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)
=> x² 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)
=> x² 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
=> y² 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