Question

Q16. Prove root8 is an irrational number?​

Answer 1

Answer:

Heya user!!

Here we begin

you would go with a proof by contradiction:

suppose √8 = a/b with integers a, b

and gcd(a,b) = 1 (meaning the ratio is simplified)

then 8 = a²/b²

and 8b² = a²

this implies 8 divides a² which also means 8 divides a.

so there exists a p within the integers such that:

a = 8p

and thus,

√8 = 8p/b

which implies

8 = 64p²/b²

which is:

1/8 = p²/b²

or:

b²/p² = 8

which implies

b² = 8p²

which implies 8 divides b² which means 8 divides b.

8 divides a, and 8 divides b, which is a contradiction because gcd (a, b) = 1

therefore, the square root of 8 is irrational.

Hope you understand

Please make it a brainlist answer

Answer 2

Answer:

let us suppose root 8 be a rational no

~ a and b are the cofficients of root 8

now,

a/b= root 8

a = root 8× b

but a and b are the integers

there fore

integer = integer × integer

but this contradict the fact that root 8 is an irrational no.

therefore,root 8 is an irrational no.