# 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

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

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
```

Please start posting anonymously - your entry will be published after you log in or create a new account.

Asked: ** 2013-04-08 11:53:32 +0200 **

Seen: **265 times**

Last updated: **Apr 08 '13**

HowTo Compute Past Largest Cython Supported Wordsize (efficiently)?

Changing number of decimal digits displayed in output

Numerical approximation of symbolic equation

Set global precision for reals?

unable to numerically solve an equation (basic question)

converting symbolic solve output to numbers [was "numerical value"]

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.