Rational & Irrational Numbers

All the Real numbers can be categorized under Rational numbers or Irrational numbers. 

Definition of Rational Numbers :

         Any number which can be represent as a ratio of two integers (p/q) where denominator(q) is not equal to zero. 

q can be one, so integers also rational numbers. 

Examples :- Converting decimal number to the form of rational number.

According to the above examples we can see that any finite or recurring decimal can be expressed in the form of 'p/q' where p and q are integers (q is not equal to zero).

Note : Above examples can be done using the following method as well.

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$1.233333...=1.2 + (\frac{3}{100}+\frac{3}{1000}+\frac{3}{10000}+...)$

$=1.2+3\left(\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}+...\right)$

The series within the brackets is a geometric series with infinite terms. So,

$=1.2+3\left(\frac{\frac{1}{100}\left(1-(\frac{1}{10})^\infty\right)}{1-\frac{1}{10}}\right)$

$=1.2+3(\frac{1}{90})$

$=\frac{12}{10}+\frac{1}{30}$

$=\frac{37}{30}$

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Exercise 01 :-

1)  1.24             2)   25.396            3)   121.12

4)   1.111...       5)   3.21414...       6)   3.123123...

*Answers of the exercise are at the bottom of the article.

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Showing $\sqrt{2}$ is irrational.

Let’s assume that $\sqrt{2}$  a rational number. Then,

$\sqrt{2}=\frac{p}{q}$       Where $p,q\in \mathbb{Z} ,q\ne0$ 

It is possible to assume that there is no any common factor to $p$ and $q$, as any rational number can be reduce to a fraction where there is no any factors common between denominator and numerator (other than one). 

So,  $p=\sqrt{2}q$

$p^2=2q^2$……………………(1)

This implies that $p^2$ is an even number.

If $p^2$ is an even number, then $p$ probably going to be even.

Then we can say,

$p=2m$ Where $m \in \mathbb{Z}$

Hence,

$p^2=4m^2$…………………(2)

By substituting this to equation (1),

$4m^2=2q^2$

$\therefore q^2=2m^2$

This final equation says that $q^2$ is also an even number. If so, we already know that $q$ must also be an even number. 

If $p$ and $q$ both are even numbers, then that means there is a common factor for these two numbers which is $2$. So this means there is a contradiction between the result and our first assumption. So there must be some mistake in the beginning.

All these things take us to a final conclusion that $\sqrt{2}$ cannot be a rational number. But we know that it is a real number, so it belongs to Irrational numbers.

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Answers for Exercise 01

1) $\frac{31}{25}$     2) $\frac{6349}{250}$     3) $\frac{3028}{25}$     4)$\frac{10}{9}$     5) $\frac{1591}{495}$     6) $\frac{1040}{333}$