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  "body": "\n## Proof\n\nWe prove existence with the well-ordering principle, then uniqueness by a size argument.\n\n1. Let $a$ and $b$ be integers with $b \\neq 0$. Put $c = |b|$, so $c$ is a positive integer.\n2. Let $S$ be the set of integers $a - c t$ with $t \\in \\Z$ and $a - c t \\ge 0$. So $S$ is a set of natural numbers.\n3. The set $S$ is nonempty. For $t = -|a|$ we get $a - c t = a + c|a| \\ge a + |a| \\ge 0$, because $c \\ge 1$.\n4. By the [[well-ordering principle|01BX15HSRR]], $S$ has a least element $r$. Write $r = a - c t_0$ with $t_0 \\in \\Z$. Then $r \\ge 0$.\n5. We show $r \u003c c$. Suppose $r \\ge c$. Then $r - c = a - c(t_0 + 1) \\ge 0$, so $r - c \\in S$. But $r - c \u003c r$, which contradicts the choice of $r$.\n6. Put $q = t_0$ if $b \u003e 0$ and $q = -t_0$ if $b \u003c 0$. In both cases $b q = c t_0$. Hence $a = b q + r$ with $0 \\le r \u003c |b|$.\n7. For uniqueness, let $(q, r)$ and $(q', r')$ both satisfy $a = bq + r = bq' + r'$ and $0 \\le r, r' \u003c |b|$.\n8. Subtracting gives $b(q - q') = r' - r$. From the bounds on $r$ and $r'$, we get $|r' - r| \u003c |b|$.\n9. If $q \\neq q'$, then $|q - q'| \\ge 1$, so $|b(q - q')| \\ge |b|$. This contradicts step 8. Hence $q = q'$.\n10. Then $r' - r = b(q - q') = 0$, so $r = r'$.\n11. So there is exactly one pair $(q, r)$ with $a = bq + r$ and $0 \\le r \u003c |b|$.\n\n## Idea\n\nSubtract multiples of $|b|$ from $a$ for as long as the result stays nonnegative. The smallest nonnegative value reached is the remainder, and it must be less than $|b|$, or one more subtraction would be possible. Two different quotients would make the remainders differ by at least $|b|$, which is impossible.\n",
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