GCD of Integers with Common Divisor/Proof 1

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Theorem

Let $a, b \in \Z$ be integers such that not both $a = 0$ and $b = 0$.

Let $k \in \Z_{>0}$ be a strictly positive integer.


Then:

$\gcd \set {k a, k b} = k \gcd \set {a, b}$

where $\gcd$ denotes the greatest common divisor.


Proof

Consider the demonstration of the operation of the Euclidean Algorithm applied to $a$ and $b$.

Let each equation be multiplied by $k$.

We have:

\(\ds a k\) \(=\) \(\ds q_1 \paren {b k} + r_1 k\) where $0 < r_1 k < b_k$
\(\ds b k\) \(=\) \(\ds q_2 \paren {r_1 k} + r_2 k\) where $0 < r_2 k < r_1 k$
\(\ds r_1 k\) \(=\) \(\ds q_3 \paren {r_2 k} + r_3 k\) where $0 < r_3 k < r_2 k$
\(\ds \cdots\) \(\) \(\ds \)
\(\ds r_{n - 2} k\) \(=\) \(\ds q_n \paren {r_{n - 1} k} + r_n k\) where $0 < r_n k < r_{n - 1} k$
\(\ds r_{n - 1} k\) \(=\) \(\ds q_{n + 1} \paren {r_n k} + 0\)

This is the operation of the Euclidean Algorithm applied to $k a$ and $k b$.

Hence the greatest common divisor is the last non-zero remainder $r_n k$.

That is:

$\gcd \set {k a, k b} = k \gcd \set {a, b}$

$\blacksquare$


Sources