Question

x= float(raw_input("enter a number: "))
ans =0
while ans*ans*ans < abs(x):
ans=ans+.01

if ans*ans*ans!= abs(x):
print x, ' is not a perfect cube'
else:
if x<0:
ans=-ans
print "the cube of "+str(x)+' is '+str(ans)
## can anyone please let me know what have i done wrong and how to fix it. I replaced 1 with .01 for the incremental. now even if i enter a number that has a perfect cube, the program cannot identify it. i still want to use .01 incremental value

Answer 1

There you go!

x = float(input("enter a number: "))
ans = 0
while (ans*ans*ans) < abs(x): ans= ans+.01
ans = round(ans, 1)
if (ans*ans*ans) == x : print('the cube of ' + str(x) + ' is ' + str(ans))
else: print(str(x)+ ' is not a perfect cube')

Answer 2

val = float(raw_input("Enter a number: "))
ans = 0
while (ans**3) < abs(val):
ans = ans + 0.01
print ans, 'is ans'
# by adding this print statement you can see the value of ans each time the loop
# executes. If you use the the value 9 as an input, you'll see that the last value of ans
# is 2.09. But 2.09 * 2.0 9 * 2.09 = 9.129...... Since this value does not == 9. Then,
# the program outputs that 9 is not a perfect square.

if ans**3 != abs(val):
print val, 'is not a perfect cube'
else:
if val < 0:
ans = -ans
print 'the cube of', val, 'is', ans

# the solution to this problem is addressed by comparing the resulting 9.129... value
# with the inputted value of 9 using an epsilon value. The comparison is: if 9.129 is
# close enough to 9, then it (9) is a perfect square. This concept is covered in
# lecture 3.