OCA certification and remainder function

Question

Hello, I'm studying for the OCA certification and need help to understand how the remainder function works.
I read that (http://www.techonthenet.com/oracle/functions/remainder.php) the remainder(m,n) is calculated as: m - (n * X) where X is the integer nearest m / n.
I ask: why remainder(12.5,5)=2.5 ?
According to me the result should be >>> remainder(12.5,5)=12.5-(5*[the integer nearest 12.5/5])=12.5-(5*3)=-2.5.
Thanks for your replies. Regards.

Answer 1

odie_63 wrote:
Hi,
I read that (http://www.techonthenet.com/oracle/functions/remainder.php) the remainder(m,n) is calculated as: m - (n * X) where X is the integer nearest m / n.
Actually X = [m/n] = floor(m / n), i.e. the integer nearest m/n rounded down.
So remainder(12.5, 5) = 12.5 - 5*2 = 2.5You sure about that?
SQL> select 12.6/5, floor(12.6/5) from dual;
    12.6/5 FLOOR(12.6/5)
      2.52             2
SQL> select 12.6-(5*2) from dual;
12.6-(5*2)
       2.6
SQL> select remainder(12.6,5) from dual;
REMAINDER(12.6,5)
             -2.4
SQL>Looks like the rounding is a little odd. Typically rounding goes upwards when it get's to .5 of a number, but remainder takes .5 downwards and anything above that upwards...
SQL> select remainder(12.50000000000001,5) from dual;
REMAINDER(12.50000000000001,5)
                          -2.5
SQL>It does seem a little odd behaviour.
Me, I just prefer using MOD(m,n), cos then I know what I'm getting. ;)