On 3 aug 2008, at 17.16, [EMAIL PROTECTED] wrote:

for nth square root: use math.sqrt n times for example

        Ehum. The OP wants to compute the nth root ( not the nth square root)

import math
num = 625
how_many_sqrt = 2
for i in range(how_many_sqrt):
..     num = math.sqrt(num)
..
num
5.0

all comparisons work fine for arbitrary floating point numbers...
For readability print them with required precision. for example
a = .2
b = .4
b = b/2
a == b
True
a, b
(0.20000000000000001, 0.20000000000000001)
'%.2f' % a, '%.2f' % b
('0.20', '0.20')


thx. Edwin

-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] ]
On Behalf Of CNiall
Sent: Sunday, August 03, 2008 10:03 AM
To: python-list@python.org
Subject: Decimals not equalling themselves (e.g. 0.2 = 0.2000000001)


I am very new to Python (I started learning it just yesterday), but I
have encountered a problem.

I want to make a simple script that calculates the n-th root of a given
number (e.g. 4th root of 625--obviously five, but it's just an example
:P), and because there is no nth-root function in Python I will do this
with something like x**(1/n).

However, with some, but not all, decimals, they do not seem to 'equal
themselves'. This is probably a bad way of expressing what I mean, so
I'll give an example:
0.5
0.5
0.25
0.25
0.125
0.125
0.2
0.20000000000000001
0.33
0.33000000000000002

As you can see, the last two decimals are very slightly inaccurate.
However, it appears that when n in 1/n is a power of two, the decimal
does not get 'thrown off'. How might I make Python recognise 0.2 as 0.2
and not 0.20000000000000001?

This discrepancy is very minor, but it makes the whole n-th root
calculator inaccurate. :\
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