determine whether sqrt 8 is rational or irrational?

Answer 1

By itself - essentially it is an irrational number

sqrt8 --- 2 sqrt 2 ---- sqrt 2 is an irrational number (multiplying it by 2 makes no difference)

An essential characteristic of differentiating irrational and rational numbers are that irrational numbers don't seem to have an end....

Answer 2

sqrt of 8 = 2 sqrt of 2 irrational

Answer 3

It's equal to 2*√2 and √2 is irrational, so √8 is also
irrational.

Answer 4

For any whole number A, unless A is a perfect square (i.e. unless A = B², with B a whole number also), √A is irrational.

8 = 2*2*2 is not a perfect square, so √8 is irrational.

Answer 5

suppose sqrt 8 is rational
then sqrt 8 = a/b , a,b=integer b is not 0 GCD(a,b)=1
square ; 8 = a^2/b^2
a^2 = 8*b^2
a^2 is an even number
if a is odd , a^2 is odd too.
therefore , a is even
so a = 2k , k=integer
a^2 = (2k)^2 =4k^2 = 8*b^2
k^2 = 2*b^2
in the same way, k is even
so k = 2n , n=integer
a^2 = 4k^2=4(2n)^2=4(4n^2)=16n^2 = 8b^2
therefore , 2*n^2=b^2
in the same way, b is even
so both a and b are even
therefore , GCD(a,b) is not 1
but GCD(a,b)=1
therefore, sqrt 8 is not rational
so sqrt 8 is irrational

Answer 6

It is irrational because the sqrt(8)=2(sqrt(2)).
the sqrt(2) is irrational

Answer 7

For an expression ax^2+bx+c, the discriminant is b^2-4ac. b^2-4ac > 0 => 2 different rational suggestions b^2-4ac = 0 => precisely one rational answer b^2-4ac < 0 => 2 different irrational suggestions right here b^2-4ac = (-12)^2 - 4*a million*34 = a hundred and forty four - 136 = 8 that's greater advantageous than 0 for this reason this equation has 2 different rational suggestions.

Answer 8

8 is not a perfect square; so the square root of 8 must be irrational

Answer 9

8 is not a perfect square so it would be irrational