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Why is it such a big deal that some numbers are irrational? It means they can't be represented as integer fractions. Cool. But almost all numbers satisfy that property. So why is it that, for example on $\pi$'s wikipedia page, already in the third line it tells us that $\pi$ is irrational?

Even if it had not told me that, I would've assumed it anyways. It's like if wikipedia had a page on a certain dog breed and told me "and you know what, this breed has a tail!".

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    $\begingroup$ being rational is a big deal, and $\pi$ is important, so knowing if $\pi$ is rational is a good idea. $\endgroup$
    – Asinomás
    Commented Jan 10, 2017 at 19:37
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    $\begingroup$ To be honest, I wouldn't be surprised if we would read about the specific tail of a breed in the beginning of an article... irrationality is just one of many properties one could name. If you continue to read the article, you'll also get to know that $\pi$ is a transcendental number (a subset of the irrational numbers) and this has some serious implications, for example the well-known squaring-the-circle problem. So, it is a somehow a big deal, ofc it depends on what you actually want to know. $\endgroup$
    – user190080
    Commented Jan 10, 2017 at 19:41
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    $\begingroup$ In a similar vein, almost all numbers are strongly normal in every base simultaneously. But we lack even a single explicit example of a number which provably has this property. (Chaitin's constant is a specific example but being inherently uncomputable, I don't count it as explicit). Number theory is replete with statements that are probabilistically unremarkable, but whose proofs required powerful breakthroughs, which are indeed remarkable. $\endgroup$
    – Erick Wong
    Commented Jan 10, 2017 at 19:49
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    $\begingroup$ Here's an analogy: it appears that the number of species of bacteria may dwarf the number of other species on Earth. So why should the Wikipedia articles on E. coli or pneumococcus mention that these organisms are bacteria? $\endgroup$
    – David K
    Commented Jan 10, 2017 at 20:13
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    $\begingroup$ The fact that $\pi$ is irrational is perhaps surprising in that it is itself the ratio of a circumference and a diameter. $\endgroup$
    – Karl
    Commented Jan 10, 2017 at 21:19

7 Answers 7

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Even if it had not told me that, I would've assumed it anyways.

Really? Would you also assume that $ A=\frac{1}{\pi^2} \sum_{n=1}^\infty \frac{1}{n^2} $ is irrational? And what about $$ B=\frac{\sum_{n=1}^\infty \frac{1}{n^4}}{\left(\sum_{n=1}^\infty \frac{1}{n^2}\right)^2} \ \ ? $$ $$ C=\sum_{n=1}^\infty \frac{1}{n(n+1)} \ \ ?$$ $$ D=\sum_{n=1}^\infty \frac{1}{n(n+2)} \ \ ?$$ $$ E=\frac{\sum_{n=1}^\infty \frac{1}{n^6}}{\left(\sum_{n=1}^\infty \frac{1}{n^3}\right)^2} \ \ ? $$ $$ F= \frac{1}{\sqrt{\pi}} \int_{-\infty}^\infty e^{-x^2}\,dx \ \ ? $$ The fact is, mathematics is full of results of the form "a given number defined in some special way is rational" - for example, the number $B$ defined above is equal to $2/5$, and $F$ is equal to 1. Usually we don't describe such results in those terms however, we call them "formulas" or "identities" and write them in a way that doesn't emphasize rationality, e.g. as $$ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $$ (a different way to state the theorem "$A$ is rational and is equal to $1/6$"). So if you assume that any number with a complicated definition that you come across is irrational just because in a statistical sense almost all real numbers are irrational, you would certainly be wrong a lot of the time. And other times you'll be lucky and get it right, and yet other times (as with the number $E$ defined above) it will be an open problem whether you're right or wrong.

To summarize, it is a big deal whether an important number like $\pi$ is rational or not because rationality/irrationality is both one of the most important attributes that real numbers possess, and an attribute that is surprisingly nontrivial both to guess and to prove. The reason you think it's so obvious that $\pi$ is irrational has more to do with psychological conditioning than with mathematical obviousness: it's just that if $\pi$ were rational then the way to write it as a ratio of integers would be one of the most famous facts in mathematics, which everyone would learn in grade school, and so from the fact that you never learned such a representation of $\pi$ in grade school or heard about it anywhere, it is easy for you to guess that $\pi$ cannot in fact be represented in such a way.

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    $\begingroup$ Could someone add some links to further reading on these equations $\endgroup$
    – HEGX64
    Commented Jan 12, 2017 at 6:54
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    $\begingroup$ Also $$G=\lim_{n\to\infty} \left( \left( \sum_{k=1}^n \frac{1}{k} \right) - \log n \right)$$ is the famous Euler–Mascheroni constant and it is not known whether it is irrational. $\endgroup$ Commented Jan 12, 2017 at 9:31
  • $\begingroup$ Isn't your calculation of $F$ off by a factor of $\sqrt{2}$? $$\int_{-\infty}^\infty e^{-x^2}dx = \sqrt{\int_{x=-\infty}^\infty e^{-x^2}dx {\int_{y=-\infty}^\infty e^{-y^2}dy}= \sqrt{ \int_r=0}^\infty \int{\theta=0}^{2\pi} e^{-r^2} r\,d\theta\,dr} = \sqrt{2\pi}$$ $\endgroup$ Commented Jan 13, 2017 at 23:33
  • $\begingroup$ @MarkFischler Seems fine according to WA. bit.ly/2kazlTb $\endgroup$
    – Jam
    Commented Jan 27, 2017 at 11:21
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    $\begingroup$ @MarkFischler That's some tasty MathJax. $\endgroup$
    – TheSimpliFire
    Commented Sep 24, 2020 at 19:24
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It is a big deal historically. When the mathematics of numbers was developed, it was natural to start with integers and progress to rationals. Rationals are so useful that became the end of what was thought about, and eventually was like a religion: Every number is rational.

So when the first quantity was discovered that was demanded by Euclidean geometry, working with only integer lengths to start with, yet could not be a rational number, those mathematicians underwent somewhat of an existential crisis.

Another point, more relevant to the wiki page, is that it is not always easy to know that a specific definable number is irrational. As an example, consider $$\zeta(3) = \sum_{n=1}^\infty \frac1{n^3}$$

Although nobody ever expected this to turn out to be rational, or even algebraic, until 1977 nobody had proven that it is irrational. In 1978, Apéry proved it is irrational and this is important information (we still don't know whether it is transcendental). So the fact that $\pi$ has been proven to be irrational and in fact has been proven to be transcendental is indeed an important fact about $\pi$.

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    $\begingroup$ Apéry proved that $\zeta(3)$ is irrational in the 1970s. However, no further odd zeta values have been proven irrational, although there are lower bounds on the number of irrational values taken by $\zeta(2n+1)$. I believe it is still open whether $\zeta(3)$ is transcendental. $\endgroup$
    – Erick Wong
    Commented Jan 10, 2017 at 19:42
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    $\begingroup$ @ErickWong Is it correct that it is still unknown today if $\frac{\zeta(3)}{\pi^3}$ is rational? It is a classical result that $\frac{\zeta(k)}{\pi^k}$ is in fact rational when $k$ is an even positive integer (cf. Basel problem). $\endgroup$ Commented Jan 12, 2017 at 9:59
  • $\begingroup$ en.wikipedia.org/wiki/Hippasus $\endgroup$
    – mvw
    Commented Jan 12, 2017 at 10:46
  • $\begingroup$ @JeppeStigNielsen That is still open as far as I know. Since both constants are available at high precision, I'm sure it's known that it isn't a rational number of moderate height (millions if not billions of digits). $\endgroup$
    – Erick Wong
    Commented Jan 12, 2017 at 13:21
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"It can be of no practical use to know that $\pi$ is irrational, but if we can know, it surely would be intolerable not to know."

— E.C.Titchmarsh

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    $\begingroup$ Extra info: A proof of the irrationality of $\pi$ was given by Lambert in 1761 (while Euler was still alive), and in 1882 Lindemann proved it was also a transcendental number. E. C. Titchmarsh lived 1899-1963. $\endgroup$ Commented Jan 12, 2017 at 16:52
  • $\begingroup$ "It can be of no practical use" - on what basis could he have made such a strong assertion? It's just hyperbole, right? $\endgroup$
    – LarsH
    Commented Jan 12, 2017 at 20:24
  • $\begingroup$ Kind of like his assertion that great mathematicians "have contributed something to human thought even more lasting than great literature, since it is independent of language" -- as if literature didn't contribute any ideas that are independent of language. $\endgroup$
    – LarsH
    Commented Jan 12, 2017 at 20:31
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    $\begingroup$ @LarsH I would semi-agree that it is hyperbole (though I strongly agree with the second part of the quotation). Titchmarsh was probably thinking of a measurement or engineering context, where it is most likely entirely reasonable to pretend that $\pi$ is exactly equal to $3.141592654$. $\endgroup$
    – David
    Commented Jan 12, 2017 at 23:49
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    $\begingroup$ @LarsH Mathematicians are not always mathematicians, sometimes they are people! At least I hope I am :) $\endgroup$
    – David
    Commented Jan 15, 2017 at 23:13
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But almost all numbers satisfy that property.

This is the key understatement. There's a lot of irrational numbers. To get a sense of how many there are, consider the task of writing down all of the real numbers between 0 and 1. There's a lot right? Infinitely many. No way you could write them all. How about all of the rational numbers between 0 and 1. Writing the numbers at a finite rate, it would take you an infinite amount of step to write down all of the rational numbers.

But note the difference in phrasing. For the rational numbers, it would take an infinite number of steps to write down all the numbers. For real numbers, I said it was impossible. I chose a different phrasing, and for good reason.

We have this concept of counting. It goes 0, 1, 2, 3, 4... and keeps going. These are called the natural numbers. The set of natural numbers is "countably infinite." If you kept adding 1 over and over, you could eventually construct every natural number. (this is basically the definition of a "countably infinite" set)

If we look at the real numbers, and pile in $\pi$ and $\sqrt 2$ and all of their irrational friends, we have more numbers than we had natural numbers. The set of all real numbers is "uncountably infinite." There's more real numbers than there are natural numbers.

Big deal right? There's more rational numbers than natural numbers too, right? Well... not exactly.

It actually turns out that the set of rational numbers is countably infinite. There is a formal way to map all of the rational numbers onto the natural numbers. It's typically done using a diagonalizing approach:

Cantor diagonlization

So, as strange as it may sound there are exactly the same number of natural numbers as rational numbers, but there are more real numbers than that. This has many interesting effects deeper into mathematics, because we can use mathematical induction to prove things as long as the set of values we're proving it over is no bigger than the natural numbers. Once we move into larger sets, like the set of irrational numbers, mathematical induction is no longer a valid tool in a proof.

And yes, the ability to use mathematical induction is a big deal =)

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    $\begingroup$ You count many of rationals multiple (infinitely many) times in your diagonalizing approach (if you don't say that you skip those already counted). $\endgroup$
    – Ruslan
    Commented Jan 11, 2017 at 7:28
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    $\begingroup$ There are some forms of induction that work even in uncountable sets, or even proper classes. The important thing is to exploit the order type in a way that allows a proof. $\endgroup$
    – J.G.
    Commented Jan 11, 2017 at 7:29
  • $\begingroup$ @J.G. Yes, see for example the response to this MSE question given by Pete L. Clark. $\endgroup$
    – Will R
    Commented Jan 11, 2017 at 12:20
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    $\begingroup$ How does this answer the question, and why do you assume that the OP is not already aware of all of this? $\endgroup$
    – Carsten S
    Commented Jan 11, 2017 at 16:06
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    $\begingroup$ Of course, "almost all" has a very specific meaning beyond the colloquial interpretation. I think OP's use of that terminology suggests that @CarstenS is correct that OP already knows this. $\endgroup$
    – Tristan
    Commented Jan 11, 2017 at 16:13
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Because if you start with the integers, with the operations of addition, subtraction, multiplication, and division, you only get the rationals.

It is only when you take square roots (via the Pythagorean theorem) that you get irrationals, and that was unexpected. There does not seem to be any a priori reason why square roots should not be rational.

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    $\begingroup$ IMO, the statement "It is only when you take square roots..." is far from being true. There are many other operations that take rational input and return irrational output. For example, cube-root, log, etc. In fact, you don't even need those. An infinite $\sum$ with the "right" expression would achieve that as well. $\endgroup$ Commented Jan 10, 2017 at 19:46
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    $\begingroup$ Historically, the irrational square roots of rationals are believed to be the first numbers that were known to be irrational. Perhaps this answer can be reworded. $\endgroup$
    – David K
    Commented Jan 10, 2017 at 19:50
  • $\begingroup$ @DavidK: There's no "historically" mentioned there, and I still have a problem with that "only". $\endgroup$ Commented Jan 10, 2017 at 19:51
  • $\begingroup$ @barakmanos Right, my comment was what the answer could have said but did not quite say. I edited my comment to suggest the answer be edited. $\endgroup$
    – David K
    Commented Jan 10, 2017 at 19:52
  • $\begingroup$ I understood Marty's answer to mean "only when" in a sense of progressively advanced operations, which, yes, follows mathematical history (at least in Greece). I think this gets to the heart of the question... irrationality is interesting, not because irrational numbers are more or less common than rationals, but because they're "harder to get to" starting from the real-life basis of mathematics. $\endgroup$
    – LarsH
    Commented Jan 20, 2017 at 15:41
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Why is it such a big deal that some numbers are irrational?

It isn't, always.

It is, sometimes, because classification of stuff is what we do as mathematicians.

Split groups of stuff into subgroups of stuff, and talk about how different rule changes alter the groups around. We love classification.

So why is it that, for example on π's wikipedia page, already in the third line it tells us that π is irrational?

If you look at any well-stocked Wikipedia article, there will be plenty of information that one may find irrelevant to their given task. For example, "I need code samples, not the history of MVC." Or, "I want Will Smith's height, not his net worth."

"π is irrational" is just a fact about π, and "facts about subjects" is exactly what the Wikipedia pages are for.

Even if it had not told me that, I would've assumed it anyways. It's like if wikipedia had a page on a certain dog breed and told me "and you know what, this breed has a tail!".

But I'd bet you didn't know Will Smith's net worth was 250 million without being told. Or that the patent for the cotton gin was granted on March 14, 1794.

Yet other people may have known that stuff at the top of their heads. So why was it included on the page?

Wikipedia doesn't alter the content of their pages based on the presumed knowledge of their visitors. It just hosts a web site full of "facts about stuff", and the editors have discussions about each page and what type of content to include.

If you're really passionate about what types of content should be included on Wikipedia pages, you should join the editor's community. You can start with π.

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    $\begingroup$ If Google ran Wikipedia they would make an estimate of the presumed knowledge of their visitors, based on other pages searched, and would alter the content of their pages besed on that estimate. As to whether than would be good or bad... $\endgroup$ Commented Jan 12, 2017 at 17:14
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Because of the very nature of irrational numbers, we can use them to construct new objects to exhibit a certain behavior.

An example which I like a lot is the following mapping:

$f : x \mapsto \begin{cases} 0 \text{ if } x \in \mathbb{R} \setminus \mathbb{Q} \text{ (irrational)} \\ \dfrac{1}{q} \text{ if } x = \dfrac{p}{q} \in \mathbb{Q} \text{ (rational)} \end{cases}$

One can show that such a mapping is discontinuous in every rational point.

Question: Is the above mapping continuous on irrational point?

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    $\begingroup$ I don't feel that this answer relates well to the original question. This is more "what are irrational numbers useful for?" rather than "why is it worth pointing out that x is irrational?". $\endgroup$
    – Erick Wong
    Commented Jan 11, 2017 at 0:12
  • $\begingroup$ @ErickWong I agree! I wanted to give another type of answer, that is: the one which points out how irrationals are useful so that it's worth to point out that they are irrationals. But definitely, it's not a direct answer. $\endgroup$
    – Raito
    Commented Jan 11, 2017 at 7:13
  • $\begingroup$ +! to forf your interesting question about whether thus cunftion is continuous at irrational points. $\endgroup$ Commented Jan 12, 2017 at 17:12

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