In the book Discrete Mathematics and Its Applications, 8e, Kenneth Rosen the quotient and remainder when $-11$ is divided by $3$ are specified as $-4$ and $1$ respectively. I would appreciate some help in understanding how we got there. It doesn't gel well with what I was taught in my elementary school.

Specifically, since $-11=(-3)\times 3+(-2)$, why do we not say that the quotient is $-3$ and the remainder is $-2$?

  • $\begingroup$ $-11= ((-4) \times 3 )+1$ $\endgroup$ Oct 20, 2021 at 6:59
  • $\begingroup$ @MauroALLEGRANZA I want to know why not ((−3)×3)-2 $\endgroup$
    – Sandeep
    Oct 20, 2021 at 7:01
  • $\begingroup$ You can: see Remainder: Examples $\endgroup$ Oct 20, 2021 at 8:28
  • $\begingroup$ In a more advanced setting one might treat "remainders" of $+1$ and $-2$ modulo $3$ as equal (literally, elements of the same equivalence class of residues). But in doing basic arithmetic one must adopt a convention to get a single result, and the simplest convention is to get the remainder $r$ on dividing by $b$ such that $0\le r \t b$. So following that convention (as many authors will), we chose the quotient that gives that nonnegative remainder. Something is wrong in your subject line BTW, as $11 \div 3$ should be quotient $3$ with remainder $2$. Apparently a minus sign was omitted? $\endgroup$
    – hardmath
    Oct 21, 2021 at 0:13

2 Answers 2


What were you taught in elementary school that goes against it? Almost everywhere, this division rule is formally introduced with the name of Euclid's Division as follows:

Given two integers $a$ and $b$, with $b ≠ 0$, there exist unique integers $q$ and $r$ such that

$$ a = bq + r $$ and

$$0 ≤ r < |b|$$, $$q = \lfloor\frac{a}{b}\rfloor$$

The above statement is taken from this Wikipedia article.

Specifically, in your case,

  • a = -11
  • b = 3
  • q = -4 = floor(a/b)
  • r = 1

I guess you hadn't noticed these constraints on the parameters of the equation. I hope this answers your question.


Dividend$=$Divisor$\times$Quotient +Remainder

Dividend $=-11$, Divisor$=3$, Quotient $=-4$, Remainder $=1$

$$-11=(3)(-4)+1$$ Edit: When you divide a number by another number, we get two different remainder values one of them being negative. In your case, remainder can be $1$ as well as $-2$(as $1-3=-2$)

  • $\begingroup$ I want to understand why not -11=(3)(−3) - 2 $\endgroup$
    – Sandeep
    Oct 20, 2021 at 7:04
  • $\begingroup$ Do you know concept of negative remainder$?$ $\endgroup$
    – user982341
    Oct 20, 2021 at 7:06

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