Question 338587: If 0 < n < 1, which of the following gives the correct ordering of
√n, n, and, n^2 ?
A. √n < n < n^2
b √n < n^2 < n
c n <√n < n^2
d n < n^2 < √n
e n^2 < n < √n
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! If 0 < n < 1, which of the following gives the correct ordering of
√n, n, and, n^2 ?
A. √n < n < n^2
b √n < n^2 < n
c n <√n < n^2
d n < n^2 < √n
e n^2 < n < √n
√n = square root (sqrt) of n, n, and n^2, 0 < n < 1
n = 0 --> sqrt(n) = 0, n^2 = 0, the values are all 0 and all equal
n = 1 --> sqrt(n) = 1, n^2 = 1, the values are all 1 and all equal
n = 0 and n = 1 are the end points of the interval (0,1) and are not included
checking 2 values of n
n= 0.9 which is close to 1
and n=10^(-6) which is close to 0
n = 0.9 = 9/10
sqrt(n) = 3/(sqrt(10)) = (3sqrt(10))/10 = 0.9487 to 4 places
n^2 = 81/100 = 0.81
order: n^2, n, sqrt(n)
n^2 < n < sqrt(n)
n = 0.000001 = 1/1000000 = 1/(10^6) = 10^(-6)
sqrt(n) = 10^(-3) = 0.001
n^2 = 10^(-12) = 0.000000000001
order: n^2, n, sqrt(n)
n^2 < n < sqrt(n)
e n^2 < n < √n is the answer when 0 < n < 1
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