Mod Function and Negative NumbersDate: 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). You can read more how Excel handles that calculation in this note from our archives: Order of Operations and Negation in Excel http://mathforum.org/library/drmath/view/69058.html 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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