Choose a least positive combination
For integers a,b not both zero, positive values au+bv form a nonempty set. The well-ordering principle gives a smallest value d, itself an integer linear combination.
Michael Penn · YouTube · 16:34
Prove Bézout’s identity from the least positive integer linear combination, then apply it to coprimality.
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For integers a,b not both zero, positive values au+bv form a nonempty set. The well-ordering principle gives a smallest value d, itself an integer linear combination.
Divide a or b by d. A positive remainder smaller than d would again be a positive linear combination, contradicting minimality. Thus d divides both; every common divisor also divides d.
If integer coefficients give ax+by=1, every common divisor divides 1, so gcd(a,b)=1. Conversely Bézout’s identity provides such coefficients when a and b are coprime.
For integers a,b not both zero, positive values au+bv form a nonempty set. The well-ordering principle gives a smallest value d, itself an integer linear combination.
Divide a or b by d. A positive remainder smaller than d would again be a positive linear combination, contradicting minimality. Thus d divides both; every common divisor also divides d.
If integer coefficients give ax+by=1, every common divisor divides 1, so gcd(a,b)=1. Conversely Bézout’s identity provides such coefficients when a and b are coprime.
For integers a,b not both zero, positive values au+bv form a nonempty set. The well-ordering principle gives a smallest value d, itself an integer linear combination. Divide a or b by d. A positive remainder smaller than d would again be a positive linear combination, contradicting minimality. Thus d divides both; every common divisor also divides d. If integer coefficients give ax+by=1, every common divisor divides 1, so gcd(a,b)=1. Conversely Bézout’s identity provides such coefficients when a and b are coprime.
The converse states that if integers a and b are coprime (i.e., their greatest common divisor is 1), then there exist integer coefficients x and y such that ax+by=1. This is provided by Bézout's identity.
Conditions: a and b are coprime integers.; gcd(a,b) = 1.
The proof demonstrates this by showing that d is a linear combination of a and b. Since any common divisor divides both a and b, it must also divide any integer linear combination of them, including d.
Conditions: d is the least positive linear combination of a and b.; c is a common divisor of a and b.
It serves as a certificate because if integer coefficients x and y satisfy ax+by=1, then any common divisor of a and b must divide 1. The only positive integer dividing 1 is 1 itself, proving that the greatest common divisor is 1.
Conditions: a and b are integers.; There exist integers x and y such that ax+by=1.
It contradicts minimality because the remainder can be expressed as a linear combination of a and b. If the remainder is positive and smaller than d, it belongs to the set of positive linear combinations, violating the definition of d as the smallest such value.
Conditions: d is defined as the least positive integer linear combination of a and b.; The remainder r is obtained by dividing a or b by d.; r is positive and .
The well-ordering principle guarantees this by ensuring that any nonempty set of positive integers has a smallest element. Since the set of positive linear combinations is nonempty, it must contain a minimum value.
Conditions: Integers a and b are not both zero.; The set consists of positive values of the form au+bv where u and v are integers.