Question 338587
If 0 < n < 1, which of the following gives the correct ordering of
&#8730;n, n, and, n^2 ? 
A. &#8730;n < n < n^2
b &#8730;n < n^2 < n
c n <&#8730;n < n^2
d n < n^2 < &#8730;n 
e n^2 < n < &#8730;n 



&#8730;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 < &#8730;n    is the answer when 0 < n < 1