Mod Function and Negative Numbers

Date: 04/28/2000 at 11:17:09

From: Anne

Subject: Using the mod() function with negative numbers

I work in IT - Technical Support. I am trying to sort out a problem

for a user who is using the MOD() function in Excel 97 as follows:

=MOD(-340,60)

He thinks it should return a value of 40, but it returns a value of

20.

It returns a value of 40 if we do the following:

=MOD(340,60)

Now I am wondering why the negative sign makes a difference. You can

see from my age that it was years ago that I had math, and I was very

good at it. But I cannot remember how negative numbers work in

division.

Can you help me?

Anne

Date: 04/28/2000 at 12:57:10

From: Doctor Peterson

Subject: Re: Using the mod() function with negative numbers

Hi, Anne.

There are different ways of thinking about remainders when you deal

with negative numbers, and he is probably confusing two of them. The

mod function is defined as the amount by which a number exceeds the

largest integer multiple of the divisor that is not greater than that

number. In this case, -340 lies between -360 and -300, so -360 is the

greatest multiple LESS than -340; we subtract 60 * -6 = -360 from -340

and get 20:

-420 -360 -300 -240 -180 -120 -60 0 60 120 180 240 300 360

--+----+----+----+----+----+----+----+----+----+----+----+----+----+--

| | | |

-360| |-340 300| |340

|=| |==|

20 40

Working with a positive number like 340, the multiple we subtract is

smaller in absolute value, giving us 40; but with negative numbers, we

subtract a number with a LARGER absolute value, so that the mod

function returns a positive value. This is not always what people

expect, but it is consistent.

If you want the remainder, ignoring the sign, you have to take the

absolute value before using the mod function.

- Doctor Peterson, The Math Forum

http://mathforum.org/dr.math/

Date: 05/02/2000 at 04:58:50

From: Norwich C.C. Network Services

Subject: Re: Using the mod() function with negative numbers

Thanks very much for sorting this out. I had worked out how it was

arriving at the result, but did not know whether it was correct. We

had also tried it in Lotus 1-2-3 and the result given was -40, which

was what the user expected. Nice to have built some user confidence in

Excel.

Thanks again for your help.

Anne

Date: 05/02/2000 at 17:16:19

From: Doctor Peterson

Subject: Re: Using the mod() function with negative numbers

Hi, Anne.

It occurs to me that I don't quite want to leave you trusting Excel

too much. A comment I had considered making before was that a program

may often use its own definitions for functions like this, and

apparently Lotus proves my point that "mod" can be taken in different

ways. You should always go by the manual, not by what a mathematician

says.

Excel has its own share of mathematical foibles; here's a copy of an

answer I gave to someone who asked about its handling of -a^b:

===========================================================

The proper rule is that negation has the same precedence as

multiplication and division. After all, negation means multiplication

by -1. So -a^b should be taken as -(a^b).

Some time ago when I was researching order of operations for another

inquiry, I ran across this page from Microsoft explaining this quirk:

XL: Order of Precedence Causes Unexpected Positive Value

http://support.microsoft.com/support/kb/articles/Q132/6/86.asp

It doesn't claim that their rule agrees with the mathematical world,

only that they make their own standards, and don't have to agree with

anyone:

This behavior is by design of Microsoft Excel. Microsoft Excel uses

an order of calculation to evaluate operators in formulas. The order

of evaluation of operators dictates that a minus sign (-) used as a

negation operator (such as -1) is evaluated before all other

operators. Because of this order, the formula =-1^2 represents the

value -1 squared, and returns the value 1, a positive value.

...

Note that this has been the standard method for evaluating formulas

since the first version of Microsoft Excel.

NOTE: This order of operation is different from the order of

operation in Lotus 1-2-3.

What I suspect is that the programmers were accustomed to the C

language, in which unary operators such as negation have higher

precedence than any binary operator; there is no exponent operator in

C. When they added the exponent operator, they may simply have

forgotten that it should have higher precedence, or they may have

found it was easier to program this way. Once the rule became

established, they couldn't change it and make customer's programs

fail.

There are many other instances where calculators or software make

their own rules; sometimes this is because of limitations of their

interface, sometimes because of a misguided desire to "improve" the

rules. In any case, we can't use any piece of software as our guide to

mathematical practice, and students should be taught not to confuse

the rules of a particular program with those of math. We wouldn't want

Microsoft to be making these decisions for us anyway, would we?

===========================================================

So the score is Microsoft 1, Lotus 1!

- Doctor Peterson, The Math Forum

http://mathforum.org/dr.math/

Date: 05/03/2000 at 08:53:33

From: Norwich C.C. Network Services

Subject: Re: using the mod() function with negative numbers

Thanks for that!

I will remember not to trust Excel too much.

Anne

Date: 07/05/2001 at 13:58:23

From: A. H. Banen

Subject: Mod Function and Negative Numbers

Dear Dr. Peterson,

Concerning the topic "Mod Function and Negative Numbers," I found it

strange you did not refer to the simple rule concerning integer

division and modulo calculation: Given the integer numbers A and B

(where B cannot be equal to zero), the following holds:

A = ( A DIV B ) * B + A MOD B

e.g. for A = 340 and B = 60

A = ( 340 DIV 60 ) * 60 + ( 340 MOD 60 ) =>

A = ( 5 ) * 60 + 40 =>

A = 340

Also, when A = -340 and B = 60

A = ( -340 DIV 60 ) * 60 + ( -340 MOD 60 )

A = ( -5 ) * 60 + ( -40 ) =>

A = -340

From this can be derived that MOD(-340, 60) should have been -40

Sincerely,

Andre Banen

Date: 07/05/2001 at 16:17:04

From: Doctor Peterson

Subject: Re: Mod Function and Negative Numbers

Hi, Andre.

You're right that this relation is relevant to the question; below

I'm going to include a more detailed answer.

But it only proves what MOD should do _if_ we know how DIV is defined;

that is, it is a statement of consistency between the mod function and

the direction of integer truncation. You're _assuming_ truncation

toward zero, so that -340/60 gives -5. But in Excel, we see that the

relation looks like this:

A = ( -340 DIV 60 ) * 60 + ( -340 MOD 60 )

-340 = -6 * 60 + 20

This is perfectly consistent if their integer division truncates

toward -infinity rather than toward zero, so that -340/60 is taken to

be -6. And that's just what I said, using words rather than the

formula:

The mod function is defined as the amount by which a number

exceeds the largest integer multiple of the divisor that is

not greater than that number. In this case, -340 lies between

-360 and -300, so -360 is the greatest multiple LESS than -340;

we subtract 60 * -6 = -360 from -340 and get 20.

So in fact I did refer to your rule.

Here's a more complete answer:

Computer languages and libraries are notoriously inconsistent, or at

least unmathematical, in their implementation of "mod" for negative

numbers. There are several ways it can be interpreted in such cases,

and the choice generally made is not what a mathematician would

probably have made. The issue is what range of values the function

should return. Mathematically, we define "modulo" not as a function,

but as a relation: any two numbers a and b are congruent modulo m if

(a - b) is a multiple of m. If we want to make a function of this, we

have to choose which number b, of all those that are congruent to a,

should be returned.

Properly, the modulus operator a mod b should be mathematically

defined as the number in the range [0,b) that is congruent to a, as

stated here:

http://mathworld.wolfram.com/ModulusCongruence.html

In many computer languages (such as FORTRAN or Mathematica), the

common residue of b (mod m) is written mod(b,m) (FORTRAN) or

Mod[b,m] (Mathematica).

http://mathworld.wolfram.com/CommonResidue.html

The value of b, where a=b (mod m), taken to be nonnegative and

smaller than m.

Unfortunately, this statement about FORTRAN, and implicitly about the

many languages that have inherited their mathematical libraries from

FORTRAN, including C++, is not quite true where negative numbers are

concerned.

The problem is that people tend to think of modulus as the same as

remainder, and they expect the remainder of, say, -5 divided by 3 to

be the same as the remainder of 5 divided by 3, namely 2, but negated,

giving -2. We naturally tend to remove the sign, do the work, and put

the sign back on, because that's how we divide. In other words, we

expect to truncate toward zero for both positive and negative numbers,

and have the remainder be what's left "on the outside," away from

zero. More particularly, computers at least since the origin of

FORTRAN have done integer division by "truncating toward zero," so

that 5/2 = 2 and -5/2 = -2, and they keep their definition of "mod" or

"%" consistent with this by requiring that

(a/b)*b + a%b = a

so that "%" is really defined as the remainder of integer division as

defined in the language.

Because FORTRAN defined division and MOD this way, computers have

tended to follow this rule internally (in order to implement FORTRAN

efficiently), and so other languages have perpetuated it as well. Some

languages have been modified more recently to include the more

mathematical model as an alternative; in fact, FORTRAN 90 added a new

MODULO function that is defined so that the sign of MODULO(a,b) is

that of b, whereas the sign of MOD(a,b) is that of a. This makes it

match the mathematical usage, at least when b is positive.

Similarly, in Ada there are two different operators, "mod" (modulus)

and "rem" (remainder). Here's an explanation with plenty of detail:

Ada '83 Language Reference Manual - U.S. Government

http://archive.adaic.com/standards/83lrm/html/lrm-04-05.html

Integer division and remainder are defined by the relation

A = (A/B)*B + (A rem B)

where (A rem B) has the sign of A and an absolute value less than

the absolute value of B. Integer division satisfies the identity

(-A)/B = -(A/B) = A/(-B)

The result of the modulus operation is such that (A mod B) has the

sign of B and an absolute value less than the absolute value of B;

in addition, for some integer value N, this result must satisfy

the relation

A = B*N + (A mod B)

...

For positive A and B, A/B is the quotient and A rem B is the

remainder when A is divided by B. The following relations are

satisfied by the rem operator:

A rem (-B) = A rem B

(-A) rem B = -(A rem B)

For any integer K, the following identity holds:

A mod B = (A + K*B) mod B

The relations between integer division, remainder, and modulus are

illustrated by the following table:

A B A/B A rem B A mod B A B A/B A rem B A mod B

10 5 2 0 0 -10 5 -2 0 0

11 5 2 1 1 -11 5 -2 -1 4

12 5 2 2 2 -12 5 -2 -2 3

13 5 2 3 3 -13 5 -2 -3 2

14 5 2 4 4 -14 5 -2 -4 1

10 -5 -2 0 0 -10 -5 2 0 0

11 -5 -2 1 -4 -11 -5 2 -1 -1

12 -5 -2 2 -3 -12 -5 2 -2 -2

13 -5 -2 3 -2 -13 -5 2 -3 -3

14 -5 -2 4 -1 -14 -5 2 -4 -4

So what's the conclusion? There are basically two models, reasonably

distinguished in Ada terms as Remainder and Mod; the C++ "%" operator

is really Remainder, not Mod, despite what it's often called.

Actually, its behavior for negative numbers is not even defined

officially; like many things in C, it's left to be processor-dependent

because C does not define how a processor should handle integer

division. Just by chance, all compilers I know truncate integers

toward zero, and therefore treat "%" as remainder, following the

precedent of FORTRAN. As _C: A Reference Manual_, by Harbison and

Steele, says, "For maximum portability, programs should therefore

avoid depending on the behavior of the remainder operator when

applied to negative integral operands."

- Doctor Peterson, The Math Forum

http://mathforum.org/dr.math/

Date: 07/06/2001 at 04:22:22

From: A. H. Banen

Subject: Re: Mod Function and Negative Numbers

Dear Dr. Peterson,

I must concur with almost of your reasoning with one exception: you state I

have a _presumption_ about truncation towards zero. But, toward the end of

the very complete answer you sent me, you quote from

references found in the Ada '83 Language Reference Manual that integer

division has to satisfy the identity (-A)/B = -(A/B) = A/(-B). When this

identity is observed there is only one correct outcome of the integer divison

-340/60 (i.e. -5).

I can only hope that the identities the above site states to be

satisfied by integer devision, remainder and modulo function, are not the

presumption used by the definition of Ada, but that these are fundamental

identities all calculus is based on. I really hope you can confirm the

latter, as the trouble I had finding valid definitions of integer division,

etc. on the Internet really bothers me.

Also, the links you quoted to MathWorld (mathworld.wolfram.com) are no

longer valid, which is very disappointing as Wolfram really has given

mathematics on the computer its rightful place.

And thank you for your answer.

Sincerely,

Andre Banen

Date: 07/06/2001 at 09:05:12

From: Doctor Peterson

Subject: Re: Mod Function and Negative Numbers

Hello again!

I am aware that MathWorld has been unavailable for some time; we have left

links to it in many places, hoping that it will come back. I was very glad

that I had quoted from it rather than just giving a link.

I called "truncation toward zero" a presumption with regard to how Excel or

any other program should work; the developers of any particular programming

language or program are free to define conversion from "real" to integer in

any way they choose, as long as they define it. There is certainly no

presumption that those developers always make the mathematically "proper"

choices, since they may have different goals. The fact is that both forms of

truncation make sense in different applications; both "remainder" and

"modulo" are useful. Therefore it would be wrong to claim that only one is

valid. As I said, "Mod" is more mathematical, and there are many

formulas for which it is necessary (causing great difficulties for C

programmers); but "Rem" is useful in other cases, particularly since it is

consistent with the most natural definition of integer division. You show a

preference for Rem, and I won't argue against you, as long as you recognize

that Mod has its place as well.

Note that the identity you refer to is used in Ada only to define integer

division, and that Rem is defined from that; Mod is defined separately, but

is consistent with the "greatest integer" formulation of integer division,

which does not satisfy that identity. There is no reason it should; it has

different purposes.

I think the provision of two formulations as in Ada is reasonable, and would

not try to prove that one is invalid. The fact is that there is no definition

of "integer division" in math; rather, we divide and apply the greatest

integer function or some variation of it to the resulting rational number,

depending on our needs. Integer division is a formulation of computer

languages, and they define it for their own use.

- Doctor Peterson, The Math Forum

http://mathforum.org/dr.math/

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