Is there a "positive" definition for irrational numbers?

Question

An irrational number is a real number that cannot be expressed as a ratio of integers.

Is it possible to formulate a "positive" definition for irrational numbers?

A few examples:

• An irrational number is a real number that can be expressed as...
• An irrational number is a real number that can be expressed only as...

I cannot find any such definition which does not include the term irrational number to begin with.

Since not all irrational numbers are computable, I believe that the answer is No.

Any idea how to prove it formally?

Thanks

Answer 1

An irrational number is a real number that can be expressed as an infinite simple continued fraction.

Answer 2

You can give one such characterization using continued fractions. An irrational number is a real number that can be expressed as an infinite continued fraction $$a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + {_\ddots} }}}$$ for a sequence $(a_n)_{n\in\mathbb{N}}$ where $a_0\in\mathbb{Z}$ and $a_n\in\mathbb{Z}_+$ for all $n>0$. (In fact, the sequence $(a_n)$ is then unique, and this correspondence is actually a homeomorphism between the space of irrational numbers and the space of such sequences in the product topology.)

Answer 3

A irrational number is a real number $x$ such that for any integer $q'$ there exists a rational number $p/q$ with $q > q'$ and

$$0 < \left|x - \frac pq \right| < \frac{1}{q^2}.$$

Answer 4

Dirichlet's approximation theorem implies that irrational numbers are those that can be well-approximated by rational numbers, in a certain precise sense. Curiously, rational numbers cannot be approximated well by rational numbers.

Answer 5

An irrational number is a real number in which every infinite contiguous subsequence of its decimal representation is unique.

Answer 6

An irrational number is one whose every representation in any base contains infinitely many digits.

Answer 7

In addition to Dirichlet's approximation theorem as given by @Eric M. Schmidt, you have Hurwitz's theorem, saying that an irrational number $\alpha$ is a real number such that there exist infinitely many relatively prime integers $p$, $q$ such that $$\displaystyle \left|\alpha -{\frac {p}{q}}\right|<{\frac {1}{{\sqrt {5}}\,q^{2}}}.$$

Here, $\sqrt {5}$ is the best possible constant. It is attained for instance by the irrational $\phi = \frac{1+\sqrt{5}}{2}$ (the golden mean or golden ratio). The reason is explained in Why is there a $\sqrt {5}$ for instance in Hurwitz's Theorem? See for instance Introduction to number theory, M. Klazar.

As a side note, what happens to mere rationals: if $\alpha = \frac{r}{s}$, ($r$ and $s$ relatively prime):

$$ \left| \frac{r}{s} - \frac{p}{q}\right| = \left| \frac{rq-ps}{sq} \right| \geq \frac{\left| \frac{q}{s}\right|}{q^2}$$

and

$$\frac{\left| \frac{q}{s}\right|}{q^2}> \frac{k}{q^2}$$ except for a finite number of $q$ such that $|q|\leq |ks|$. Nice graphic versions with circles of Hurwitz's theorem (and some more on fractions) can be found in Ford, Fractions, 1938, American Mathematical Monthly. You can check that out, for instance, at Ford circles:

Answer 8

Some of the answers here are trying to define an irrational number via some of their non-obvious properties which are not shared by a rational number. This strikes me as quite counter-intuitive and perhaps not so helpful for a person trying to study about irrational numbers.

To give an analogy let's compare the following definitions of a prime number:

1) A natural number $p > 1$ is said to be prime if it can't be expressed as product $p = ab$ where $a, b$ are natural numbers greater than $1$. This one according to you is a negative definition.

2) A natural number $p > 1$ is said to be prime whenever $p \mid ab$ implies $p \mid a$ or $p \mid b$ where $a, b$ are natural numbers. This is a slightly complicated but positive definition.

3) A natural number $p > 1$ is said to be prime whenever $(p - 1)! + 1$ is divisible by $p$. This is again a very very complicated but positive definition.

Out of these I prefer the first definition because it is much much simpler than others and can be explained to a kid of 7-8 years who knows how to multiply small numbers. I have written a post on definitions which highlights other important qualities of a good definition but I don't see the positivity as a sufficiently desirable quality to trump simplicity.

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Coming back to the definition of irrational number the simplest possible answer is that defining irrational numbers on the basis of existing knowledge of rational numbers is a reasonably hard problem which was solved only in 19th century by the likes of Cantor, Dedekind and Weierstrass. Some amount of simplicity can be achieved (at the cost of having difficult proofs for properties of irrational numbers) by defining irrational numbers as numbers with non-repeating non-terminating decimal representation.

But the really useful approach here is to define real numbers in terms of rationals (like a rational number is defined in terms of integers) and then define an irrational number as a real number which is not rational. This is the most frequently used definition in practice (say while proving that some particular number is irrational). The name irrational itself has negative connotation because of ir-prefix and hence it is somewhat natural to expect a definition with negative connotation. The real problem with this definition is not its negative connotation, but rather the fact that almost always whenever we encounter this definition, it is accompanied by a total lack of definition of "a real number" and in general this leads to a circularity in the definition game:

• An irrational number is a real number which is not rational.
• A real number is either a rational number or an irrational number.

And the blame for this circularity is entirely on textbook authors. It is better to be intellectually honest and say: Real numbers/irrational numbers are difficult to define in terms of rationals and henceforth we will take their existence for granted and assume some of their properties which are almost similar to rationals. You will study a proper definition when you have attained reasonable mathematical maturity. Unfortunately this attitude is very rare in textbooks authors and instead student is expected to be content with the circularity involved above.

Answer 9

If you want simple definition that's not based on Dirichlet's or Hurwitz's theorem, try this:

A real number $x$ is irrational if and only if for all positive integers $n$ there exists an integer $m$ such that $0\lt nx-m\lt1$.

The underlying theorem here is that for all real numbers $x$ and all positive integers $n$ there is a unique integer $m$ (namely, $m=\lfloor nx\rfloor$) such that $0\le nx-m\lt1$. If $0=nx-m$, then $x=m/n$ is rational, and vice versa.

Answer 10

Whether a definition is positive or negative seems to be rather subjective and not significant to the elegance of a definition (as @ParamanandSingh pointed out). In fact you can formulate any idea positive as well as negative. Starting with your originally negative definition:

An irrational number is a real number that is not rational.

The Idea "real and not rational" can also be formulated positively:

An irrational number is a real number $x$ such that $\vert \{x\} \cup \mathbb Q \vert > \vert \mathbb Q \vert$

Or alternatively:

An irrational number is a real number $x$ such that $\forall q \in \mathbb Q: \vert x-q \vert > 0$

Which is of course the "positivised" formulation of:

An irrational number is a real number $x$ such that $\forall q \in \mathbb Q: x-q \neq 0$ or equivalently $\forall q \in \mathbb Q: x \neq q$

Generally speaking, we can always create a negative formulation from a positive one by negating its negation and a positive formulation from a negative by introducing new symbols. Since all correct definitions are logically equivalent, what is negative and what is positive is only a question of our language of describing things. (Looking forward to discussing this)

Answer 11

I think the fundamental distinction is not positive/negative, but that irrationals cannot be defined without quantifying over the set of rationals (or something equivalent to that, such as the set of all integers). They are of "higher type" than rationals.

More precisely, irrationality cannot (it seems) be defined without alternating at least two quantifiers, $\forall \exists$, while rationality can be defined with an $\exists$ quantifier, where the quantifiers range over integers.

It is similar, as is hinted at in the question, to the asymmetry of defining halting and non-halting Turing machines. One definition quantifies over inputs and the other has alternating quantification over inputs and computation times.

Answer 12

If you want a definition that starts with something like this,

An irrational number is a real number that ...

then first you have to define what a real number is.

It's possible to define the real numbers as Dedekind cuts, where by definition a Dedekind cut is a partition of the set of rational numbers, $\mathbb Q$, into an ordered pair of sets $(L,U)$ such that $L \cup U = \mathbb Q$, $\forall a, b: a \in L \land b \in U \implies a < b$, and $\forall a : a \in L \implies \exists b : b \in L \land a < b$. (The last condition is usually expressed, "$L$ has no largest element," but I am trying to avoid the word "no".)

If you have done all the steps to establish that it makes sense to say that every real number is a Dedekind cut, then an irrational number is a Dedekind cut $x = (L,U)$ such that $$\forall a : a \in U \implies \exists b : b \in U \land b < a. \tag1$$ If $x$ were rational, then $U$ would have a smallest element (namely, the element of $\mathbb Q$ corresponding to $x$) and condition $(1)$ would be false.

Answer 13

I think the notion of "positive" in this question indicates that a constructive definition is requested, as opposed to an existence proof. The unsatisfying part of the traditional definition is that given some particular value $x$, I cannot easily tell whether it's irrational or not. Furthermore, if I want to start enumerating the irrationals, the definition helps me not at all.

For this reason, I think that Eric Wofsey's answer is the only one which satisfies the spirit of the question. It both constructs instances of irrational numbers and offers a method for enumerating them!

From a computational perspective, all definitions require an infinite amount of work to verify the irrationality of $x$, due to the nature of irrational numbers to begin with. But with the continued fraction definition, one can compute increasingly longer approximations which get closer and closer to $x$, which are the convergents of $x$. With the other definitions, it is not clear how to proceed in a manner which increases one's confidence in the irrationality of $x$.

That is, if $x$ is actually rational, then there must be some $n$ at which the sequence equals $x$, and it should always be clear how to adjust the sequence to get closer to $x$. That's because increasing $a_i$ will always produce a bigger change than increasing $a_j$ when $i < j$. Therefore, searching for the sequence $a_0...a_n$ which is closest to $x$ should be linear in $n$.

Of course, this is hand-waving a bit, since this process requires a computer which can represent arbitrary reals to begin with, even though there can be no such thing. But the argument still works if we bound the representation by saying that we can only inspect the first $B$ digits of each value.

More importantly, for rational values of $x$, we know that computing the best rational approximation will terminate in finite time, and that this time is proportional to the "irrationality" of $x$, where I define "irrationality" to be the number of values $n$ in the continued fraction sequence which exactly equals $x$ (meaning, all successive values are 0). When $n = \infty$, $x$ is "completely irrational" or just "irrational". Of course, when $n = 0$, then $x$ is "completely rational", or "integer". When $n$ is small, $x$ is only "slightly irrational". ;)

On the other hand, it is not at all apparent to me how one could use Dirichlet's or Hurwitz's theorems to determine the rationality of $x$ in finite time, let alone an algorithm which is linear in the irrationality of $x$.

Answer 14

Just look at this algorithm which may gets you a continued fraction:

You start with a real number $\alpha_0$, then define $a_i$ recusively by $a_i=\frac{1 }{\alpha_{i-1} -n_i}$ where $n_i$ is the greatest integer strictly smaller than $\alpha_{i-1}$.

Now my claim is:

An irrational number is a real number that can be expressed as continued fraction fraction using this algorithm.

By expressed I mean $$ \limsup_i [n_1;n_2, \dots ,n_i]= \alpha_0 = \liminf_i [n_1;n_2, \dots ,n_i].$$

I used the axiomatic definition of the real numbers,(which to me is somehow positive) that is why I write the limit like this.

Answer 15

An irrational number is a number $c$ such that $\{cn \text{ mod } 1 \mid n \in \mathbb{N}\}$ is dense in $[0, 1]$.

More strongly, an irrational number is a number $c$ such that for every continuous function $f : [0, 1] \rightarrow \mathbb{R}$,

$$\lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^N f(nc \text{ mod } 1) = \int_0^1 f \text{d}x.$$

Answer 16

I think the spirit of "positive" in the original question is some way to test if a number is irrational in a finite time.

So assume(*) we have a computing engine (CE) that has all the language elements of C, common functions like floor and pow, but which can handle Real numbers with no loss of precision.

With this CE, can you write a function isIrrational(X) that gets the right answer in a finite time, and with a finite number of Real variables (i.e. infinite set of all rationals not allowed)? If not, can you prove that it is impossible with this CE?

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(*) I expect this CE is impossible to implement with a Turing machine. But maybe it could be implemented by an analog computer in a non digital Universe, which could store the Real value as some continuous physical property.

Answer 17

There is a somewhat trivial answer:

An irrational number is a real number $x$ such that for every rational number $r$ we have $|x−r|>0$.

Of course, this is equivalent to $x$ being not rational - that is the goal, after all. However, I think the statement has no element that is obviously "negative".

This definition can be phrased in terms of positive knowledge: for each rational $r$, there is a pair of disjoint open intervals, one containing $x$ and one containing $r$. This is to say: $x$ is irrational if and only if it is 'separated' from each rational number.

Answer 18

This question is discussed in some detail on a different answer of mine: Are there real numbers that are neither rational nor irrational? See also What's an example of a number that is neither rational nor irrational?.

A small quote, with the actual definition:

Thus there is a measurable gap between the irrationality measures of rational and irrational numbers, and this yields an alternative "constructive" definition of irrational: let $x\in\Bbb I$, read "$x$ is irrational", if $|x-p/q|<q^{-2}$ has infinitely many coprime solutions.

Let me add, to forestall confusion about the word "infinite" in the above: I mean that the set of solutions is denumerable, which is to say that there is a function which takes each solution $(p,q)$ to another solution $(p',q')$ such that $(p,q)<(p',q')$ under some suitable total order (such as lexicographic order).

The definition is equivalent to Eric Schmidt's, but the extra comments about application to intuitionistic logic are worth a read if that is your interest.

Answer 19

A real number $\alpha$ is irrational precisely when the equation $y=\alpha x$ has one solution in integers, namely $(0,0)$. The number $\alpha$ is rational if and only there are and infinite number of integral solutions. Here is the picture from C. D. Olds' book Continued Fractions: