3 ms·
GCD(0,0) exists in any commutative ring by definition (and is not unique) :) d is a common divisor of a,b if there exists x,y such that dx = ay, and d is a GCD
by aruss 6y ago
GCD(0,0) exists in any commutative ring by definition (and is not unique) :)
d is a common divisor of a,b if there exists x,y such that dx = ay, and d is a GCD of a,b if all divisors c divide d. So there exist many such x where GCD(0,0) = x (including x = 0).
- zodiac 6y agoIf you define GCD this way then writing gcd(a, b) = c is an abuse of notation :)
- thaumasiotes 6y ago> d is a common divisor of a,b if there exists x,y such that dx = ay d = 9 a = 3 b = 537 x = 1 y = 3 dx = 9(1) = 9 ay = 3(3) = 9 = dx You can't actually have meant this? You're claiming that 9 is a common divisor of the pair (3, b), where b is any value. It's not even a divisor of 3.