```Issue #18018 has been updated by sawa (Tsuyoshi Sawada).

With:

> g's decimal string representation has at most n digits

I think you mean:

> g's decimal string representation has at most n decimal (or fractional) d=
igits

(By the way, note here that the word "decimal" is used twice under differen=
t meanings: 1. base ten, and 2. the fractional part)

----------------------------------------
Bug #18018: Float#floor / truncate sometimes result that is too small.
https://bugs.ruby-lang.org/issues/18018#change-92728

* Author: marcandre (Marc-Andre Lafortune)
* Status: Open
* Priority: Normal
* Target version: 3.1
* Backport: 2.6: UNKNOWN, 2.7: UNKNOWN, 3.0: UNKNOWN
----------------------------------------

```ruby
291.4.floor(1) # =3D> 291.4 (ok)
291.4.floor(2) # =3D> 291.39 (not ok)
291.4.floor(3) # =3D> 291.4 (ok)
291.4.floor(4) # =3D> 291.4 (ok)
291.4.floor(5) # =3D> 291.39999 (not ok)
291.4.floor(6) # =3D> 291.4 (ok)
```

`g =3D f.floor(n)`, for `n > 0` must return the highest float that has the =
correct properties:
* `g` <=3D `f`
* `g`'s decimal string representation has at most `n` digits

I'll note that `floor` should be stable, i.e. `f.floor(n).floor(n) =3D=3D f=
.floor(n)` for all `f` and `n`.

Same idea for `truncate`, except for negative numbers (where `(-f).truncate=
(n) =3D=3D -(f.floor(n))` for positive `f`).

Noticed by Eust=E1quio Rangel but posted on the mailing list.

Please do not reply that I need to learn how floats work. Note that example=
given in doc `(0.3/0.1).floor =3D=3D 2` is not this issue, since `0.3/0.1 =
#=3D> 2.9999999999999996`

-- =

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