# CLASS-9SOME THEOREM IN IRRATIONAL NUMBER

Some Theorem In The Irrational Number

Theorem 1.) Prove that √2 is an irrational number.

Proof:-  If possible, let √2 be rational.

Lets its the simplest form be √2 = x/y, where ‘x’ & ‘y’ are integers having no common factor, other than 1 and y ≠ 0.

x

Then, √2 = ------

Y

Then, √2 y = x

Or,     x² = (√2 y)²

Or,     x² = 2y² …………………(1)

=> is even

=> x is even  [ so, only squares of even numbers are even]

Let x = 2m for some integer m

Then, x = 2m  =>  x² =  (2m)²  = 4m²

=>  2y² =  4m²

=>   y² = 2m²

=>  y²  is even

=>  y is even

Thus, ‘x’ is even and ‘y’ is even.

This shows that 2 is a common factor of ‘x’ and ‘y’.

This contradicts the hypothesis that ‘x’ and ‘y’ have no common factor, other than 1.

So, √2 is not rational and hence it is irrational.

Theorem 2.) Prove that √3 is irrational.

Proof.) If possible, let √3 be rational.

Let its simplest form be √3 = x / y, where ‘x’ & ‘y’ are integers, having no common factors, other than 1 and y ≠ 0

Then, √3 = x / y

Or, √3 y = x

Or,  x² = (√3 y)²

Or,   x² = 3y²…………………….(1)

=>  is a multiple of 3

=>  x is multiple of 3

Let, x = 3m for some positive integer m

Then, x = 3m =>  x² = (3m)² = 9m²

=> 3y² = 9m²         [from equation (1), x² = 3y²]

=>  y² = 3m²

=>  is a multiple of 3

=>  y is a multiple of 3

Thus, x as well as y is a multiple of 3,

This shows that 3 is a common factor of x & y. this contradicts the hypothesis that x & y have no common factor, other than 1.

So, √3 is not a rational number and hence it is irrational.

Similarly, we can prove that each of the numbers √5, √6, √7, √8, √10, √11, √12,………., etc. is irrational