Euclid's Lemma/Proof 2

From ProofWiki
Jump to: navigation, search


Let $a, b, c \in \Z$.

Let $a \divides b c$, where $\divides$ denotes divisibility.

Let $a \perp b$, where $\perp$ denotes relative primeness.

Then $a \divides c$.


Let $a, b, c \in \Z$.

We have that $a \perp b$.

That is:

$\gcd \set {a, b} = 1$

where $\gcd$ denotes greatest common divisor.

From Bézout's Lemma, we may write:

$a x + b y = 1$

for some $x, y \in \Z$.

Upon multiplication by $c$, we see that:

$c = c \paren {a x + b y} = c a x + c b y$

Now note that $c a x + c b y$ is an integer combination of $a c$ and $b c$.

So, since:

$a \divides a c$


$a \divides b c$

it follows from Common Divisor Divides Integer Combination that:

$a \divides \paren {c a x + c b y}$


$c a x + c b y = c \paren {a x + b y} = c \cdot 1 = c$


$a \divides c$


Source of Name

This entry was named for Euclid.

Also see