# Revision history [back]

When you write sqrt(2.), you define a floating-point approximation of $sqrt(2.)$ with 53 bits of precision. So, since $sqrt(2.)$ is not rational (hence not of the form p/2^53), there is a rounding, so there is no reason that at the end of the computation you will get an exact result. However, you can see that the error is very small:

sage: two = 2.
sage: two.parent()
Real Field with 53 bits of precision
sage: sqrt(2)
sqrt(2)
sage: sqrt(2).parent()
KeyboardInterrupt
sage: two.parent()
Real Field with 53 bits of precision
sage: two = 2.
sage: two.parent()
Real Field with 53 bits of precision
sage: sqrt(two)
1.41421356237310
sage: sqrt(two).parent()
Real Field with 53 bits of precision
sage: sqrt(two)*sqrt(two)/2
1.00000000000000
sage: sqrt(two)*sqrt(two)/2 - 1
2.22044604925031e-16


If you want exact computation, you can define 2 as an integer or as an algebraic number:

sage: two = 2
sage: two.parent()
Integer Ring
sage: sqrt(two)
sqrt(2)
sage: sqrt(two).parent()
Symbolic Ring
sage: sqrt(two)*sqrt(two)/2
1
sage: sqrt(two)*sqrt(two)/2 == 1
1 == 1
sage: bool(sqrt(two)*sqrt(two)/2 == 1)
True

sage: two = AA(2)
sage: two.parent()
Algebraic Real Field
sage: sqrt(two)
1.414213562373095?
sage: sqrt(two).parent()
Algebraic Real Field
sage: sqrt(two)*sqrt(two)/2
1.000000000000000?
sage: sqrt(two)*sqrt(two)/2 == 1
True


If you still want to work with floating-point numbers, but want a guarantee on the error, rou can work with real intervals:

sage: two = RIF(2)
sage: two.parent()
Real Interval Field with 53 bits of precision
sage: sqrt(two).parent()
Real Interval Field with 53 bits of precision
sage: sqrt(two)*sqrt(two)/2
1.000000000000000?
sage: sqrt(two)*sqrt(two)/2 == 1
False
sage: 1 in sqrt(two)*sqrt(two)/2
True
sage: (sqrt(two)*sqrt(two)/2).endpoints()
(0.999999999999999, 1.00000000000001)


Iy tyou want to increase the precision with more bits you can do:

sage: two = RealField(100)(2)
sage: two
2.0000000000000000000000000000
sage: two.parent()
Real Field with 100 bits of precision
sage: sqrt(two).parent()
Real Field with 100 bits of precision

sage: two = RealIntervalField(100)(2)
sage: two.parent()
Real Interval Field with 100 bits of precision
sage: sqrt(two).parent()
Real Interval Field with 100 bits of precision


When you write sqrt(2.), you define a floating-point approximation of $sqrt(2.)$ with 53 bits of precision. So, since $sqrt(2.)$ is not rational (hence not of the form p/2^53), there is a rounding, so there is no reason that at the end of the computation you will get an exact result. However, you can see that the error is very small:

sage: two = 2.
sage: two.parent()
Real Field with 53 bits of precision
sage: sqrt(2)
sqrt(2)
sage: sqrt(2).parent()
KeyboardInterrupt
sage: two.parent()
Real Field with 53 bits of precision
sage: two = 2.
sage: two.parent()
Real Field with 53 bits of precision
sage: sqrt(two)
1.41421356237310
sage: sqrt(two).parent()
Real Field with 53 bits of precision
sage: sqrt(two)*sqrt(two)/2
1.00000000000000
sage: sqrt(two)*sqrt(two)/2 - 1
2.22044604925031e-16


If you want exact computation, you can define 2 as an integer integer:

sage: two = 2
sage: two.parent()
Integer Ring
sage: sqrt(two)
sqrt(2)
sage: sqrt(two).parent()
Symbolic Ring
sage: sqrt(two)*sqrt(two)/2
1
sage: sqrt(two)*sqrt(two)/2 == 1
1 == 1
sage: bool(sqrt(two)*sqrt(two)/2 == 1)
True


or as an algebraic number:

sage: two = 2
sage: two.parent()
Integer Ring
sage: sqrt(two)
sqrt(2)
sage: sqrt(two).parent()
Symbolic Ring
sage: sqrt(two)*sqrt(two)/2
1
sage: sqrt(two)*sqrt(two)/2 == 1
1 == 1
sage: bool(sqrt(two)*sqrt(two)/2 == 1)
True

sage: two = AA(2)
sage: two.parent()
Algebraic Real Field
sage: sqrt(two)
1.414213562373095?
sage: sqrt(two).parent()
Algebraic Real Field
sage: sqrt(two)*sqrt(two)/2
1.000000000000000?
sage: sqrt(two)*sqrt(two)/2 == 1
True


If you still want to work with floating-point numbers, but want a guarantee on the error, rou can work with real intervals:

sage: two = RIF(2)
sage: two.parent()
Real Interval Field with 53 bits of precision
sage: sqrt(two).parent()
Real Interval Field with 53 bits of precision
sage: sqrt(two)*sqrt(two)/2
1.000000000000000?
sage: sqrt(two)*sqrt(two)/2 == 1
False
sage: 1 in sqrt(two)*sqrt(two)/2
True
sage: (sqrt(two)*sqrt(two)/2).endpoints()
(0.999999999999999, 1.00000000000001)


Iy tyou want to increase the precision with more bits you can do:

sage: two = RealField(100)(2)
sage: two
2.0000000000000000000000000000
sage: two.parent()
Real Field with 100 bits of precision
sage: sqrt(two).parent()
Real Field with 100 bits of precision

sage: two = RealIntervalField(100)(2)
sage: two.parent()
Real Interval Field with 100 bits of precision
sage: sqrt(two).parent()
Real Interval Field with 100 bits of precision


When you write sqrt(2.), you define a floating-point approximation of $sqrt(2.)$ with 53 bits of precision. So, since $sqrt(2.)$ is not rational (hence not of the form p/2^53), there is a rounding, so there is no reason that at the end of the computation you will get an exact result. However, you can see that the error is very small:

sage: two = 2.
sage: two.parent()
Real Field with 53 bits of precision
sage: sqrt(2)
sqrt(2)
sage: sqrt(2).parent()
KeyboardInterrupt
sage: two.parent()
Real Field with 53 bits of precision
sage: two = 2.
sage: two.parent()
Real Field with 53 bits of precision
sage: sqrt(two)
1.41421356237310
sage: sqrt(two).parent()
Real Field with 53 bits of precision
sage: sqrt(two)*sqrt(two)/2
1.00000000000000
sage: sqrt(two)*sqrt(two)/2 - 1
2.22044604925031e-16


If you want exact computation, you can define 2 as an integer:

sage: two = 2
sage: two.parent()
Integer Ring
sage: sqrt(two)
sqrt(2)
sage: sqrt(two).parent()
Symbolic Ring
sage: sqrt(two)*sqrt(two)/2
1
sage: sqrt(two)*sqrt(two)/2 == 1
1 == 1
sage: bool(sqrt(two)*sqrt(two)/2 == 1)
True


or as an algebraic number:

sage: two = AA(2)
sage: two.parent()
Algebraic Real Field
sage: sqrt(two)
1.414213562373095?
sage: sqrt(two).parent()
Algebraic Real Field
sage: sqrt(two)*sqrt(two)/2
1.000000000000000?
sage: sqrt(two)*sqrt(two)/2 == 1
True


If you still want to work with floating-point numbers, but want a guarantee on the error, rou can work with real intervals:

sage: two = RIF(2)
sage: two.parent()
Real Interval Field with 53 bits of precision
sage: sqrt(two).parent()
Real Interval Field with 53 bits of precision
sage: sqrt(two)*sqrt(two)/2
1.000000000000000?
sage: sqrt(two)*sqrt(two)/2 == 1
False
sage: 1 in sqrt(two)*sqrt(two)/2
True
sage: (sqrt(two)*sqrt(two)/2).endpoints()
(0.999999999999999, 1.00000000000001)


Iy tyou want to increase the precision with more bits you can do:

sage: two = RealField(100)(2)
sage: two
2.0000000000000000000000000000
sage: two.parent()
Real Field with 100 bits of precision
sage: sqrt(two).parent()
Real Field with 100 bits of precision

sage: two = RealIntervalField(100)(2)
sage: two.parent()
Real Interval Field with 100 bits of precision
sage: sqrt(two).parent()
Real Interval Field with 100 bits of precision


When you write sqrt(2.), you define a floating-point approximation of $sqrt(2.)$ with 53 bits of precision. So, since $sqrt(2.)$ is not rational (hence not of the form p/2^53), there is a rounding, so there is no reason that at the end of the computation you will get an exact result. However, you can see that the error is very small:

sage: two.parent()
Real Field with 53 bits of precision
sage: two = 2.
sage: two.parent()
Real Field with 53 bits of precision
sage: sqrt(two)
1.41421356237310
sage: sqrt(two).parent()
Real Field with 53 bits of precision
sage: sqrt(two)*sqrt(two)/2
1.00000000000000
sage: sqrt(two)*sqrt(two)/2 - 1
2.22044604925031e-16


If you want exact computation, you can define 2 as an integer:

sage: two = 2
sage: two.parent()
Integer Ring
sage: sqrt(two)
sqrt(2)
sage: sqrt(two).parent()
Symbolic Ring
sage: sqrt(two)*sqrt(two)/2
1
sage: sqrt(two)*sqrt(two)/2 == 1
1 == 1
sage: bool(sqrt(two)*sqrt(two)/2 == 1)
True


or as an algebraic number:

sage: two = AA(2)
sage: two.parent()
Algebraic Real Field
sage: sqrt(two)
1.414213562373095?
sage: sqrt(two).parent()
Algebraic Real Field
sage: sqrt(two)*sqrt(two)/2
1.000000000000000?
sage: sqrt(two)*sqrt(two)/2 == 1
True


If you still want to work with floating-point numbers, but want a guarantee on the error, rou can work with real intervals:

sage: two = RIF(2)
sage: two.parent()
Real Interval Field with 53 bits of precision
sage: sqrt(two).parent()
Real Interval Field with 53 bits of precision
sage: sqrt(two)*sqrt(two)/2
1.000000000000000?
sage: sqrt(two)*sqrt(two)/2 == 1
False
sage: 1 in sqrt(two)*sqrt(two)/2
True
sage: (sqrt(two)*sqrt(two)/2).endpoints()
(0.999999999999999, 1.00000000000001)


Iy tyou If you want to increase the precision with more bits you can do:

sage: two = RealField(100)(2)
sage: two
2.0000000000000000000000000000
sage: two.parent()
Real Field with 100 bits of precision
sage: sqrt(two).parent()
Real Field with 100 bits of precision

sage: two = RealIntervalField(100)(2)
sage: two.parent()
Real Interval Field with 100 bits of precision
sage: sqrt(two).parent()
Real Interval Field with 100 bits of precision

 5 No.5 Revision slelievre 8962 ●6 ●91 ●175 http://carva.org/samue...

When you write sqrt(2.), you define a floating-point approximation of $sqrt(2.)$ $\sqrt{2}$ with 53 bits of precision. So, since $sqrt(2.)$ $\sqrt{2}$ is not rational (hence not of the form p/2^53), there is a rounding, so there is no reason that at the end of the computation you will get an exact result. However, you can see that the error is very small:

sage: two.parent()
Real Field with 53 bits of precision
sage: two = 2.
sage: two.parent()
Real Field with 53 bits of precision
sage: sqrt(two)
1.41421356237310
sage: sqrt(two).parent()
Real Field with 53 bits of precision
sage: sqrt(two)*sqrt(two)/2
1.00000000000000
sage: sqrt(two)*sqrt(two)/2 - 1
2.22044604925031e-16


If you want exact computation, you can define 2 as an integer:

sage: two = 2
sage: two.parent()
Integer Ring
sage: sqrt(two)
sqrt(2)
sage: sqrt(two).parent()
Symbolic Ring
sage: sqrt(two)*sqrt(two)/2
1
sage: sqrt(two)*sqrt(two)/2 == 1
1 == 1
sage: bool(sqrt(two)*sqrt(two)/2 == 1)
True


or as an algebraic number:

sage: two = AA(2)
sage: two.parent()
Algebraic Real Field
sage: sqrt(two)
1.414213562373095?
sage: sqrt(two).parent()
Algebraic Real Field
sage: sqrt(two)*sqrt(two)/2
1.000000000000000?
sage: sqrt(two)*sqrt(two)/2 == 1
True


If you still want to work with floating-point numbers, but want a guarantee guarantee on the error, rou you can work with real intervals:

sage: two = RIF(2)
sage: two.parent()
Real Interval Field with 53 bits of precision
sage: sqrt(two).parent()
Real Interval Field with 53 bits of precision
sage: sqrt(two)*sqrt(two)/2
1.000000000000000?
sage: sqrt(two)*sqrt(two)/2 == 1
False
sage: 1 in sqrt(two)*sqrt(two)/2
True
sage: (sqrt(two)*sqrt(two)/2).endpoints()
(0.999999999999999, 1.00000000000001)


If you want to increase the precision with more bits you can do:

sage: two = RealField(100)(2)
sage: two
2.0000000000000000000000000000
sage: two.parent()
Real Field with 100 bits of precision
sage: sqrt(two).parent()
Real Field with 100 bits of precision

sage: two = RealIntervalField(100)(2)
sage: two.parent()
Real Interval Field with 100 bits of precision
sage: sqrt(two).parent()
Real Interval Field with 100 bits of precision