Question

Prove that root 8 is an irrational no.

Answer 1

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.

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Answer 2

Answer:

Let be assumed that √8 is rational no.

then it can be written in form of..

p/q

where q is not equals to 0

take any integer b..

a/b =√8

a=√8b

squaring both side..

a2= 8b2

then ,

if a2 divides 8 then a will also divides √8 ..

so it will rational no.

but we assumed it wrong it will irrational no.

so, √8 will be irrational no.