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

[Edwin.Madari]

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

>>import math
num = 625
how_many_sq rt = 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

(0.200000000000 00001, 0.2000000000000 0001)

>>'%.2f' % a, '%.2f' % b

('0.20', '0.20')

>>

thx. Edwin

-----Original Message-----
From: py************* *************** *************** *******@python. org
[mailto:py****** *************** *************** **************@ python.org]
On Behalf Of CNiall
Sent: Sunday, August 03, 2008 10:03 AM
To: py*********@pyt hon.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

0.2000000000000 0001

>>0.33

0.3300000000000 0002

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 0.2000000000000 0001?

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