In order to justify calling quotients and remainders "the quotient" or "the remainder", we should check that they are unique. This is justified by the following claim:

For any

[math]a \href{/cs2800/wiki/index.php/%E2%88%88}{∈} \href{/cs2800/wiki/index.php/%E2%84%A4}{ℤ}
[/math] and

[math]b \neq 0 \href{/cs2800/wiki/index.php/%E2%88%88}{∈} \href{/cs2800/wiki/index.php/%E2%84%A4}{ℤ}
[/math], if

[math]q
[/math] and

[math]r
[/math] are a

quotient and

remainder of

[math]a
[/math] and

[math]b
[/math] (that is, if

[math]a = qb + r
[/math] and

[math]0 \leq r \lt b
[/math]), and if

[math]q'
[/math] and

[math]r'
[/math] are also a quotient and remainder of

[math]a
[/math] over

[math]b
[/math], then

[math]q = q'
[/math] and

[math]r = r'
[/math].

Proof:

Assume that

[math]a = qb + r
[/math] and

[math]a = q'b + r'
[/math] and that both

[math]r
[/math] and

[math]r'
[/math] are between 0 and

[math]b-1
[/math]. We

want to show that

[math]q = q'
[/math] and

[math]r = r'
[/math].

We have [math]qb + r = a = q'b + r'
[/math], so [math](r - r') = (q' - q)b
[/math]. This means that [math]r - r'
[/math] is a multiple of [math]b
[/math]; it could be any of [math]\dots, -2b, -b, 0, b, 2b, \dots
[/math]. However, since we have [math]0 \leq r \lt b
[/math] and [math]0 \leq r' \lt b
[/math], we have [math]-b \lt r - r' \lt b
[/math]. Thus, the only possible value of [math]r - r'
[/math] is 0, so [math]r = r'
[/math].

Plugging this back in to the equation

[math]0 = r - r' = (q - q')b
[/math] shows that either

[math]q - q' = 0
[/math] or

[math]b = 0
[/math]. Since we have defined the quotient and remainder by

[math]b
[/math], we must have

[math]b \neq 0
[/math], so

[math]q - q' = 0
[/math] and thus

[math]q = q'
[/math].