# How to Get a Remainder Using MOD() in PostgreSQL, MS SQL Server, and MySQL

Database:

Operators:

#### Problem:

You want to find the (non-negative) remainder.

#### Example:

In the table

, you have two columns of integers: **numbers**`a`

and `b`

.

a | b |
---|---|

9 | 3 |

5 | 3 |

2 | 3 |

0 | 3 |

-2 | 3 |

-5 | 3 |

-9 | 3 |

5 | -3 |

-5 | -3 |

5 | 0 |

0 | 0 |

You want to compute the remainders from dividing `a`

by `b`

. Each remainder should be a non-negative integer value smaller than `b`

.

#### Solution 1 (not entirely correct):

SELECT a, b, MOD(a, b) AS remainder FROM numbers;

The result is:

a | b | remainder |
---|---|---|

9 | 3 | 0 |

5 | 3 | 2 |

2 | 3 | 2 |

0 | 3 | 0 |

-2 | 3 | -2 |

-5 | 3 | -2 |

-9 | 3 | 0 |

5 | -3 | 2 |

-5 | -3 | -2 |

5 | 0 | error |

0 | 0 | error |

#### Discussion:

This solution works correctly if a is non-negative. However, when it is negative, it doesn’t follow the mathematical definition of the remainder.

Conceptually, a remainder is what remains after an integer division of ```
a
```

by `b`

. Mathematically, a remainder of two integers is a
non-negative integer that is smaller than the divisor `b`

. More precisely,
it is a number r∈{0,1,...,*b* - 1} for which there exists some integer k such
that *a = k * b + r*. E.g.:

`5 = 1 * 3 + 2`

, so the remainder of 5 and 3 equals `2`

.

`9 = 3 * 3 + 0`

, so the remainder of 9 and 3 equals `0`

.

`5 = (-1) * (-3) + 2`

, so the remainder of 5 and -3 equals `2`

.

This is how `MOD(a, b)`

works for the non-negative dividends in the column `a`

. Obviously, an error is shown if the divisor `b`

is `0`

, because you can't divide by `0`

.

Getting the correct remainder is problematic when the dividend a is a negative number. Unfortunately, `MOD(a, b)`

can return a negative value when a is negative. E.g.:

`MOD(-2, 5)`

returns `-2`

when it should return `3`

.

`MOD(-5, -3)`

returns `-2`

when it should return `1`

.

#### Solution 2 (correct for all numbers):

SELECT a, b, CASE WHEN MOD(a, b) >= 0 THEN MOD(a, b) ELSE MOD(a, b) + ABS(b) END AS remainder FROM numbers;

The result is:

a | b | remainder |
---|---|---|

9 | 3 | 0 |

5 | 3 | 2 |

2 | 3 | 2 |

0 | 3 | 0 |

-2 | 3 | 1 |

-5 | 3 | 1 |

-9 | 3 | 0 |

5 | -3 | 2 |

-5 | -3 | 1 |

5 | 0 | error |

0 | 0 | error |

#### Discussion:

To compute the remainder of a division between **any** two integers (negative or non-negative), you can use the `CASE WHEN`

construction. When `MOD(a, b)`

is non-negative, the remainder is simply `MOD(a, b)`

. Otherwise, we have to correct the result returned by `MOD(a, b)`

.

How do you get the correct remainder when `MOD()`

returns a negative value? You should add the absolute value of the divisor to `MOD(a, b)`

. That is, make it `MOD(a, b) + ABS(b)`

:

`MOD(-2, 5)`

returns `-2`

when it should return `3`

. You can fix this by adding `5`

.

`MOD(-5, -3)`

returns `-2`

when it should return `1`

. You can fix this by adding `3`

.

When `MOD(a, b)`

returns a negative number, the `CASE WHEN`

result should be `MOD(a, b) + ABS(b)`

. This is how we get Solution 2. If you need a refresher on how the `ABS()`

function works, take a look at the cookbook How to compute an absolute value in SQL.

Of course, you still can't divide any number by `0`

. So, if `b = 0`

, you'll get an error.

#### Solution 3 (correct for all numbers):

SELECT a, b, MOD(a, b) + ABS(b) * (1 - SIGN(MOD(a, b) + 0.5)) / 2 AS remainder FROM numbers;

The result is:

a | b | remainder |
---|---|---|

9 | 3 | 0 |

5 | 3 | 2 |

2 | 3 | 2 |

0 | 3 | 0 |

-2 | 3 | 1 |

-5 | 3 | 1 |

-9 | 3 | 0 |

5 | -3 | 2 |

-5 | -3 | 1 |

5 | 0 | error |

0 | 0 | error |

#### Discussion:

There is another way to solve this problem. Instead of a `CASE WHEN`

, use a more complex one-line mathematical formula:

`MOD(a, b) + ABS(b) * (1 - SIGN(MOD(a, b) + 0.5)) / 2`

In Solution 2, `MOD(a, b) + ABS(b)`

was returned for cases when `MOD(a, b) < 0`

. Note that `MOD(a, b) + ABS(b) = MOD(a, b) + ABS(b) * 1 when MOD(a, b) < 0`

.

In contrast, you return `MOD(a, b)`

when `MOD(a, b) >= 0`

. Note that `MOD(a, b) = MOD(a, b) + ABS(b) * 0 when MOD(a, b) >= 0`

.

So, we can multiply `ABS(b)`

by an expression that equals 1 for a negative `MOD(a, b)`

and `0`

for a non-negative `MOD(a, b)`

. Since `MOD(a, b)`

is always an integer, the expression `MOD(a, b) + 0.5`

is always positive for `MOD(a, b) ≥ 0`

and negative for `MOD(a, b) < 0`

. You can use any positive number less than `1`

instead of `0.5`

.

The sign function `SIGN()`

returns `1`

if its argument is strictly positive, `-1`

if it is strictly negative, and `0`

if it equals `0`

. However, you need something that returns only `0`

and `1`

, not `1`

and `-1`

. Here is how you fix this:

`(1 - 1) / 2 = 0`

`(1 - (-1)) / 2 = 1`

Then, the correct expression by which you multiply `ABS(b)`

is:

`(1 - SIGN(MOD(a, b) + 0.5)) / 2`

So, the entire formula is:

`MOD(a, b) + ABS(b) * (1 - SIGN(MOD(a, b) + 0.5)) / 2`