Common Divisor Divides Integer Combination/General Result

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Theorem

Let $c$ be a common divisor of a set of integers $A := \set {a_1, a_2, \dotsc, a_n}$.

That is:

$\forall x \in A: c \divides x$


Then $c$ divides any integer combination of elements of $A$:

$\forall x_1, x_2, \dotsc, x_n \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_n x_n}$


Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:

$\forall x \in \set {a_1, a_2, \dotsc, a_n}: c \divides x \implies \forall x_1, x_2, \dotsc, x_n \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_n x_n}$


Basis for the Induction

$\map P 2$ is the case:

$\forall x \in \set {a_1, a_2}: c \divides x \implies \forall x_1, x_2 \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2}$

This is demonstrated to be true in Common Divisor Divides Integer Combination.

Thus $\map P 2$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\forall x \in \set {a_1, a_2, \dotsc, a_k}: c \divides x \implies \forall x_1, x_2, \dotsc, x_k \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k}$


from which it is to be shown that:

$\forall x \in \set {a_1, a_2, \dotsc, a_k, a_{k + 1} }: c \divides x \implies \forall x_1, x_2, \dotsc, x_k, x_{k + 1} \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k + a_{k + 1} x_{k + 1} }$


Induction Step

This is the induction step:

Let :

$\forall x \in \set {a_1, a_2, \dotsc, a_k, a_{k + 1} }: c \divides x$

We have that:

$c \divides a_{k + 1} \implies \forall x_{k + 1} \in \Z: c \divides a_{k + 1} x_{k + 1}$

and we have that:

$\forall x \in \set {a_1, a_2, \dotsc, a_k}: c \divides x \implies \forall x_1, x_2, \dotsc, x_k \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k}$


Let $x_1, x_2, \dotsc, x_k \in \Z$ be arbitrary.

Let $d = a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k$.

Then:

\(\ds c\) \(\divides\) \(\ds cd\)
\(\, \ds \land \, \) \(\ds c\) \(\divides\) \(\ds a_{k + 1}\)
\(\ds \leadsto \ \ \) \(\ds c\) \(\divides\) \(\ds \paren {1 \times d + x_{k + 1} a_{k + 1} }\)
\(\ds \) \(=\) \(\ds \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k + a_{k + 1} x_{k + 1} }\)

But $x_1, x_2, \dotsc, x_k \in \Z$ are arbitrary, and so:

$\forall x \in \set {a_1, a_2, \dotsc, a_k, a_{k + 1} }: c \divides x \implies \forall x_1, x_2, \dotsc, x_k, x_{k + 1} \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k + a_{k + 1} x_{k + 1} }$


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{\ge 2}: \forall x \in \set {a_1, a_2, \dotsc, a_n}: c \divides x \implies \forall x_1, x_2, \dotsc, x_n \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_n x_n}$

$\blacksquare$


Sources

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