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  "body": "\n## Proof\n\nWe show that the least positive integer of the form $ax + by$ is the [[greatest common divisor|02WF878BH1]].\n\n1. Let $a$ and $b$ be integers and let $d = \\gcd(a, b)$.\n2. If $a = b = 0$, then $d = 0$ by convention, and $x = y = 0$ give $a x + b y = 0 = d$. From now on, $a$ and $b$ are not both $0$.\n3. Then $1$ is a common divisor, so $d \\ge 1$.\n4. Let $S$ be the set of positive integers of the form $a x + b y$ with $x, y \\in \\Z$. It contains $a \\cdot a + b \\cdot b = a^2 + b^2 \u003e 0$, so $S$ is nonempty.\n5. By the [[well-ordering principle|01BX15HSRR]], $S$ has a least element $e$. Write $e = a x_0 + b y_0$ with $x_0, y_0 \\in \\Z$.\n6. By the [[division algorithm|0CHQKZFK5S]], $a = e q + r$ with integers $q$, $r$ and $0 \\le r \u003c e$.\n7. Then $r = a - q(a x_0 + b y_0) = a(1 - q x_0) + b(-q y_0)$, so $r$ has the form $ax + by$.\n8. If $r \u003e 0$, then $r \\in S$ and $r \u003c e$, which contradicts the choice of $e$. So $r = 0$, and $e$ [[divides|096PW7Q014]] $a$.\n9. The same argument with $b$ in place of $a$ shows that $e \\mid b$. So $e$ is a common divisor, and $e \\le d$ because $d$ is the largest one.\n10. Since $d \\mid a$ and $d \\mid b$, write $a = d a'$ and $b = d b'$ with integers $a'$, $b'$. Then $e = d(a' x_0 + b' y_0)$.\n11. Here $e \u003e 0$ and $d \u003e 0$, so the integer $a' x_0 + b' y_0$ is positive. Hence it is at least $1$, and $e \\ge d$.\n12. By steps 9 and 11, $e = d$. So $a x_0 + b y_0 = d$, and $x = x_0$, $y = y_0$ are the required integers.\n\n## Idea\n\nThe integers $ax + by$ form a set that is closed under subtracting multiples of each other. Its least positive element divides $a$ and $b$, because the remainder of a division stays in the set and is smaller. Every common divisor of $a$ and $b$ divides this element, so it is the greatest common divisor.\n",
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  "name": "Proof of Bézout's identity",
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