“Irrational numbers are those real numbers which are not rational numbers!”

## Def.1: Rational Number

A rational number is a real number which can be expressed in the form of $\frac{a}{b}$ where $a$ and $b$ are both integers relatively prime to each other and $b$ being non-zero.
Following two statements are equivalent to the definition 1.
1. $x=\frac{a}{b}$ is rational if and only if $a$ and $b$ are integers relatively prime to each other and $b$ does not equal to zero.
2.   $x=\frac{a}{b}&space;\in&space;\mathbb{Q}&space;\iff&space;\mathrm{g.c.d.}&space;(a,b)&space;=1,&space;\&space;a&space;\in&space;\mathbb{Z},&space;\&space;b&space;\in&space;\mathbb{Z}&space;\setminus&space;\{0\}$.

## Def. 2: Relatively Prime Numbers

Two integers $a$ and $b$ are said to be relatively prime to each other if the greatest common divisor of $a$ and $b$ is $1$ .
For example: The pairs (2, 9); (4, 7) etc. are such that each element is relatively prime to other.

## Def. 3: Irrational Number

A real number, which does not fit well under the definition of rational numbers is termed as an irrational number.

A silly question: Let, in the definition of a rational numbers, $a=0$ and $b=8$ , then, as we know $\frac{0}{8}=0$ is a rational number, however $8$ can divide both integers $0$ and $8$ , i.e., $\mathrm{g.c.d.} (0,8) =8$ . (Why?) $\Box$

# Primary ways to prove the irrationality of a real number

It is all clear that any real, if not rational, is irrational. So, in order to prove a (real) number irrational, we need to show that it is not a rational number (i.e., not satisfying definition 1). Most popular method to prove irrationality in numbers, is the Proof by Contradiction, in which we first assume the given (irrational) number to be ‘almost’ rational and later we show that our assumption was untrue. There are many more ways to prove the irrational behavior of numbers but all those are more or less derived from the proof by contradiction.
Some methods which I’ll discuss here briefly are:

1. Pythagorean Approach
2. Using Euclidean Algorithm
3. Power series expansion of special numbers
4. Continued Fraction representation of irrational numbers.

## (1) Pythagorean Approach

This proof is due to Pythagoras and thus called Pythagorean Approach to irrationality. In this approach, we assume a number to be first. Later using the fundamental rules of arithmetic, we make sure whether or not our assumption was true. If our assumption was true, the number we took was rational, otherwise irrational.
For example:

Prove that the number $\sqrt{2}$ is irrational.$Proof: Suppose, to the contrary, that$ \sqrt{2}$is a rational number. Then as according to the definition 1, we can write$ \sqrt{2}=\frac{a}{b} \ldots (1)$where$ a$and$ b$are both integers with$ \mathrm{g.c.d.} (a,b) =1$and$ b \ne 0$. Squaring equation (1),$ 2=\frac{a^2}{b^2}$or,$ a^2=2b^2 \ldots (2)$From the equation (2), we can proceed our proof into two ways: Way I:$ \sqrt{2}$is a positive number, therefore we can assume$ a$and$ b$both to be positive. Since,$ a^2=2b^2$then$ a^2=2b \cdot b$or,$ b|a^2$(read as b divides a squared). Since,$ b$is positive integer,$ b \ge 1$. However,$ b=1$is impossible since corresponding$ a=\sqrt{2}$is not an integer. Thus,$ b > 1$and then according to fundamental theorem of arithmetic, there exists at least one prime$ p> 1$which divides$ b$. Mathematically,$ p|b$but as$ b|a^2$. It is clear that$ p|a^2$. This implies that$ p|a$. Since$ p|a$and$ p|b$, therefore$ \mathrm{g.c.d.}(a,b) \ge p$. So for given number,the greatest common divisor of$ a$and$ b$is not$ 1$, but another prime larger than$ 1$. Thus, it fails to satisfy the definition 1. Thus our claim that$ \sqrt{2}$is rational, is untrue. Therefore,$ \sqrt{2}$is an irrational number. Way II: As a deviation, we can proceed our proof from equation (2) by taking the fact into mind that$ \sqrt{2}$is positive. The number (natural number)$ a$can either be odd or even. Let$ a$be odd, i.e.,$ a=2k+1$where$ k\in \{0,1,2,3, \ldots \}$. Therefore$ a^2$would also be odd. Which contradicts (2), since$ 2b^2$is always even and that equals to$ a^2$. Therefore,$ a$must be an even number. Let$ a=2k$. Putting this into (2) we get,$ 4k^2=2b^2$or,$ b^2=2k^2$or,$ b=\sqrt{2} k \mathrm{g.c.d.}(a,b)=\sqrt{2} \ne 1$. Which is contradiction to our claim. Thus$ \sqrt{2}$is an irrational number. In similar ways, one can prove$ \sqrt{3}$,$ \sqrt{5}$,$ \sqrt{7}$etc. to be irrationals. ## (2) Using Euclidean Algorithm This is an interesting variation of Pythagorean proof. Let$ \sqrt{2}=\frac{a}{b}$with$ \mathrm{g.c.d.}(a,b)=1$, then according to Euclidean Algorithm, there must exist integers$ r$and$ s$, satisfying$ ar+bs=1$. or,$ \sqrt{2} \cdot 1 =\sqrt{2}(ar+bs)$or,$ \sqrt{2}=\sqrt{2}ar +\sqrt{2}bs$or,$ \sqrt{2}=(\sqrt{2}a)r+ (\sqrt{2}b)s$or,$ \sqrt{2}=2br +as$. (From$ \sqrt{2}=a/b$we put$ a=\sqrt{2} b$.) This representation of$ \sqrt{2}$leads us to conclude that$ \sqrt{2}$is an integer, which is completely false. Hence our claim that$ \sqrt{2}$can be written in form of$ \frac{a}{b}$is untrue. Thus,$ \sqrt{2}$is irrational. Similarly, we can use other numbers to prove so. ## (3) Power Series Expansion Some irrational numbers, like$ e$, can be proved to be irrational by expanding them and arranging the terms. Over all, it is another form of proof by contradiction but different from the Pythagorean Approach.$ e$can be defined by the following infinite series:$ e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!}+\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\ldots$. Suppose, to the contrary, that$ e$is rational, and$ e=\frac{a}{b}$(say) where$ a$and$ b$are positive integers. Then for any$ n>b$and also$ n>1$,$ N=n! \left({e-(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!})}\right)$is positive, since$ \left({e-(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!})}\right)$is positive. Or,$ N=n! \left({\dfrac{1}{(n+1)!}+\dfrac{1}{(n+2)!}+\dfrac{1}{(n+3)!}+\ldots}\right)$or,$ N=\left({\dfrac{n!}{(n+1)!}+\dfrac{n!}{(n+2)!}+\dfrac{n!}{(n+3)!}+\ldots}\right)$or,$ N=\left({\dfrac{1}{(n+1)}+\dfrac{1}{(n+1)(n+2)}+\dfrac{1}{(n+1)(n+2)(n+3)}+\ldots}\right) > 0$. It is clear that$ N$is less than$ \left({\dfrac{1}{(n+1)}+\dfrac{1}{(n+1)(n+2)}+\dfrac{1}{(n+2)(n+3)}+\ldots}\right)$. And$ \left({\dfrac{1}{(n+1)}+\dfrac{1}{(n+1)(n+2)}+\dfrac{1}{(n+2)(n+3)}+\ldots}\right) =\left({\dfrac{1}{(n+1)}+\dfrac{1}{(n+1)}-\dfrac{1}{(n+2)}+\dfrac{1}{(n+2)}-\dfrac{1}{(n+3)}+\ldots}\right) =\dfrac{2}{(n+1)}$. Thus,$ N < \dfrac{2}{(n+1)}<1$. So,$ N$being positive integer is less than$ 1$? It is impossible for any integer. Thus our claim is not true and hence$ e$is irrational. ## (4) Continued Fractions Any number, that can be expressed in form of an infinite continued fraction is always irrational. For example: 1)$ e$can be represented in form of infinite continued fractions, thus$ e$is irrational. 2) Similarly$ \pi$is irrational. 3)$ \sqrt{2}$is also irrational.$ \Box$### Posted by Gaurav Tiwari A designer by profession, a mathematician by education but a Blogger by hobby. With an experience of over seven years with WordPress, PHP and CSS3, Gaurav is capable of doing almost anything related to these. Beyond that, He is a mathematics graduate & a civil service aspirant. 1. Math has never been my strong point but I hugely admire those who excel at it. 2. Heya Gaurav, I was looking for explanations on how Ramnujuan derived his nested radicals problem when I found your blog, and thought I would tell you I have read a few of your articles and think they are fantastic! Your explanation of the nested radicals problem was very clear for me to understand, however even if I dont understand all of the working in your articles (I have not started university maths yet), I can usually follow them and at the very least find them interesting. The only addition I can make to this article is that ‘Way 1’ of how you have proved the irrationality of root 2 is an example of proof by infinite descent which you did not mention, and it would be interesting to know how you can derive those continued fractions as I have never seen them before. Please keep up the writing and I will be sure to continue reading your blog 🙂 1. Hi and Thanks for reading this article Alex. I appreciate your comments and suggestions. I think this paper on modifying irrational numbers as continued fractions would be very useful for you. Download 3. Hello Gaurav, While going through this post I found a serious error in the fourth method of irrationality proof based on continued fractions. The idea that rationals have a finite continued fraction representation and irrationals have infinite ones is not fully correct. It applies only to simple continued fractions where the numerators in the continued fraction are all equal to$ 1$. Note that to use continued fractions for proving irrationality we need to apply special logic in each particular case. I have presented the proofs of irrationality of$ \pi$based on continued fractions in my posts here and here. Please go through them and you will understand that technique of continued fractions requires great skill and is not applicable in many places generally. In this connection it is also of interest to read the proof of irrationality of$ \zeta(3)$based on continued fraction. This can be found by searching on google. Apart from that you may also like to have a look at irrationality proofs related to$ e, e^{2}, e^{4}\$ given by Liouville which I have presented here, here and here.

Bye
Paramanand

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5. Thank you for this detailed analysis on irrational numbers

6. To find correct answer in Maths is not such an easy task. lol 😀 😀