# Proving Irrationality

How is it possible to prove a number is irrational?

First part of that question: How it possible to know that a number will go on infinitely?

Second part: How is it possible to know that no repetition will occur during the infinite sequence of digits?

Any examples of proofs of irrationality?

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See this question and its answers for an example. –  Jyrki Lahtonen Sep 8 '11 at 16:36
It can be hard in general. For example, nobody's been able to show that the Euler-Mascheroni constant is irrational. –  Ｊ. Ｍ. Sep 8 '11 at 16:36
...and it took quite a while before $\zeta(3)=\sum\limits_{k=1}^\infty \frac1{k^3}$ was proven irrational. –  Ｊ. Ｍ. Sep 8 '11 at 16:37
As an example, en.wikipedia.org/wiki/Proof_that_e_is_irrational –  Alex Sep 8 '11 at 16:37
You don't prove "goes on indefinitely" (since that is not the definition of irrational); instead you prove "not a quotient of integers". –  GEdgar Sep 8 '11 at 18:18

There are many proofs of irrationality, and some of them are quite different from each other. The simplest that I know is a proof that $\log_2 3$ is irrational. Here it is: remember that to say that a number is rational is to say that it is $a/b$, where $a$ and $b$ are integers (e.g. $5/7$, etc.). So suppose $\log_2 3 = a/b$. Since this is a positive number, we can take $a$ and $b$ to be positive. Then $$2^{a/b} = 3.$$ $$2^a = 3^b.$$ But that says an even number equals an odd number. That is impossible. Hence $\log_2 3$ cannot be rational.

The most well known and oldest proof of irrationality is a proof that $\sqrt{2}$ is irrational. I see that that's already posted here. Here's another proof of that same result:

Suppose it is rational, i.e. $\sqrt{2} = n/m$. We can take $n$ and $m$ to be positive and the fraction to be in lowest terms. Then a bit of algebra shows that $(2m-n)/(n-m)$ is also equal to $\sqrt{2}$ but is in even lower terms. That is a contradiction. Hence it is impossible for $\sqrt{2}$ to be rational.

It is also not very hard to show that $e$, the base of natural logarithms, is irrational.

To show that $\pi$ is irrational is much harder---in fact so hard that it was not done until the 18th century.

Another proof of irrationality begins by proving that when you divide an integer by another integer, if the decimal expansion does not terminate, then it must repeat. I posted an explanation of that here. Once you've done that, you can construct a non-repeating decimal. For example: $$0.10110111011110111110\ldots$$ (a 1, then a 0, then two 1s, then a 0, then three 1s, then a 0, then four 1s, then a 0, and so on).

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"To show that $\pi$ is irrational is much harder---in fact so hard that it was not done until the 18th century." Just for the record (and picking a rather minor nit), $e$ was first proved irrational in 1737 (by Euler) and $\pi$ was first proved irrational in 1767 (by Lambert), both in the 18th century, so the part about "not done until the 18th century" is a little misplaced (but not the part about $\pi$ being harder to prove irrational, of course). The proof for $\log_{2}3$ is a nice example, by the way. Peter, you should look at Ivan Niven's book Numbers: Rational and Irrational. –  Dave L. Renfro Sep 8 '11 at 17:34
Well, $\pi$ was thought about in ancient times---in the third century BC by Archimedes, who showed that $3 + 10/71 < \pi < 3 + 1/7$. It was two millennia after that that $\pi$ was shown to be irrational. I suspect that $e$, on the other hand, might not have been considered until the 1500s or 1600s. So it took a much shorter time. –  Michael Hardy Sep 8 '11 at 17:56

Proving a number is irrational may or may not be easy. For example, nobody knows whether $\pi+e$ is rational.

On the other hand, there are properties we know rational numbers have and only rational numbers have, and properties we know irrational numbers have and only irrational numbers have. If we can show a given number have one of the former, we can guarantee it is rational; if we can show it has the latter, we can guarantee it is irrational. There are also properties that rational numbers, among others, have; if we can prove a given $x$ does not have such a property, then it cannot be rational. Or there are properties that only some rational numbers have (like terminating decimal expansion). If you number does have such a property, then it must be rational. Etc.

And sometimes it is possible to simply prove it "directly."

First: remember that the definition of "rational number" is not about its decimal expansion, but rather:

A real number $r$ is rational if there exist integers $a$ and $b$, $b\neq 0$, such that $\displaystyle r= \frac{a}{b}$.

It is a consequence of this definition that, if you write down a decimal expansion for a rational number, then it will be periodic (it will eventually repeat, perhaps with $0$s).

So it's not about numbers "going on to infinity". Or about decimal expansions. It's about being able to express the number as a ratio of two integers (hence "rational": a ratio).

(As a matter of fact, "most" numbers have non-terminating decimal expansions; not only do all irrationals have nonterminating decimal expansion, but a rational number $\frac{a}{b}$, with $a$ and $b$ relatively prime, has terminating decimal expansion if and only if no prime other than $2$ or $5$ divides $b$).

For example, the ancient greeks proves that $\sqrt{2}$ was not rational by contradiction:

Assume that $\sqrt{2}$ is rational, and write $\sqrt{2}=\frac{a}{b}$ with $a$ and $b$ integers. By cancelling, we may assume that $a$ and $b$ are not both even (if they are, we can simply keep cancelling powers of $2$ until one of them is not). Squaring we get that $2 = \frac{a^2}{b^2}$. Then $2b^2=a^2$. Since the left hand side is even, $a^2$ is even; but for a square to be even, we must have $a$ even. So $a=2k$ for some $k$. That means that $2b^2 = a^2 = (2k)^2 = 4k^2$. From $2b^2 = 4k^2$ we conclude that $b^2 = 2k^2$, so $b$ must be even. But this contradicts our assumption that $a$ and $b$ were not both even. The contradiction arises from assuming $\sqrt{2}$ is rational, therefore $\sqrt{2}$ is irrational.

We did not need to find the decimal expansion of $\sqrt{2}$, or prove it never repeated; we simply proved that it is impossible for $\sqrt{2}$ to be expressible as a ratio of two integers.

Likewise, one can show that for every positive integer $n$ and every positive integer $m$, $\sqrt[m]{n}$ is either an integer, or it is irrational (the proof uses either unique factorization of integers into primes or something similar).

Here's another example of something we know about rationals and irrationals: it is a corollary to a theorem of Hurwitz from 1891:

If $x$ is irrational, then there are infinitely many integers $p$ and $q$, $q\neq 0$, with $p$ and $q$ sharing no common factors other than $1$ and $-1$, such that $$\left| x- \frac{p}{q}\right| \lt \frac{1}{\sqrt{5}q^2}.$$

If you can show that for a given $x$, the inequality has only finitely many solutions, then the conclusion is that $x$ must be rational.

Likewise, there are theorems that tell us about algebraic numbers (roots of polynomials with integer coefficients). Every rational number is algebraic (since $\frac{a}{b}$ is the root of $bx-a$); if you can prove a number is not algebraic, then it must be irrational. For example, one can prove that $e$ and that $\pi$ are transcendental, but showing that they cannot be roots of any polynomial with integer coefficients; in particular, they cannot be rational either.

So, most of the time we aren't looking at the "decimal expansion" to decide if the number is rational or not (thought sometimes we do, for some very special numbers like the one Austin Mohr mentions). Instead, we look at the properties the number has to see if it has the properties of a rational number or of an irrational number.

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Regarding some comments you made (e.g. after the Greeks' proof that $\sqrt{2}$ is irrational and your last paragraph), I think you'll be amused by Richard Askey's "FFF #120" comment on p. 286 of The College Mathematics Journal, Vol. 28 No. 4 (Sept. 1997). –  Dave L. Renfro Sep 8 '11 at 17:45
@Bill: Problem. Explain why $4^{\frac{1}{2}}$ is rational while $5^{\frac{1}{2}}$ is irrational. Solution. $4^{\frac{1}{2}}=2$ which is rational. $5^{\frac{1}{2}}$, in its decimal form, does not terminate or repeat and therefore cannot be written as an integer over an integer. Richard Askey of the University of Wisconsin in Madison found this problem in a 1996 school textbook; the solution is from the teacher's edition. He wonders how many students or teachers would be able prove that the decimal expansion of $5^{\frac{1}{2}}$ is nonrepeating without first proving that it is irrational. –  Dave L. Renfro Sep 8 '11 at 19:27
While I'm at it, I'll mention that Askey has a book review to appear in the October 2011 issue of Amer. Math. Monthly that contains additional alarming stories such as this one. (I haven't seen a copy of the review yet.) –  Dave L. Renfro Sep 8 '11 at 19:43
@Dave That reminds me of a remark by H. Davenport on p. 10 of his Higher Arithmetic. While discussing the history of the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorization), he remarks that a "schoolbook ... (still in use) describes it as a 'law of thought', which it certainly is not". The mind boggles... –  Bill Dubuque Sep 8 '11 at 20:10
Isn't Hurwitz's theorem if and only if? –  user7530 Sep 9 '11 at 0:37

It's very easy to construct an infinite set of irrationals that, furthermore, are $\rm\mathbb Q$-linear independent, namely $\,\rm \{\log_2 p_{\,i}\}$ with $\rm\{p_{i}\} =$ all odd primes. For if $\rm\ c_1\log_2p_1+\cdots+\ c_{n}\log_2p_{n} =\, c_o,\,$ $\rm\, c_{i}\in \mathbb Q,$ then, by scaling by a common denominator of all $\rm\,c_{i},\,$ we can assume that all $\,\rm c_{i}\in\mathbb Z.\$ Exponentiating $\,\rm x\to 2^{x}$ yields $\rm\ p_1^{c_1}\cdots p_n^{c_n} = 2^{c_o},\,$ hence all $\rm\,c_{i} = 0\,$ by uniqueness of prime factorizations (all the primes being distinct). The special case $\rm\,n=1\,\Rightarrow\,\log_2p_{i}\not\in\mathbb Q.\ \$ QED

Note $\$ See also the similar example in my post here.

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+1. From your example it seems to me that it is easier to construct irrationals than sometimes proving a particular constant is an irrational. –  Américo Tavares Sep 8 '11 at 20:59
I wouldn't show this as a first example to a person who didn't know that it's possible to prove some numbers are irrational, unless maybe his name is Ramanujan. –  Michael Hardy Sep 9 '11 at 1:12
@Mic I've presented variations on this to many such students with much success. In fact the variation I link to above is currently my most popular post here, with over $60$ upvotes. The above proof uses nothing deeper than the log proof in your answer, except for linear dependence (which can be omitted if so desired). And if you think the above has anything to do with Ramanujan-like mathematics then I suggest you take a second look at his work, because it is generally far less trivial. –  Bill Dubuque Sep 9 '11 at 3:10

I give the following example.

Proposition. If there exist two integer sequences $a_{n}$ and $b_{n}$ such that $$0<|b_{n}\alpha -a_{n}|\rightarrow 0,$$ then $\alpha$ is an irrational number.

Proof. Assume $\alpha =p/q\in\mathbb{Q}$. For $n$ large enough the integer sequence $\left\vert pb_{n}-qa_{n}\right\vert <1$ and $pb_{n}-qa_{n}\neq 0$, which is impossible.

In his proof of the irrationality of $\zeta (3)$ Roger Apéry introduced the following sequences (see this article by van der Poorten and this one, in French, by Stéphane Fischler):

$$v_{n}=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2}$$ and $$u_{n}=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2}\left( \sum_{n=1}^{n} \frac{1}{m^{3}}+\sum_{m=1}^{k}\frac{(-1)^{m-1}}{2m^{3}\binom{n}{m}\binom{n+m }{m}}\right) .$$

The sequences $a_{n}=2d_{n}^{3}u_{n}$ and $b_{n}=2d_{n}^{3}v_{n}$, where $d_{n}=\text{lcm}(1,2,\ldots ,n)$, are integer sequences whose ratio converges to $\zeta (3)$

$$\zeta (3)=\lim_{n\rightarrow \infty }\frac{a_{n}}{b_{n}}=\lim_{n\rightarrow \infty }\frac{2d_{n}^{3}u_{n}}{2d_{n}^{3}v_{n}}=\lim_{n\rightarrow \infty }\frac{u_{n}}{v_{n}},$$

and

$$0< b_{n}\zeta (3)-a_{n}=O\left(\beta^n\right) ,$$ with $\beta=\left( 1-\sqrt{2}\right) ^{4}e^{3}<1.$

See examples 1 and 2 for proofs of the irrationality of $\sqrt{2}$ and $e$ in this entry of The Tricky

To prove that a number is irrational, show that it is almost rational

Loosely speaking, if you can approximate $\alpha$ well by rationals, then $\alpha$ is irrational. This turns out to be a very useful starting point for proofs of irrationality.

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the same answer was given by me,below see it once,but i have read it from a book ,not the website –  Iyengar Sep 8 '11 at 18:00
@iyengar: I provided a link to The Tricky. –  Américo Tavares Sep 8 '11 at 18:03
i think there must be only one question for the sake of preventing redundancy,so shall i delete my answer??,i read the both answers were same ,i think i delete my answer as you are senior to me –  Iyengar Sep 8 '11 at 18:04
senior both in age,and knowledge –  Iyengar Sep 8 '11 at 18:05
@iyengar: If someone has to delete his answer it's me, because you posted yours before. –  Américo Tavares Sep 8 '11 at 18:08

An elegant proof that $\sqrt[n]{2}$ is irrational for integers $n \ge 2$.

There are many solutions when $n = 2$. An example is this (from wiki)

When $n > 2$, we can use Fermat's Last Theorem (!)

Suppose $\sqrt[n]{2}$ is rational. Then, $\sqrt[n]{2} = \frac{a}{b}$ for some relatively prime positive integers $a, b$. Then, $a^n = 2b^n \Rightarrow a^n = b^n + b^n$, contradicting Fermat's Last Theorem!

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Actually, you should get $2(a+k)^2 = a^2$. Your proposed solutions simplify to $-k \pm k\sqrt{2}$. Just what justification you have for saying these cannot be integers is unclear. –  Michael Hardy Sep 9 '11 at 18:11
They come fairly close if $k=29$. Of course, we know by other means that this can't be exact, but only by other means. –  Michael Hardy Sep 9 '11 at 18:17
@Michael, you are right, I have a mistakes in the proof. Thanks! –  Tomas Sep 9 '11 at 18:54
@Michael, I deleted my wrong proof and rather put a link to a correct one. Thanks again. –  Tomas Sep 9 '11 at 19:05

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