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Section 3.1 Quotients and Remainders

To determine how often a natural number \(b\) goes into a natural number \(a\) we subtract \(b\) from \(a\text{,}\) then subtract \(b\) from the result, subtract \(b\) from the result of this computation, and so on. We repeat this process until \(b\) does not go into the result of the previous subtraction, that is when the result of the previous subtraction is less than \(b\text{.}\) The number of times we have subtracted \(b\) is called the quotient (of the division of \(a\) by \(b\)) The number that is leftover from \(a\) after subtracting \(b\) as often as possible is called the remainder (of the division of \(a\) by \(b\)).
We illustrate the process described above with an example.

Example 3.1. How often does \(7\) go into \(26\) ?

To find out how often \(7\) goes into \(26\) we repeatedly subtract \(7\text{.}\) We stop this process when \(7\) does not go into the result of the previous subtraction anymore.. In other words we stop when we get to a number that is less than seven. We get:
\begin{align*} 26 - 7 \amp = 19\\ 19 - 7 \amp = 12\\ 12 - 7 \amp = 5 \end{align*}
We stop here because \(7\) does not go into \(5\text{.}\) We have subtracted \(7\) three times, which means that \(7\) goes into \(26\) three times. Thus the quotient is \(3\text{.}\) We have \(5\) left over. This is the remainder. We have found that
\begin{equation*} 26 - 3\cdot 7 = 5 \mbox{ or } 26 = 3\cdot 7+5\text{.} \end{equation*}
When we divide a positive integer \(a\) by an integer \(b\) and \(q\) is the number of times \(b\) goes into \(a\) then \(q\) is the quotient of the division of \(a\) by \(b\text{.}\) As the number of times \(b\) goes into \(a\) is \(q\) and \(b\) does not go into \(a\) more often than \(q\) times we have
\begin{equation*} a-q\cdot b\lt b \end{equation*}
So the remainder is \(r:=a-q\cdot b\) with \(0\le r \lt b\text{.}\) We have
\begin{equation*} a=q\cdot b+r. \end{equation*}
If \(a\) is given in the form \(a=q\cdot b+r\) with \(0\le r \lt b\) then it is particularly easy to read of the quotient and remainder from the division of \(a\) by \(b\text{.}\)

Example 3.2.

We have
\begin{equation*} 283416405986=490687\cdot 577590+501656. \end{equation*}
We find the quotient and remainder of the division of \(283416405986\) by \(577590\text{.}\)
To match the notation from above we set \(a:=283416405986\) and \(b:=577590\text{.}\) With this notation the equation above becomes
\begin{equation*} a=490687\cdot b+501656. \end{equation*}
As \(501656 \lt 577590\) this is of the form \(a=q\cdot b+r\) with \(0\le r \lt b\text{.}\) We now read of the quotient \(q=490687\) and the remainder \(r=501656\text{.}\)
When we divide a positive integer \(a\) by an integer \(b\) that is greater than \(a\) then \(b\) does not go into \(a\text{.}\) So \(b\) goes into \(a\) zero times and what is left over, the remainder, is \(a\text{.}\)

Example 3.3. How often does \(7\) go into \(3\) ?

Clearly \(7\) does not go into \(3\text{,}\) that is, \(7\) goes into \(3\) zero times. We have \(3\) left over. So the remainder is \(3\text{.}\)
Watch the video in Figure 3.4 for a review of what we have done so far and to see more examples.
Figure 3.4. Quotients and Remainders by Matt Farmer and Stephen Steward
We formulate the process described above as an algorithm called the Division Algorithm and then generalize it to negative integers. We introduce the operations \(\fdiv\) and \(\fmod\) as notation for quotients and remainders and describe how integer division can be performed with a calculator. We investigate properties of the operation \(\fmod\text{,}\) and give an example for its application in the validation of ISBN numbers.