Proving Irrationality

Solution 1:

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 $1$s, then a $0$, then three $1$s, then a $0$, then four $1$s, then a $0$, and so on).

Solution 2:

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.

Solution 3:

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.

Solution 4:

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.