Why does 10.0/3 return 3.3333333333333335?

Question

Why does 10.0/3 return 3.3333333333333335? It does not seem correct, does it? Is it a bug?

10.0/3
# => 3.3333333333333335

This isn't Ruby specific. Most languages map a float or double type onto the machine's native floats/doubles — John La Rooy, Mar 03 '16 at 06:16

score 1 · Answer 1 · answered Mar 03 '16 at 05:47

1

This is not a bug, this is how floating point operations work. One should not care about mantissa too much, while

10.0 / 3.0 * 3.0 == 10.0
#⇒ true

answered Mar 03 '16 at 05:47

Charles Owen · Answer 2 · 2016-03-03T06:39:55.453

0

Float objects represent inexact real numbers using the native architecture's double-precision floating point representation.

If you look at the Ruby docs, you'll see that float is double precision and is considered an approximate number.http://ruby-doc.org/core-1.9.3/Float.html

Floating point has a different arithmetic and is an inexact number. You should know its esoteric system. See the following:

You could perform a round on the division to get a more accurate answer.

(10.0/3).round(5)

edited Mar 03 '16 at 06:39

answered Mar 03 '16 at 05:47

Charles Owen

How can `round` give a precise answer? It cuts off information. – sawa Mar 03 '16 at 06:26
I believe the precision in the above case is only accurate up to 15 or 16 digits. That's why you see the trailing 5. – Charles Owen Mar 03 '16 at 06:39
The original had the error less than 10^(-15) * 2. Rounding it at a higher digit 10^(-5) makes the error even larger. – sawa Mar 03 '16 at 06:39
Not true. I get 0.33333. There are no 5s in the result – Charles Owen Mar 03 '16 at 12:58
Of course you get that result. Of course it is true. – sawa Mar 03 '16 at 13:28

2 Answers2