeneberg) that every rational number has precisely one finitely-long continued fraction of the form $1-\dfrac1{2^{m_0}-\dfrac1{2^{m_1}-\dfrac1{2^{m_2-\frac1{\ldots}}}}}$

provided the terms are measured by letting $\lvert x\rvert_2$ be the 2-adic value of $x$ and iterate $1-\dfrac1{x\lvert x\rvert_2}$ until you get to zero. The number of steps taken is the length of the continued fraction and the measure.

What is more, if you wish more resolution, you can assign every rational number (up to bijection) an ordinal less than $\omega^\omega$. To do this, simply identify $m_0, m_1,\ldots$ by counting the number of divisions by $2$ required at each step. Then use any bijection $f$ from $\Bbb Z\to \Bbb N$ to send $m_i\to n_i$ and:

$1-\dfrac1{2^{m_0}-\dfrac1{2^{m_1}-\dfrac1{2^{m_2-\frac1{\ldots}}}}}\mapsto \displaystyle\sum_{i=0}^\infty\omega^i\cdot n_i=\displaystyle\sum_{i=0}^\infty\omega^i\cdot f(m_i)$

is a bijection from the rationals into the ordinal $\omega^\omega$.