# 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
```

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Asked: ** 2013-04-08 11:53:32 +0100 **

Seen: **224 times**

Last updated: **Apr 08 '13**

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