# 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

0

`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
```

0

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 -0500
**

Seen: **82 times**

Last updated: **Apr 08 '13**

Change output to numerical values globally

Why is 3e1 not equivalent to 30?

HowTo Compute Past Largest Cython Supported Wordsize (efficiently)?

How to keep my real numbers rounded

adding real literal and real number of high precision

Incorrect result for comparison (precision issues?)

Errors when plotting zeta function parametrically

Unable to Simplify to Float Approximation for a Numerical Integral

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.