by fox fox deer
Theorem: If d = GCD (a, b), then there is an integer such that s and t d = as + bt, and d is the smallest positive integer that can be represented by such.
Proof:
Suppose S is the set a, b of the linear combination. S = {s = k1a + k2b | k1, k2 is an integer, s> 0}, apparently a, b itself is S, and S is not an empty set
below by contradiction:
Suppose d̸ | a, then a can be easily expressed as :
a = dq + r, 0 < r < d
Since a, d is a linear combination of a, b, then r = a - dq is a, b, linear combination, then r belonging to the set S and is smaller than d, and d is the smallest element in the set S contradiction. It can be shown that d | a.
Same reason d | b.
The above has been demonstrated d are a, b of a common factor, the following is proved the greatest common divisor d.
For all positive integers c, if c | A c and | b is divisible c a, b, all linear combinations, so c | d.
Therefore, d is the greatest common factor a, b's. which is:
d=GCD(a,b)
Using the above method, we can easily prove the following theorem:
For b1, b2 on E, ..., bn, bi if not all zeros, then there k1, k2, ... kn∈E, such that gcd (b1, b2, ..., bn) = k1b1 + ... + knbn
Thinking: define the set S = {k1b1 + ... knbn | ki∈E}, the gcd = d, d∈S, and g (d) is the minimum. Then proof of (1) d is a common factor, (2) d is the greatest common divisor
ref https://www.math.fsu.edu/~pkirby/mad2104/SlideShow/s5_2.pdf