The Quotient-Remainder Theorem
In the last topic we discussed what it means for one integer \(n\) to be divisible by another integer \(m\). What if you don’t have \(m|n\)? The next theorem, in a strange sense, indicates how much we are off by. Additionally, it gives us a unique way of writing a number as product plus a remainder.
In the equation above, we usually refer to the \(q\) as the quotient and \(r\) as the remainder.
What this theorem is saying is that no matter what pair of numbers \(n\) and \(m\) you have, you can ALWAYS write \(n\) as \(m\) times some other number, plus some number strictly less than \(m\).
When \(n\) and \(m\) are positive, \(q\) comes from the number of times \(m\) divides \(n\) evenly, and \(r\) is the remainder of that division.
Since \(n\) and \(m\) are positive, \(q\) comes from the number of times \(m\) evenly divides \(n\), and \(r\) is found to be the remainder of that same division.
\(q=6\) and \(r=4\). So we have that \(34=5q+r=5\cdot 6+4\).
\(5\) evenly divides \(34\) exactly \(6\) times, with a remainder of \(4\). So therefore, \(34=5\cdot 6+4\) and your \(q=6\) and \(r=4\).
Since \(n\) and \(m\) are positive, \(q\) comes from the number of times \(m\) evenly divides \(n\), and \(r\) is found to be the remainder of that same division.
\(q=311\) and (r=1\). So we have that \(623=2\cdot q+r=2\cdot 311+1\).
\(2\) evenly divides \(623\) exactly \(311\) times, with a remainder of \(1\). So therefore, \(624=2\cdot 311+4\) and your \(q=311\) and \(r=1\).
\(q=-5\) and \(r=9\). So, \(-56=13q+r=13\cdot -5+9\). Note carefully that the remainder is always positive in the Quotient Remainder Theorem.
Note that when we try to divide \(-56\) by \(13\) evenly, we get \(-4\) with a remainder of \(-4\). The problem with this is that the remainder is negative, whereas in the Quotient Remainder Theorem, we must have \(0\leq r<m\). So, in order to attain this nonnegative remainder, we “overshoot” in our division, dividing by \(13\) \(q=-5\) times so that our remainder is \(r=9\), meeting the specification \(0\leq r<m\).
\(q=-12\) and \(r=4\). So, \(-128=11q+r=11\cdot -12+4\). Note carefully that the remainder is always positive in the Quotient Remainder Theorem.
Note that when we try to divide \(-128\) by \(11\) evenly, we get \(-11\) with a remainder of \(-7\). The problem with this is that the remainder is negative, whereas in the Quotient Remainder Theorem, we must have \(0\leq r<m\). So, in order to attain this nonnegative remainder, we “overshoot” in our division, dividing by \(11\) \(q=-12\) times so that our remainder is \(r=4\), meeting the specification \(0\leq r<m\).
Is the remainder you get when dividing \(505\) by \(5\) “not allowed’ or unreasonable according to the Quotient Remainder Theorem?
\(q=101\) and \(r=0\).
Note that when we divide \(505\) be \(5\) we clearly get \(101\) and so have no remainder! The Quotient Remainder Theorem allows \(r=0\) according to left part of the given inequality \(0\leq r<m\).