Why is a==b False?
Please see the following code. Why is a==b False? Thanks!
Input:
a=n(1/2*sqrt(5) - 1/2,digits=15)
b=n(1-a^2,digits=15)
print a
print b
a==b
Output:
0.618033988749895
0.618033988749895
False
Please see the following code. Why is a==b False? Thanks!
Input:
a=n(1/2*sqrt(5) - 1/2,digits=15)
b=n(1-a^2,digits=15)
print a
print b
a==b
Output:
0.618033988749895
0.618033988749895
False
This comes from working with numerical approximations.
In the computer, a
and b
are represented in binary.
They are very close but not equal.
sage: a.sign_mantissa_exponent()
(1, 11133510565745312, -54)
sage: b.sign_mantissa_exponent()
(1, 11133510565745310, -54)
But the difference is very small and their decimal expansions to 15 digits coincide.
Sage lets you work with exact algebraic numbers, either by creating a number field:
sage: K.<a> = NumberField(x^2 + x - 1, embedding=0.6)
sage: a.numerical_approx()
0.618033988749895
sage: 1 - a^2
a
or by using the field of algebraic number (QQbar
in Sage):
sage: a = QQbar(1/2*sqrt(5) - 1/2)
sage: a
0.618033988749895?
sage: 1 - a^2 == a
True
sage: a.minpoly()
x^2 + x - 1
sage: a.numerical_approx()
0.618033988749895
sage: a.radical_expression()
1/2*sqrt(5) - 1/2
Note that even without defining the corresponding number field or working in QQbar
, SageMath correctly finds that a == b
if you are not working with approximations but with exact symbolic expressions¹:
sage: a = 1/2*sqrt(5) - 1/2
sage: b = 1 - a^2
sage: bool(a == b)
True
¹ Note the necessity of calling bool(a==b)
since a == b
is kept as an expression by default, and one has to explicitly asks whether this equality holds.
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Asked: 2017-12-09 00:45:51 +0200
Seen: 316 times
Last updated: Dec 09 '17
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