Question

Show that 2√2 is an irrational number​

Answer 1

Answer:

LET US ASSUME THAT 2√2 IS A RATIONAL NUMBER.

Step-by-step explanation:

THEN, 2√2 = a/b {where a and b are co-prime positive integers}

2√2 = a/b

a/2b = √2

now, a and b are integers, therefore, a/2b must be a rational number.

so, √2 should also be a rational number.

but it is not possible because √2 is an irrational number.

therefore, our assumption was wrong,

2√2 is an irrational number.

HENCE, PROVED....

HOPE IT HELPS....

PLEASE MARK BRAINIEST...

Answer 2

Step-by-step explanation:

let us assume to the contrary that √2 is a rational number, then there exist co-prime numbers a and b, such that

√2=a/b

squaring on both side

(√2) = a² / b²

then root and square will cancel

2= a²/ b²

a²=2b²

=> 2 divides a² => 2 divides a

a²=2c

sub in (i)

(2c)²=2b²

4c²=2b²

2c²=b²

=> 2 divides b²=> 2 divides b

so, 2 is common factor of both a and b which is contraction.

therefore our assumption wrong

hence √2 is an irrational number

2+√2= a/b

√2= a/b - 2

√2 = a-2b/b

Ans : a-2b / b is also an irrational number