**REVISITING IRRATIONAL NUMBERS :
**

We have studied in the previous class about rational numbers as well as irrational numbers. Both are the members of the REAL NUMBERS family. We have also studied how to locate irrational numbers on the number line.

Any rational number is represented by ; where whereas irrational number is represented by ‘s’

0.10110111011110….. etc. are the examples of irrational numbers.

**Theorem : **Let p be a prime number. If p divides a^{2}, then p divides a, where a is a positive integer.

**Proof :
**

Let the prime factorization of a be as follows :

a = p_{1}, p_{2}, p_{3}, p_{4}, p_{5}, …., …. p_{n}

a^{2} = (p_{1}, p_{2}, p_{3}, p_{4}, …. P_{n}) (p_{1}, p_{2}, p_{3}, p_{4}, …. P_{n})

= p_{1}^{2}, p_{2}^{2}, p_{3}^{2}, p_{4}^{2}, …………… p_{n}^{2}.

It is given that p divides a^{2}, so from the Fundamental Theorem of Arithmetic p is one of the prime factors of a^{2}. Also a^{2} having prime factors p_{1}, p_{2}, p_{3}, ….. p_{n}.

Hence, p is one of p_{1}, p_{2}, p_{3}, …… p_{n}.

** Example 6: **To prove is irrational.

** Solution. **Let us assume that is rational.

Now, we can find integers r and such that .

Again let r and s have a common factor other than 1. Then we can divide by common factor to get, where a and b are coprime.

…….. (i)

[squaring both sides]

2b^{2} = a^{2}

2 divides a^{2}.

Now, by above theorem

2 divides a also.

So, we can write a = 2c for some integer c.

[From (i)]

2b^{2} = 4c^{2} [Squaring both sides]

b^{2} = 2c^{2} ……. (ii)

i.e., 2 divides b^{2}

So, 2 divides b [by using above theorem with p = 2]

Again, a and b have at least 2 as a common factor.

But this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that is rational.

Hence, we conclude that is irrational.