sage: def f(n):
... if abs(n-n.integer_part())<1*e-10:
... return n.integer_part()
... else:
... return n
sage: n=5.0000000000000001
sage: f(n)
5.000000000000000
sage: type(f(n))
<type 'sage.rings.real_mpfr.RealLiteral'>
sage: n.integer_part()
5
sage: n.integer_part()==5
True
sage: f(n)==5
False
The fact that f(n) shows 5.000000000000000 is just a printing artifact, there is no rounding, you can check that f(n) is equal to n and different from 5:
sage: n == f(n)
True
Even n itself is represented as 5.000000000000000, not 5.0000000000000001:
sage: n=5.0000000000000001
sage: n
5.000000000000000
sage: n-5
1.110223024625157e-16
EDIT Here is a way to print more digits of n
sage: n.str(truncate=False)
'5.000000000000000111'
You might be scary with the two additional 1 at the end. This is because internally, numbers are represented in binary, and you proposed a number which is exact in base 10, but not in base 2, hence some rounding (not only in the printing, but also in the representation of n in the machine).
You can get the exact value of the rounded n as follows:
sage: n.exact_rational()
45035996273704961/9007199254740992
And even see its internal representation:
sage: n.sign_mantissa_exponent()
(1, 90071992547409922, -54)
And check
sage: 90071992547409922*2^(-54)
45035996273704961/9007199254740992
You can print the non-truncated expression of n as follows:
Asked: 2016-01-12 16:20:28 +0200
Seen: 498 times
Last updated: Jan 15 '16
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Did you mean 1e-10 instead of 1*e-10?