# rounding error .n() and N()

Why does 1.414 - sqrt(2).n(digits=4) not evaluate to zero?

sage: sqrt(2).n(digits=4)

1.414

sage: 1.414 - sqrt(2).n(digits=4)

-0.0002136

rounding error .n() and N()

Why does 1.414 - sqrt(2).n(digits=4) not evaluate to zero?

sage: sqrt(2).n(digits=4)

1.414

sage: 1.414 - sqrt(2).n(digits=4)

-0.0002136

add a comment

1

`sqrt(2).n(digits=4)`

returns a number which agrees with `sort(2)`

to 4 digits of precision, but it need not be obtained by truncating `sqrt(2)`

after 4 digits. I'm guessing that it's obtained by truncating `sqrt(2)`

after the necessary number of binary digits:

```
sage: a = sqrt(2).n(digits=4)
sage: a.n(digits=30)
1.41421508789062500000000000000
sage: b = sqrt(2).n(16)
sage: b.n(digits=30)
1.41421508789062500000000000000
sage: a == b
True
```

1

Hi,

This is because the two numbers do not live in the same ground field as you can see

```
sage: 1.414.parent()
Real Field with 53 bits of precision
sage: sqrt(2).n(digits=4).parent()
Real Field with 17 bits of precision
```

1.414 is not exactly represented in memory because it is not a diadic number. Two different diadic approximations with different precisions are different.

Vincent

Asked: **
2013-04-08 04:53:32 -0600
**

Seen: **168 times**

Last updated: **Apr 08 '13**

Precision plots - How do I do those?

Rounding entries of a random vector

Undeterministic numerical approximation

numerical approximations of complex_embedding

Difference between RealField and numerical_approx

4-Digit Rounding Arithmetic:System of Equations

Copyright Sage, 2010. Some rights reserved under creative commons license. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license.