# Decimals not equalling themselves (e.g. 0.2 = 0.2000000001)

Sun Aug 3 17:16:20 CEST 2008

```for nth square root: use math.sqrt n times for example
>>> 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: python-list-bounces+edwin.madari=verizonwireless.com at python.org
On Behalf Of CNiall
Sent: Sunday, August 03, 2008 10:03 AM
To: python-list at 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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