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

A rational number is a real number which can be expressed in the form of 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. .

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

e 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}

e 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$

26.740278 83.888889

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